🧠:[<think>Okay, so I need to solve this problem about a sinusoidal alternating current voltage. Let me start by reading the problem again carefully.The voltage v is given by the function v = 120√2 sin(100πt - π/6), where t is time in seconds starting from 0 to infinity. There are two parts to the problem. First, part (1) asks for the period, frequency, and amplitude of the voltage. Second, part (2) is about a neon lamp that lights up only when the voltage is greater than 84V. We need to find the time during half a period when the neon lamp is on. They also mention to take √2 ≈ 1.4, which might be useful for part (2).Starting with part (1). Let's recall some basic properties of sinusoidal functions. The general form is usually written as v(t) = A sin(Bt + C) + D, where A is the amplitude, the period is 2π/B, the frequency is 1/period, and C is the phase shift. D is the vertical shift, but in this case, since there's no DC offset, D should be zero.Looking at the given equation: v = 120√2 sin(100πt - π/6). Let's parse this.Amplitude: The coefficient in front of the sine function. That's 120√2. So the amplitude should just be 120√2 volts. But wait, sometimes in AC voltage, they mention RMS voltage. But the problem specifically says "amplitude," so that's the peak voltage. So the amplitude is 120√2 volts. Let me confirm: in the equation v = V_peak sin(ωt + φ), V_peak is the amplitude. So yes, here it's 120√2. So that's straightforward.Frequency: The angular frequency ω is given by the coefficient of t inside the sine function. Here, ω is 100π. The relation between angular frequency and frequency f is ω = 2πf. So solving for f, we get f = ω / (2π) = 100π / (2π) = 50 Hz. That's the frequency.Period: The period T is the reciprocal of the frequency. So T = 1/f = 1/50 seconds, which is 0.02 seconds. Alternatively, since the period of sin(Bt + C) is 2π/B, here B is 100π, so period is 2π/(100π) = 2/100 = 0.02 seconds. Yep, same result.So for part (1):Amplitude: 120√2 volts (but maybe they want the numerical value? Wait, 120√2 is already simplified. Let's check the problem statement. It says "Find the period, frequency, and amplitude". Since they didn't specify to approximate, we can leave it in exact form. So amplitude is 120√2 V, frequency is 50 Hz, period is 0.02 seconds. That should be part (1) done.Now part (2). The neon lamp lights up when the voltage is greater than 84V. We need to find the time during half a period when the lamp is on. So during half a period, which is T/2 = 0.01 seconds, we need to find the duration when v > 84V.First, let's recall that in a sinusoidal wave, the voltage exceeds a certain threshold twice per cycle: once when rising above and once when falling below. But since we are looking at half a period, which is half of the full cycle, we need to see how the threshold intersects the sine wave in that interval.Wait, but actually, in half a period of the sine wave, which is a half-cycle (either from 0 to peak back to zero, or depending on the phase shift). But given the general sine function here, we need to consider the phase shift. The given function is sin(100πt - π/6). Let's perhaps first sketch the waveform or find the points where the voltage crosses 84V.First, let's express the equation when v = 84V:120√2 sin(100πt - π/6) = 84We can solve for t in this equation. Let's first solve for the angle θ where sinθ = 84/(120√2). Let's compute 84/(120√2). Let's compute that:First, 84 divided by 120 is 0.7. So 0.7 / √2. Since √2 is approx 1.4, 0.7 / 1.4 = 0.5. So sinθ = 0.5. Therefore, θ = π/6 or 5π/6 (plus multiples of 2π). Wait, but let's check:Wait, 84/(120√2) = (84/120)/√2 = (7/10)/√2 ≈ 0.7 / 1.414 ≈ 0.4949, which is approximately 0.5. But if they tell us to take √2 ≈ 1.4, so 0.7 / 1.4 = 0.5 exactly. So sinθ = 0.5. Therefore, θ = π/6 or 5π/6 in the interval [0, 2π). So the general solution would be θ = π/6 + 2πk or 5π/6 + 2πk for integer k.But in our equation, θ is 100πt - π/6. Therefore:100πt - π/6 = π/6 + 2πkOR100πt - π/6 = 5π/6 + 2πkSolving for t:First case:100πt = π/6 + π/6 + 2πk = π/3 + 2πkt = (π/3 + 2πk)/(100π) = (1/3 + 2k)/100Second case:100πt = 5π/6 + π/6 + 2πk = π + 2πkt = (π + 2πk)/(100π) = (1 + 2k)/100So the solutions are t = (1/3 + 2k)/100 and t = (1 + 2k)/100 for integer k.Now, since we need to consider the time during half a period, which is T/2 = 0.01 seconds. Let's first figure out the interval of half a period. The period is 0.02 seconds, so half a period is 0.01 seconds. But we need to know over which interval. Since the function is periodic, but the phase shift is -π/6, so the starting point might affect where the half period is. Wait, but perhaps it's easier to consider one full period and then take half of that. Wait, but maybe we need to find within a specific half period.Wait, actually, when dealing with time intervals for neon lamps lighting up, typically, the half period is from one zero crossing to the next zero crossing with the same slope, but in this case, perhaps they just want any half period. However, the problem states "the time during half a period when the neon lamp is on", so maybe they mean in each half period, the time during which the voltage is above 84V. But since the waveform is symmetric over each half period? Wait, but due to the phase shift, maybe the first half period is shifted?Alternatively, maybe since the function is sinusoidal, we can consider the time between two consecutive instances where the sine function starts a new half cycle. Let me think.Alternatively, perhaps it's simpler to find all solutions where v > 84V in a full period and then divide by 2? But the problem says "during half a period". Maybe in each half period, the time when it's on is the same. So maybe the duration per half period is half the total time per full period. Wait, but actually, in a full period, the voltage is above 84V twice: once when increasing and once when decreasing. So the total time above 84V in a full period is two intervals. Therefore, in half a period, it should be one interval. But let's check.Wait, let's suppose that in a full period, the sine wave goes up to the peak, then down, crosses zero, goes negative, etc. But since the neon lamp is only concerned with positive voltages (assuming it's a voltage that can be positive or negative, but maybe the neon lamp is connected in a way that only considers the absolute value? Wait, the problem says "voltage across it is greater than 84V". So if the voltage is negative, then it's -v, but the problem just says "voltage greater than 84V", so perhaps we need to consider absolute value? Wait, but in the given function, the voltage is sinusoidal, so it goes both positive and negative. However, neon lamps typically have a breakdown voltage, and they conduct when the voltage exceeds a certain threshold in either polarity. But the problem doesn't specify, it just says "voltage greater than 84V". So if the voltage is negative, is that considered? The problem statement says "the voltage across it is greater than 84V". So if the voltage is, say, -100V, is that greater than 84V? No, because -100V is less than 84V. So maybe the neon lamp only conducts when the voltage is greater than 84V in the positive half cycle. Wait, but if that's the case, then in each full period, the neon lamp is on during the time when the sine wave is above 84V in the positive half cycle. But the problem says "the time during half a period when the neon lamp is on". So maybe they are considering only the positive half cycle as half a period? Wait, but half a period is 0.01 seconds, which is half of 0.02 seconds. However, the positive half cycle is actually half the period, but depending on the phase shift. Hmm.Wait, let's clarify. In a sinusoidal AC voltage, the period is the time for a full cycle. A half period is half that time, so 0.01 seconds here. However, a half cycle (positive or negative) is also 0.01 seconds. But depending on the phase shift, the starting point might not be at zero. So in our function, the voltage is 120√2 sin(100πt - π/6). Let's analyze the waveform.At t=0, the voltage is 120√2 sin(-π/6) = 120√2 * (-1/2) = -60√2 V. So it starts at -60√2 V. Then, as t increases, the argument of the sine function increases, so the voltage will go from -60√2 V, cross zero, reach the positive peak, come back down, cross zero again, and go to the negative peak, etc.But since the neon lamp lights up only when the voltage is greater than 84V, which is a positive value, so we need to consider the portions of the sine wave where it is above 84V in the positive half cycles. Each positive half cycle is 0.01 seconds long (half a period). But due to the phase shift, the starting point of the half period might not be at zero.But maybe the problem is just asking for the duration during each half period (i.e., each 0.01 seconds interval) where the voltage is above 84V. So regardless of where the half period starts, the duration should be the same due to the periodicity. Wait, but actually, the phase shift would affect when the threshold is crossed. But since we have a steady-state sinusoidal function, the duration above 84V in each half period should be the same. So perhaps we can just compute the time between the two points in a half period where the voltage crosses 84V.But let's proceed step by step.First, let's find all t in [0, +infty) where v = 84V. We already started that earlier.We have t = (1/3 + 2k)/100 and t = (1 + 2k)/100.So let's compute these for k=0,1,2,...For k=0:t1 = (1/3)/100 ≈ 0.003333... secondst2 = 1/100 = 0.01 secondsFor k=1:t3 = (1/3 + 2)/100 = (7/3)/100 ≈ 0.023333... secondst4 = (1 + 2)/100 = 3/100 = 0.03 secondsSimilarly, for k=2: t5 ≈ 0.043333..., t6=0.05, etc.So the solutions are at t ≈ 0.003333, 0.01, 0.023333, 0.03, etc.Wait, but in the equation sin(theta) = 0.5, the solutions are theta = pi/6 + 2pi k and theta = 5pi/6 + 2pi k. But in our case, theta = 100pi t - pi/6. So:First solution: 100pi t - pi/6 = pi/6 + 2pi kSo 100pi t = pi/6 + pi/6 + 2pi k = pi/3 + 2pi kt = (pi/3 + 2pi k)/(100 pi) = (1/3 + 2k)/100Second solution: 100pi t - pi/6 = 5pi/6 + 2pi k100pi t = 5pi/6 + pi/6 + 2pi k = pi + 2pi kt = (pi + 2pi k)/(100 pi) = (1 + 2k)/100So yes, that's correct.So between t=0 and t=0.02 (one full period), the solutions are at t=1/300 ≈0.003333, t=0.01, t=1/300 + 0.02 ≈ 0.023333, t=0.03, etc. Wait, no. For k=0, t1=1/300≈0.003333 and t2=1/100=0.01. For k=1, t3=(1/3 +2)/100=7/300≈0.023333 and t4=3/100=0.03. So in the first full period [0,0.02], we have two crossings at ~0.003333 and 0.01, then again at ~0.023333 and 0.03. Wait, but 0.03 is beyond 0.02. Wait, 0.02 is 1/50=0.02. So in the interval [0,0.02), the crossings are at t=1/300≈0.003333, t=0.01, t=7/300≈0.023333. Wait, 0.023333 is still less than 0.02? Wait, 7/300 is approximately 0.023333, which is 0.023333 seconds, but 0.02 seconds is 1/50=0.02. So 0.023333 is actually beyond the first full period. Therefore, within the first period [0,0.02], the crossings are at t=1/300 and t=0.01. Then the next crossings would be at t=7/300≈0.023333 and t=0.03, which are in the next period.Wait, perhaps I made an error here. Let me check.Wait, the period is 0.02 seconds. So the first period is from t=0 to t=0.02. Let's compute the exact values:First solution for k=0:t = (1/3 + 2*0)/100 = 1/300 ≈0.003333 secondsSecond solution for k=0:t = (1 + 2*0)/100 = 1/100 =0.01 secondsSo in the first period, 0 to 0.02, the first two solutions are at 0.003333 and 0.01. Then for k=1:First solution with k=1:t=(1/3 +2*1)/100=(1/3 +2)/100=(7/3)/100≈0.023333 secondsSecond solution with k=1:t=(1 +2*1)/100=3/100=0.03 secondsBut 0.023333 and 0.03 are beyond 0.02, so they belong to the next period.So within the first period, 0 to 0.02, we have two times when the voltage crosses 84V: once going up at t=1/300 (~0.003333) and once going down at t=0.01. Then, after that, it goes below 84V until t=0.01, then the next crossing is at t=7/300 (~0.023333) which is in the next period.Wait, hold on, this seems contradictory. Wait, let's think about the sine wave. The function is v=120√2 sin(100πt - π/6). Let's consider the behavior over time.At t=0: v=120√2 sin(-π/6)= -60√2 ≈-84.85V. So it starts at -84.85V. Then as t increases, the argument of the sine function increases, so the voltage increases. The first time it crosses zero is when 100πt - π/6 =0 => t=(π/6)/(100π)=1/600≈0.001666 seconds. Then it reaches the positive peak at t when 100πt - π/6=π/2 => 100πt=2π/3 => t=2/(300)=1/150≈0.006666 seconds. Then it goes back down to zero at t when 100πt - π/6=π => t=(7π/6)/(100π)=7/600≈0.011666 seconds. Then it goes to the negative peak at t when 100πt - π/6=3π/2 => t=(10π/6)/(100π)=10/600=1/60≈0.016666 seconds. Then back to zero at t=13π/6/(100π)=13/600≈0.021666 seconds, which is the end of the first full period at 0.02 seconds? Wait, no. Wait, period is 0.02 seconds, so 0 to 0.02 is the first period. But 13/600≈0.021666 is beyond 0.02.Wait, perhaps I need to check the period again. The period T=2π/(100π)=0.02 seconds. So the first period is from t=0 to t=0.02. The function at t=0 is -60√2, then it goes up, crosses zero at t=1/600≈0.001666, reaches positive peak at t=1/150≈0.006666, crosses zero again at t=7/600≈0.011666, goes to negative peak at t=1/60≈0.016666, and crosses zero again at t=13/600≈0.021666, which is outside the first period.Therefore, in the first period, the voltage starts at -84.85V, crosses zero at ~0.001666, peaks at ~0.006666, crosses zero again at ~0.011666, goes to -84.85V at ~0.016666, and then would cross zero again at ~0.021666, which is in the next period.Therefore, in the first period, the voltage is above zero from ~0.001666 to ~0.011666, which is a duration of ~0.01 seconds (half period). Then below zero from ~0.011666 to ~0.021666, another ~0.01 seconds.But we are interested in when the voltage is above 84V. Let's see. The positive peak is 120√2≈169.68V. So the sine wave goes from -169.68V to +169.68V. The threshold is 84V.So in the positive half cycle (~0.001666 to ~0.011666), the voltage exceeds 84V at some point after rising above zero, stays above 84V until it starts to decrease below 84V, then goes down to zero.Similarly, in the negative half cycle, the voltage is negative, so it doesn't reach 84V. So the neon lamp is only on during the time in the positive half cycle when voltage is above 84V.Therefore, in each period, the neon lamp is on twice? Wait, no. Wait, in each positive half cycle, the voltage crosses 84V twice: once when rising and once when falling. So the time when voltage is above 84V is between those two crossings. So in each positive half cycle, there's a single interval where voltage is above 84V. Similarly, in the negative half cycle, if the threshold was -84V, but since we are only considering >84V, which is positive, only the positive half cycle matters.But the problem says "the time during half a period when the neon lamp is on". So since the period is 0.02 seconds, half a period is 0.01 seconds. But the positive half cycle is 0.01 seconds. So within that 0.01 seconds (the duration of the positive half cycle), how much time is the voltage above 84V.Wait, but actually, the positive half cycle isn't exactly 0.01 seconds. Wait, the half period is 0.01 seconds, but the positive half cycle might be shifted due to the phase. Let's see.Wait, the period is 0.02 seconds, so half period is 0.01 seconds. The positive half cycle is the interval during which the voltage is positive, which, due to the phase shift, might not be exactly half the period. Wait, but in a sine wave, the positive and negative half cycles are each half the period, regardless of phase. Because shifting the phase doesn't change the fact that the positive and negative parts each occupy half the period.Wait, actually, no. Wait, phase shift affects where the starting point is, but over a full period, the positive and negative parts are each half the period. However, if the function is phase-shifted, the positive half cycle might start earlier or later, but its duration is still half the period.Wait, for example, sin(theta - phi) has the same period as sin(theta), but shifted. So in each period, the amount of time the function is positive is still half the period.Therefore, the positive half cycle is 0.01 seconds, regardless of phase shift. So in our case, the positive half cycle is from t=1/600≈0.001666 to t=7/600≈0.011666 seconds. Wait, that duration is 6/600=0.01 seconds. So yes, 0.01 seconds. Similarly, the negative half cycle is from t=7/600≈0.011666 to t=13/600≈0.021666, which is also 0.01 seconds.But within the positive half cycle (0.001666 to 0.011666), the voltage exceeds 84V from t1 to t2, where t1 is when the voltage crosses 84V on the way up, and t2 is when it crosses 84V on the way down.We found earlier that the solutions for v=84V are at t=1/300≈0.003333 and t=0.01 seconds. Wait, 1/300 is 0.003333, and 0.01 is 1/100. So within the positive half cycle (0.001666 to 0.011666), the crossings are at t=0.003333 and t=0.01. So the time when the voltage is above 84V is from t=0.003333 to t=0.01, which is a duration of 0.01 - 0.003333=0.006666 seconds.But the problem says "the time during half a period when the neon lamp is on". So half a period is 0.01 seconds. But the time when the lamp is on is 0.006666 seconds. Let me check the calculations again.First, solving 120√2 sin(theta) = 84:sin(theta) = 84 / (120√2) = 7/(10√2) ≈0.7/1.4=0.5. So theta = pi/6 and 5pi/6.But theta is 100pi t - pi/6.So:For theta=pi/6: 100pi t - pi/6 = pi/6 => 100pi t = 2pi/6 => t= (pi/3)/(100pi)=1/300≈0.003333.For theta=5pi/6: 100pi t - pi/6 =5pi/6 =>100pi t=6pi/6=pi => t=pi/(100pi)=1/100=0.01.So between t=0.003333 and t=0.01, the voltage is above 84V.But the positive half cycle is from t=1/600≈0.001666 to t=7/600≈0.011666. So within this interval, the voltage is above 84V from t=0.003333 to t=0.01. So the duration is 0.01 - 0.003333=0.006666 seconds≈0.006666 seconds.But 0.006666 seconds is 2/300 seconds=1/150≈0.006666...Expressed as a fraction, that's (0.01 - 1/300)= (3/300 -1/300)=2/300=1/150≈0.006666 seconds.But the problem says to take sqrt2≈1.4. So let's check if our approximation affected the result.Wait, when we calculated sin(theta)=0.5, we approximated 84/(120√2)=0.7/1.4=0.5. But actually, 84/(120*1.4)=84/168=0.5 exactly. So with sqrt2≈1.4, the calculation is exact. Therefore, sin(theta)=0.5, so theta=30 degrees (pi/6) and 150 degrees (5pi/6). Therefore, our solutions are exact under this approximation.Therefore, the time when the voltage is above 84V is from t=1/300 to t=1/100, which is a duration of 1/100 -1/300= (3/300 -1/300)=2/300=1/150≈0.006666 seconds.But the problem says "the time during half a period when the neon lamp is on". The half period is 0.01 seconds. So 1/150 seconds is approximately 0.006666 seconds, which is 2/3 of the half period. Wait, 1/150 divided by 0.01 is (1/150)/(1/100)=100/150=2/3. So the lamp is on for 2/3 of the half period. Therefore, the answer is 1/150 seconds, or 2/300 seconds, which is approximately 0.006666 seconds.But to present the answer, they might want it in terms of fractions or using the approximate value. Let's compute 1/150 seconds. 1/150 is approximately 0.006666..., which is 6.666... milliseconds. If they want the exact value, it's 1/150 seconds. If they want a decimal, approximately 0.00667 seconds. But let's see how they want it.But let me check if there's another way to compute this duration. In a sinusoidal wave, the time during which the voltage is above a certain level can be found using the arcsin function. The formula for the duration above a threshold V_threshold in a sinusoidal waveform V_peak sin(theta) is:Δt = (2/ω) * (pi/2 - arcsin(V_threshold / V_peak))But wait, let's see. For a sine wave V(t) = V_peak sin(ωt + phi), the times when V(t) = V_threshold are at t1 and t2, where t1 = (arcsin(V_threshold / V_peak) - phi)/ω and t2 = (pi - arcsin(V_threshold / V_peak) - phi)/ω. Then the duration is t2 - t1.In our case, the function is V(t)=120√2 sin(100πt - pi/6). So V_peak=120√2, V_threshold=84V. So as before, sin(theta)=84/(120√2)=0.5. Then theta=pi/6 and 5pi/6.But theta=100πt - pi/6. Therefore, solving for t gives us t=(theta + pi/6)/(100π).For theta=pi/6: t=(pi/6 + pi/6)/(100π)= (pi/3)/(100π)=1/300≈0.003333.For theta=5pi/6: t=(5pi/6 + pi/6)/(100π)= (6pi/6)/(100π)=pi/(100pi)=1/100=0.01.Thus, the duration is 0.01 - 0.003333=0.006666 seconds=1/150 seconds.So this confirms our previous result.Alternatively, the angular width between the two crossings is 5pi/6 - pi/6=4pi/6=2pi/3. Therefore, the time difference is (2pi/3)/omega= (2pi/3)/(100pi)=2/(300)=1/150 seconds.Yes, this is another way to compute it. The central angle between the two crossings is 2pi/3 radians, so dividing by the angular velocity 100pi rad/s gives the time.So 1/150 seconds is the correct duration. Therefore, during half a period (0.01 seconds), the neon lamp is on for 1/150 seconds≈0.006666 seconds.But let's check if the problem expects the answer in fractions or decimal. Since they mentioned to take sqrt2≈1.4, but in the calculation, we ended up with exact fractions. So perhaps 1/150 seconds is acceptable. But let's compute 1/150 seconds in terms of fractions multiplied by pi or something? Wait, 1/150 is just a fraction. Alternatively, we can write it as (2/300)=2/(3*100)=1/150. Alternatively, if they want milliseconds, 1/150 seconds is approximately 6.666... milliseconds, which can be written as 20/3 milliseconds. But since the problem didn't specify units, just "time", and the original units are in seconds, so 1/150 seconds is appropriate. But maybe they want it in terms of pi? Unlikely, since our calculation didn't involve pi in the final duration.Alternatively, let's see:The angular frequency ω=100π rad/s.The phase angle between the two crossings is 5pi/6 - pi/6=2pi/3.Therefore, the time between crossings is Δt= (2pi/3)/ω= (2pi/3)/(100pi)=2/(300)=1/150.So yes, same result. So the answer is 1/150 seconds.Alternatively, 1/150 s is approximately 0.0067 seconds. If they want it as a decimal, maybe 0.0067 seconds. But since they mentioned to take sqrt2≈1.4, which made our calculation exact, 1/150 is exact, so we can present that.Alternatively, if they want the answer in terms of the period, 1/150 divided by 0.02 (the period) is (1/150)/(1/50)=50/150=1/3. So 1/3 of the period? But the question is about half a period. So 1/150 is 2/3 of half a period, since half a period is 1/100. 1/150 / (1/100)=2/3. So 2/3 of the half period. But the question says "the time during half a period when the neon lamp is on". So the answer is 1/150 seconds, or 2/3 of the half period.But let me check once more. Suppose we have a sine wave V(t)=V_peak sin(ωt + φ). The time during which V(t) > V_threshold in each half period can be calculated as:Δt = (1/ω) * [pi - 2 arcsin(V_threshold / V_peak)]Wait, is that the formula? Let's see.The total time above V_threshold in a half cycle (assuming positive half cycle) would be the time from when the sine wave crosses V_threshold on the way up until it crosses it on the way down. The angular interval between these two points is pi - 2*theta, where theta=arcsin(V_threshold / V_peak). Therefore, the time interval is (pi - 2 theta)/omega.In our case, theta=arcsin(0.5)=pi/6. So angular interval=pi - 2*(pi/6)=pi - pi/3=2pi/3. Then Δt=(2pi/3)/omega=(2pi/3)/(100pi)=2/(300)=1/150 seconds. So yes, same result. So that's another way to remember it.Therefore, the answer for part (2) is 1/150 seconds, which is approximately 0.0067 seconds. But since they told us to take sqrt2≈1.4, and our calculation was exact under that approximation, we can confidently say 1/150 seconds.But let me verify with the approximate value. If we compute 84/(120√2)=84/(120*1.4)=84/168=0.5, as given. Then arcsin(0.5)=30 degrees=pi/6. So the calculation holds.Therefore, the time during half a period when the neon lamp is on is 1/150 seconds. To write this as a box:boxed{dfrac{1}{150}text{ s}}Alternatively, if they prefer decimals, approximately 0.0067 seconds, but since the problem allows using √2≈1.4 which makes the calculation exact, 1/150 is exact.Therefore, summarizing:Part (1):Amplitude: 120√2 voltsFrequency: 50 HzPeriod: 0.02 secondsPart (2):Time during half a period when neon lamp is on: 1/150 secondsFinal Answer(1) The period is boxed{0.02} seconds, the frequency is boxed{50} Hz, and the amplitude is boxed{120sqrt{2}} volts.(2) The time during half a period when the neon lamp is on is boxed{dfrac{1}{150}} seconds.</think>The voltage (v) (in volts) of a certain sinusoidal alternating current varies with time (t) (in seconds) according to the function (v = 120sqrt{2}sin(100pi t - frac{pi}{6})). Part (1)1. Amplitude: The coefficient in front of the sine function is the amplitude. Therefore, the amplitude is (120sqrt{2}) volts.2. Angular Frequency: The coefficient of (t) inside the sine function is (100pi). The frequency (f) is given by (f = frac{omega}{2pi} = frac{100pi}{2pi} = 50) Hz.3. Period: The period (T) is the reciprocal of the frequency, (T = frac{1}{f} = frac{1}{50} = 0.02) seconds. Part (2)The neon lamp lights up when the voltage is greater than 84V. We need to find the time during half a period when the voltage exceeds 84V.1. Solving for (t) when (v = 84V): [ 120sqrt{2} sin(100pi t - frac{pi}{6}) = 84 ] Simplifying, we get: [ sin(100pi t - frac{pi}{6}) = frac{84}{120sqrt{2}} = frac{0.7}{1.4} = 0.5 ] The solutions for (sin(theta) = 0.5) are (theta = frac{pi}{6}) and (theta = frac{5pi}{6}).2. Finding (t) values: - For (theta = frac{pi}{6}): [ 100pi t - frac{pi}{6} = frac{pi}{6} implies 100pi t = frac{pi}{3} implies t = frac{1}{300} ] - For (theta = frac{5pi}{6}): [ 100pi t - frac{pi}{6} = frac{5pi}{6} implies 100pi t = pi implies t = frac{1}{100} ]3. Duration during half a period: - The time interval when the voltage is above 84V is from (t = frac{1}{300}) to (t = frac{1}{100}). - The duration is (frac{1}{100} - frac{1}{300} = frac{3}{300} - frac{1}{300} = frac{2}{300} = frac{1}{150}) seconds. Final Answers(1) The period is (boxed{0.02}) seconds, the frequency is (boxed{50}) Hz, and the amplitude is (boxed{120sqrt{2}}) volts.(2) The time during half a period when the neon lamp is on is (boxed{dfrac{1}{150}}) seconds.,<think>Okay, so I have this problem about a sinusoidal alternating current voltage. The voltage is given by the function v = 120√2 sin(100πt - π/6). There are two parts to the problem. The first part asks for the period, frequency, and amplitude of this voltage. The second part is about a neon lamp that lights up only when the voltage is greater than 84V, and I need to find the time during half a period when the lamp is on. Let me start with part (1).First, let me recall the standard form of a sinusoidal function. Usually, it's written as v = A sin(ωt + φ), where A is the amplitude, ω is the angular frequency, φ is the phase shift, and t is time. The amplitude is straightforward—it's the coefficient in front of the sine function. In this case, the coefficient is 120√2. So, the amplitude should be 120√2 volts. Wait, but sometimes in AC voltage, the amplitude is given as the peak voltage, and the RMS voltage is something else. But the problem specifically asks for the amplitude, which is the peak value. Since the formula is given as sin(...), the amplitude is indeed 120√2. Hmm, but wait, let me check. If the standard RMS voltage for a sine wave is V_rms = V_peak / √2, then maybe they are giving the RMS voltage as 120V, and the amplitude is 120√2? Because in some cases, the voltage is given as RMS. But here, the equation is written as v = 120√2 sin(...), so the amplitude is 120√2. So that's the peak voltage. So I think the amplitude here is 120√2 volts.Next, the angular frequency ω is the coefficient in front of t inside the sine function. In this case, it's 100π. So ω = 100π rad/s. The period T is related to ω by the formula T = 2π / ω. So substituting ω = 100π, we get T = 2π / (100π) = 2 / 100 = 1 / 50 seconds. So the period is 1/50 seconds, which is 0.02 seconds. Then the frequency f is the reciprocal of the period, so f = 1 / T = 50 Hz. That makes sense because standard AC in many countries is 50 Hz or 60 Hz. So here, it's 50 Hz.So summarizing part (1):- Amplitude: 120√2 V- Period: 1/50 s- Frequency: 50 HzI think that's straightforward. Let me double-check. The general form is A sin(ωt + φ). Here, A is 120√2, ω is 100π, so period T = 2π / ω = 2π / 100π = 1/50. Yes. Frequency is 1/T = 50. So that's correct.Now part (2). The neon lamp lights up when the voltage is greater than 84V. We need to find the time during half a period when the lamp is on. The problem mentions to take √2 ≈ 1.4. Let me note that 84V is the threshold. So we need to find all times t in half a period where 120√2 sin(100πt - π/6) > 84. Since the period is 1/50 seconds, half a period is 1/100 seconds, which is 0.01 seconds. So we need to find the interval(s) within a 0.01-second window where the sine function exceeds 84 / (120√2). Let me compute that ratio first.First, compute 84 divided by (120√2). Let's approximate √2 as 1.4 as given. So 120√2 ≈ 120 * 1.4 = 168. So 84 / 168 = 0.5. So essentially, the neon lamp is on when sin(100πt - π/6) > 0.5. So we can write:sin(θ) > 0.5, where θ = 100πt - π/6.We need to find the values of θ where sinθ > 0.5. The sine function is greater than 0.5 in the intervals (π/6, 5π/6) within each period of 2π. But since we are dealing with half a period here, which is π in terms of θ? Wait, let's see.Wait, the original function has angular frequency ω = 100π. The period T is 1/50 seconds, so half a period is 1/100 seconds. Over half a period, the argument of the sine function, θ = 100πt - π/6, will increase by ω*(T/2) = 100π*(1/100) = π. So over half a period, θ increases by π. Therefore, we need to solve for θ in an interval of length π where sinθ > 0.5.But we need to adjust for the phase shift. Let me think. The voltage function is v = 120√2 sin(100πt - π/6). Let me consider the half-period starting at t=0. But depending on the phase shift, the interval where sinθ > 0.5 might cross over into the next half-period. Wait, but the problem says "during half a period", so probably we can consider any half-period. However, since the function is periodic, the time during each half-period when the voltage is above 84V should be the same. Therefore, we can solve it in general.Alternatively, maybe the safest way is to solve the inequality over a half-period interval. Let's consider t from 0 to 1/100 seconds (half a period). Then θ = 100πt - π/6. When t=0, θ = -π/6. When t=1/100, θ = 100π*(1/100) - π/6 = π - π/6 = (5π/6). So θ ranges from -π/6 to 5π/6 over the half-period. So we need to find all θ in (-π/6, 5π/6) where sinθ > 0.5.The sine function is greater than 0.5 between π/6 and 5π/6 in each period. However, since our interval starts at θ = -π/6 and ends at θ = 5π/6, we need to see where in this interval sinθ > 0.5.From θ = -π/6 to θ = 5π/6, the sine function starts at sin(-π/6) = -0.5, increases to sin(π/2) = 1 at θ = π/2, then decreases back to sin(5π/6) = 0.5. Wait, sin(5π/6) is 0.5. So sinθ > 0.5 occurs between θ = π/6 and θ = 5π/6. But our interval starts at θ = -π/6. So the portion of θ where sinθ > 0.5 is from θ = π/6 to θ = 5π/6. Therefore, within the interval θ ∈ (-π/6, 5π/6), the values of θ where sinθ > 0.5 are θ ∈ (π/6, 5π/6). The length of this interval is 5π/6 - π/6 = 4π/6 = 2π/3.But θ is related to t by θ = 100πt - π/6. So solving for t when θ = π/6 and θ = 5π/6.First, when θ = π/6:π/6 = 100πt - π/6Adding π/6 to both sides:π/3 = 100πtDivide both sides by π:1/3 = 100tSo t = 1/(300) ≈ 0.003333... seconds.Second, when θ = 5π/6:5π/6 = 100πt - π/6Adding π/6 to both sides:5π/6 + π/6 = 100πt6π/6 = 100πtπ = 100πtDivide both sides by π:1 = 100tt = 1/100 = 0.01 seconds.Wait, that's the end of the half-period. So the interval when sinθ > 0.5 is from t = 1/300 to t = 1/100. But we need to check if θ = 5π/6 is indeed at t = 1/100. Let's confirm:At t = 1/100, θ = 100π*(1/100) - π/6 = π - π/6 = 5π/6. Correct.But θ = π/6 occurs at t = (π/6 + π/6)/(100π) = (2π/6)/(100π) = (π/3)/(100π) = 1/300. Wait, hold on, let's do the calculation again.Wait, θ = 100πt - π/6. So solving for θ = π/6:π/6 = 100πt - π/6Add π/6 to both sides:π/3 = 100πtDivide by 100π:t = (π/3)/(100π) = 1/(300) seconds.Similarly, θ = 5π/6:5π/6 = 100πt - π/6Add π/6:5π/6 + π/6 = 6π/6 = π = 100πtt = π / 100π = 1/100 seconds.So the time interval when the voltage is above 84V is from t = 1/300 to t = 1/100. The duration is 1/100 - 1/300 = (3/300 - 1/300) = 2/300 = 1/150 seconds ≈ 0.006666... seconds.But wait, is that correct? Let's see. If θ ranges from -π/6 to 5π/6 over the half-period, and sinθ > 0.5 occurs between π/6 and 5π/6 in θ, which corresponds to t between 1/300 and 1/100. Therefore, the duration is 1/100 - 1/300 = 2/300 = 1/150 seconds. So 1/150 seconds is approximately 0.006666... seconds, which is 6.666... milliseconds.But let me check again. When θ = π/6, sinθ = 0.5. So the voltage is exactly 84V at that point, and the problem says the neon lamp lights up when the voltage is greater than 84V. So the interval should be (1/300, 1/100). But we need to make sure that between these two times, the sine function is above 0.5. Since the sine function reaches 0.5 at θ = π/6 and θ = 5π/6, and in between, it's higher. So yes, between those two times, the voltage is above 84V. Therefore, the time during half a period when the lamp is on is 1/150 seconds.But let me verify with another approach. Let's solve the inequality directly.We have:120√2 sin(100πt - π/6) > 84Divide both sides by 120√2:sin(100πt - π/6) > 84 / (120√2) = 7 / (10√2) ≈ 7 / 14 ≈ 0.5 (since √2 ≈ 1.4, 10*1.4=14). So yes, 84/(120√2) = 0.5. Therefore, sin(100πt - π/6) > 0.5.The general solution for sinθ > 0.5 is θ ∈ (π/6 + 2πk, 5π/6 + 2πk) for integer k. However, since we are looking within a specific half-period, which spans θ from -π/6 to 5π/6, we need to find the intersection of the general solution with this interval.In the interval θ ∈ (-π/6, 5π/6), the values where sinθ > 0.5 are θ ∈ (π/6, 5π/6). Therefore, as before, θ must be between π/6 and 5π/6.Converting back to t:θ = 100πt - π/6So,π/6 < 100πt - π/6 < 5π/6Add π/6 to all parts:2π/6 < 100πt < 6π/6Simplify:π/3 < 100πt < πDivide all parts by 100π:1/300 < t < 1/100Therefore, the time interval is from 1/300 to 1/100 seconds, which is 1/150 seconds duration. Therefore, during half a period (which is 1/100 seconds), the lamp is on for 1/150 seconds. Wait, but 1/150 is less than 1/100? Wait, 1/150 is approximately 0.006666..., and 1/100 is 0.01. So 0.006666... is indeed less than 0.01. Wait, but the duration is 1/100 - 1/300 = 2/300 = 1/150. So yes, 1/150 ≈ 0.006666 seconds.But let me check with another perspective. Let's graph the sine function over the half-period. Starting at t=0, θ = -π/6, which is sin(-π/6) = -0.5, so the voltage is 120√2*(-0.5) = -60√2 ≈ -84V. But voltage is negative here. Then the voltage increases, crosses zero at θ=0, which is when t = (π/6)/(100π) = 1/600 ≈ 0.001666 seconds. Then it reaches the peak at θ = π/2, which is sin(π/2) = 1, so voltage is 120√2*1 ≈ 168V. Then it decreases back to 0.5 at θ = 5π/6, which is at t = 1/100 seconds. So between t = 1/300 and t = 1/100, the voltage is above 84V. So the time when it's above is 1/150 seconds. That seems correct.But just to ensure, let's convert 1/150 seconds to milliseconds: 1/150 ≈ 0.006666... seconds ≈ 6.666... milliseconds. And half a period is 10 milliseconds (since period is 20 milliseconds, half is 10). So in 10 milliseconds, the lamp is on for approximately 6.666 milliseconds. That seems reasonable, as the sine wave is above 0.5 for a third of the period? Wait, no. In a full period, the sine wave is above 0.5 for two intervals each of 1/3 of the period. Wait, in a full period of 2π, sinθ > 0.5 occurs in two intervals: (π/6, 5π/6) and (7π/6, 11π/6), but since sine is positive only in the first and second quadrants. Wait, no, actually, in each period, sinθ > 0.5 occurs in two intervals: (π/6, 5π/6) and then again in the negative half, but since we're dealing with a half-period here. Wait, perhaps in a full period, the total time above 0.5 is 2*(5π/6 - π/6)/ω? Wait, maybe better to think in terms of the proportion.But since we are dealing with half a period here, and in that half-period, the sine wave goes from -π/6 to 5π/6. So the portion above 0.5 is from π/6 to 5π/6, which is 4π/6 = 2π/3 in terms of θ. Since the total θ range in half a period is π (from -π/6 to 5π/6, which is 6π/6, so 6π/6 - (-π/6) = 7π/6? Wait, hold on. Wait, θ starts at -π/6 and ends at 5π/6. The total range is 5π/6 - (-π/6) = 6π/6 = π. So the θ interval is π radians. So the portion where sinθ > 0.5 is (5π/6 - π/6) = 4π/6 = 2π/3. Therefore, the proportion is (2π/3)/π) = 2/3. Therefore, the time duration is 2/3 of the half-period. But wait, the half-period is 1/100 seconds, so 2/3 * 1/100 = 2/300 = 1/150 seconds. So that's consistent with previous calculation. Therefore, the answer is 1/150 seconds. So 1/150 is approximately 0.006666... seconds.But the problem says "take √2 ≈ 1.4". Did we use that? When we computed 84/(120√2), we approximated √2 as 1.4, so 120*1.4=168, and 84/168=0.5. That's correct. So that step used the given approximation. Therefore, the calculation is valid.Therefore, the time during half a period when the neon lamp is on is 1/150 seconds, or in fractional form, boxed{dfrac{1}{150}} seconds.Wait, but maybe the problem expects the answer expressed as a decimal using the approximation. Let's see. 1/150 is approximately 0.006666..., which is approximately 0.0067 seconds. But the problem says "take √2 ≈ 1.4", but we already used that in simplifying 84/(120√2) to 0.5. So maybe we can present the answer as 1/150 seconds, which is approximately 0.00667 seconds. But the problem might prefer the exact fraction. Let me check the question again."Find the time during half a period when the neon lamp is on. (Take √2 ≈ 1.4)"They didn't specify to approximate the final answer, so maybe we can leave it as 1/150 seconds. Alternatively, compute it using their approximation. Wait, 1/150 is approximately 0.006666..., which is 6.666... milliseconds. If we use the approximation √2 ≈ 1.4, but we already used that in the calculation. So since we arrived at 1/150 exactly, perhaps we can leave it as 1/150 s. Alternatively, if they want it in terms of fractions with denominator 100, but 1/150 is already simplified. So I think the exact answer is 1/150 seconds. Let me check with another approach.Alternatively, maybe I should express it in terms of π. Wait, no, the time is 1/150 seconds. So unless there's another step, but I think we're done.So summarizing part (2): The time during half a period when the neon lamp is on is 1/150 seconds.Therefore, the final answers are:(1) Period: 1/50 s, Frequency: 50 Hz, Amplitude: 120√2 V;(2) Time on during half a period: 1/150 s.But let me confirm once more. Let's compute the exact value without approximating √2. Wait, the problem says "Take √2 ≈ 1.4", so we are supposed to use that approximation. When we calculated 84 / (120√2), using √2 ≈1.4, so 120*1.4=168, 84/168=0.5 exactly. Therefore, the equation sin(θ) > 0.5 is exact under this approximation. Therefore, the answer 1/150 is exact given the approximation. So yes, the answer is 1/150 seconds.Alternatively, if we didn't approximate √2, then 84/(120√2) = 7/(10√2) = 7√2/20 ≈ 0.4949..., which is slightly less than 0.5. Therefore, the inequality sinθ > 7√2/20. However, the problem instructs us to take √2≈1.4, so 7√2/20 ≈7*1.4/20=9.8/20=0.49. But the problem says to take √2≈1.4, so perhaps 84/(120√2)=84/(120*1.4)=84/168=0.5 exactly. Therefore, the threshold is exactly 0.5. Therefore, the calculation is correct. So the answer is 1/150 seconds.Therefore, I think that's solid.Final Answer(1) The period is boxed{dfrac{1}{50}} s, the frequency is boxed{50} Hz, and the amplitude is boxed{120sqrt{2}} V.(2) The time during half a period when the neon lamp is on is boxed{dfrac{1}{150}} s.</think>The voltage (v) (in volts) of a certain sinusoidal alternating current varies with time (t) (in seconds) according to the function (v = 120sqrt{2}sin(100pi t - frac{pi}{6})).(1) To find the period, frequency, and amplitude of the voltage:- Amplitude: The coefficient in front of the sine function is (120sqrt{2}), so the amplitude is (120sqrt{2}) volts.- Angular Frequency: The coefficient of (t) inside the sine function is (100pi), so the angular frequency (omega = 100pi) rad/s.- Period: The period (T) is calculated as (T = frac{2pi}{omega} = frac{2pi}{100pi} = frac{1}{50}) seconds.- Frequency: The frequency (f) is the reciprocal of the period, so (f = frac{1}{T} = 50) Hz.(2) To find the time during half a period when the neon lamp is on (voltage > 84V):- The neon lamp lights up when (120sqrt{2}sin(100pi t - frac{pi}{6}) > 84).- Simplifying, we get (sin(100pi t - frac{pi}{6}) > 0.5) (using (sqrt{2} approx 1.4)).- The solutions to (sin(theta) > 0.5) are (theta in (frac{pi}{6}, frac{5pi}{6})).- Converting back to (t), solve (frac{pi}{6} = 100pi t - frac{pi}{6}) and (frac{5pi}{6} = 100pi t - frac{pi}{6}): - For (theta = frac{pi}{6}): (t = frac{1}{300}) seconds. - For (theta = frac{5pi}{6}): (t = frac{1}{100}) seconds.- The duration when the voltage is above 84V is (frac{1}{100} - frac{1}{300} = frac{2}{300} = frac{1}{150}) seconds. Final Answer(1) The period is (boxed{dfrac{1}{50}}) s, the frequency is (boxed{50}) Hz, and the amplitude is (boxed{120sqrt{2}}) V.(2) The time during half a period when the neon lamp is on is (boxed{dfrac{1}{150}}) s.]

🔑:Solution: ((1)) The period (T= frac{2pi}{100pi}= frac{1}{50}), the frequency (f= frac{1}{T}=50), and the amplitude (A=120sqrt{2}). ((2)) From (120sqrt{2}sin(100pi t- frac{pi}{6}) > 84) and (sqrt{2} approx 1.4), we get (sin(100pi t- frac{pi}{6}) > frac{1}{2}). Considering the sine graph, for half a period, we have (frac{pi}{6} < 100pi t- frac{pi}{6} < frac{5pi}{6}), solving this gives (frac{1}{300} < t < frac{1}{100}). Therefore, the time during half a period when the neon lamp is on is (frac{1}{100}- frac{1}{300}= frac{2}{300}= frac{1}{150}(s)). So, the final answers are: ((1)) Period (T = boxed{frac{1}{50}}) s, Frequency (f = boxed{50}) Hz, Amplitude (A = boxed{120sqrt{2}}) V. ((2)) The time during half a period when the neon lamp is on is (boxed{frac{1}{150}}) s.

🧠:[<think>Alright, so I have this problem here: "The tangents of the angles of a triangle are in the ratio 1:2:3. How do the sides of the triangle relate to each other?" Hmm, okay. Let me try to figure this out step by step.First, I know that in any triangle, the sum of the angles is 180 degrees. So if the tangents of the angles are in the ratio 1:2:3, then we can denote the angles as A, B, and C such that tan A : tan B : tan C = 1:2:3. Let's assign tan A = k, tan B = 2k, and tan C = 3k for some positive real number k.But wait, since A, B, and C are angles of a triangle, each of them must be between 0 and 180 degrees. Also, since the tangent function isn't defined at 90 degrees, none of the angles can be right angles. Moreover, tangent is positive in the first and third quadrants, but since angles are between 0 and 180, they must all be acute angles (between 0 and 90) or one obtuse angle (between 90 and 180) and two acute angles. However, if one angle is obtuse, then the other two must be acute, but the tangent of an obtuse angle is negative. However, the given ratios are 1:2:3, which are all positive. Therefore, all angles must be acute. So, we have an acute-angled triangle with angles A, B, C such that tan A = 1k, tan B = 2k, tan C = 3k.Since all angles are acute, their tangents are positive. Now, we need to relate the sides of the triangle. In a triangle, the sides are proportional to the sines of the opposite angles (Law of Sines). So, perhaps if we can find the angles, then we can relate the sides via the sine of these angles.But to find the angles, we need to solve for k such that A + B + C = 180 degrees, where tan A = k, tan B = 2k, tan C = 3k. So, maybe we can use some trigonometric identities to relate these tangents. There is the tangent addition formula, but since we have three angles, perhaps we can use the formula for the tangent of the sum of three angles.Alternatively, since A + B + C = 180°, we can consider A + B = 180° - C. Then, tan(A + B) = tan(180° - C) = -tan C. Let's try that.Using the formula for tan(A + B):tan(A + B) = (tan A + tan B) / (1 - tan A tan B) = (k + 2k) / (1 - k*2k) = 3k / (1 - 2k²)But tan(A + B) = -tan C = -3kTherefore, we have:3k / (1 - 2k²) = -3kAssuming k ≠ 0 (which it can't be, because tan A = k, and A is an acute angle, so k > 0), we can divide both sides by 3k:1 / (1 - 2k²) = -1Therefore:1 = - (1 - 2k²)1 = -1 + 2k²Adding 1 to both sides:2 = 2k²Divide by 2:1 = k²So k = 1 (since k > 0)Therefore, tan A = 1, tan B = 2, tan C = 3So now, angles A, B, C have tangents 1, 2, 3 respectively. So we can find each angle.If tan A = 1, then A = 45°, since tan 45° = 1.If tan B = 2, then B is arctan(2). Similarly, tan C = 3, so C is arctan(3).But let's verify that these angles add up to 180°. Let's compute A + B + C.A = 45°, B = arctan(2), C = arctan(3). Let's check if 45° + arctan(2) + arctan(3) = 180°.I remember that arctan(1) + arctan(2) + arctan(3) = 180°, which is a well-known identity. Wait, but here A is arctan(1) = 45°, so indeed, 45° + arctan(2) + arctan(3) = 180°. So that checks out. So the angles are 45°, arctan(2), and arctan(3). Good.Now, we need to relate the sides of the triangle. Let's denote the sides opposite angles A, B, C as a, b, c respectively.By the Law of Sines:a / sin A = b / sin B = c / sin C = 2R (where R is the circumradius)Therefore, the sides are proportional to sin A, sin B, sin C.So, if we can compute sin A, sin B, sin C, then the sides will be in the ratio sin A : sin B : sin C.Given that A = 45°, sin A = √2 / 2 ≈ 0.7071.For angle B, which is arctan(2). Let's compute sin(arctan(2)).If tan θ = 2, then we can imagine a right triangle where the opposite side is 2, adjacent is 1, so hypotenuse is √(1² + 2²) = √5. Therefore, sin θ = opposite / hypotenuse = 2 / √5.Similarly, for angle C = arctan(3). Then, tan θ = 3, so opposite = 3, adjacent = 1, hypotenuse = √(1 + 9) = √10. Therefore, sin θ = 3 / √10.Therefore, sin A : sin B : sin C = (√2 / 2) : (2 / √5) : (3 / √10)But we need to express this ratio in simplest form, possibly rationalizing denominators or scaling appropriately.First, let's write all terms with rational denominators.√2 / 2 is already okay.2 / √5 can be rationalized as 2√5 / 5.3 / √10 can be rationalized as 3√10 / 10.So the ratio becomes:(√2 / 2) : (2√5 / 5) : (3√10 / 10)To simplify ratios with different denominators, it's helpful to express them with a common denominator. The denominators are 2, 5, and 10. The least common denominator is 10.Multiply each term by 10 to eliminate denominators:(√2 / 2) * 10 : (2√5 / 5) * 10 : (3√10 / 10) * 10Which simplifies to:5√2 : 4√5 : 3√10So the sides are in the ratio 5√2 : 4√5 : 3√10.But perhaps we can simplify this ratio further by factoring out common terms or expressing in terms of square roots. Let's check each term:5√2 is 5√24√5 is 4√53√10 is 3√10Alternatively, note that √10 = √2 * √5, so maybe express all terms in terms of √2 and √5.But I don't think that would combine them. Let's see:5√2 : 4√5 : 3√2√5But that might not help. Alternatively, perhaps express the ratio as multiples of √2 and √5.But actually, the ratio is 5√2 : 4√5 : 3√10. Maybe this is as simplified as it can get. But let me check if there's a different approach.Alternatively, since the sides are proportional to the sines of the angles, which we found as √2/2, 2/√5, and 3/√10. Let's compute these as numerical values to check their ratios.Compute sin A = √2 / 2 ≈ 0.7071sin B = 2 / √5 ≈ 2 / 2.236 ≈ 0.8944sin C = 3 / √10 ≈ 3 / 3.162 ≈ 0.9487So the ratio is approximately 0.7071 : 0.8944 : 0.9487But these are approximate values. To get the exact ratio, as before, we have:Multiply each term by 10 to get rid of denominators:5√2 : 4√5 : 3√10Alternatively, factor out √2 and √5:First term: 5√2Second term: 4√5Third term: 3√10 = 3√2√5So the ratio can be written as 5√2 : 4√5 : 3√2√5But that might not necessarily make it simpler. Alternatively, we can note that all terms have either √2 or √5 or both, but perhaps we can express them in terms of a common factor.Alternatively, if we square the ratios, we can relate them as:(5√2)^2 : (4√5)^2 : (3√10)^2 = 25*2 : 16*5 : 9*10 = 50 : 80 : 90 = 5 : 8 : 9Wait, that's interesting! Squaring the terms gives a ratio of 5:8:9. So the squares of the sides are in the ratio 5:8:9, which is a much simpler ratio.But wait, hold on. If the sides are in the ratio 5√2 : 4√5 : 3√10, then the squares would be (25*2) : (16*5) : (9*10) = 50 : 80 : 90, which simplifies to 5:8:9 by dividing each term by 10. So 50/10=5, 80/10=8, 90/10=9. So the squares of the sides are in the ratio 5:8:9.But does this help us? Well, in a triangle, if the squares of the sides are in a particular ratio, that relates to the Law of Cosines. For example, if a^2 : b^2 : c^2 = 5:8:9, then we can say something about the angles. But in this case, the problem is asking how the sides relate to each other, not their squares.However, the problem might accept the ratio of the squares as the answer, but I think the problem expects the ratio of the sides themselves, not their squares. But the sides are in the ratio 5√2 : 4√5 : 3√10, which is the simplified form.Alternatively, perhaps there is a more simplified ratio. Let's see:Let me check if 5√2, 4√5, and 3√10 have a common factor. Well, they all have a radical, but different ones. If we factor each term:5√2 = 5 * √24√5 = 4 * √53√10 = 3 * √10 = 3 * √2 * √5So, the common factors among the terms are √2 and √5. But since each term has different combinations, there isn't a common factor that can be factored out. Therefore, the ratio 5√2 : 4√5 : 3√10 is the simplest form in terms of radicals.Alternatively, we can rationalize the original ratio of sines. Wait, the original ratio of sines was (√2 / 2) : (2 / √5) : (3 / √10). To make these ratios have a common denominator or something.But maybe multiplying each term by √10 to eliminate denominators:(√2 / 2) * √10 : (2 / √5) * √10 : (3 / √10) * √10Simplify each term:First term: (√2 * √10) / 2 = √20 / 2 = (2√5)/2 = √5Second term: (2 * √10) / √5 = 2√(10/5) = 2√2Third term: 3So, after multiplying each term by √10, we get √5 : 2√2 : 3Wait, that's a much simpler ratio! So if we multiply each term in the original sine ratio by √10, we get √5 : 2√2 : 3. Therefore, the sides are proportional to √5 : 2√2 : 3.But why does this work? Because ratios can be scaled by any constant factor. So, if we have the ratio (√2 / 2) : (2 / √5) : (3 / √10), and we multiply each term by √10, we get:First term: (√2 / 2) * √10 = √(2*10)/2 = √20 / 2 = (2√5)/2 = √5Second term: (2 / √5) * √10 = 2√(10/5) = 2√2Third term: (3 / √10) * √10 = 3Therefore, the ratio simplifies to √5 : 2√2 : 3.Hence, the sides of the triangle are in the ratio √5 : 2√2 : 3.But let me verify this. Let's check if √5 : 2√2 : 3 corresponds to the sines of the angles.Given that:sin A = √2 / 2 ≈ 0.7071sin B = 2 / √5 ≈ 0.8944sin C = 3 / √10 ≈ 0.9487Now, the ratio √5 : 2√2 : 3, when normalized, should be proportional to the sines.Compute √5 ≈ 2.236, 2√2 ≈ 2.828, 3 ≈ 3.But if we divide each by their respective denominators from the sine expressions:Wait, maybe I'm overcomplicating. Let's take the ratio √5 : 2√2 : 3 and divide each term by √10 to get back to the original sine terms:√5 / √10 = √(5/10) = √(1/2) = 1/√2 ≈ 0.70712√2 / √10 = 2√2 / (√2 * √5) = 2 / √5 ≈ 0.89443 / √10 ≈ 0.9487Which matches the sines of the angles. Therefore, multiplying the ratio √5 : 2√2 : 3 by 1/√10 gives the sine values. Therefore, the sides are indeed proportional to √5 : 2√2 : 3.Alternatively, since ratios are defined up to a constant multiple, the answer can be presented as √5 : 2√2 : 3. Alternatively, if they need to be in the form with combined radicals, but I think √5 : 2√2 : 3 is acceptable.But let me check if there's another way to approach this problem. Maybe using the fact that in a triangle, if tan A : tan B : tan C = 1:2:3, perhaps there's a relationship between the sides via the Law of Tangents or something else. But I think the approach I took is solid.Alternatively, let's suppose that the triangle is such that the sides are proportional to a, b, c, and we need to find the relationship between a, b, c. Since the Law of Sines relates sides to sines of angles, and the Law of Cosines relates sides to cosines of angles.But we know the tangents of the angles, which are ratios of sine over cosine. So, perhaps we can express sines and cosines in terms of the given tangents.Given tan A = 1, so sin A = 1 / √(1 + 1²) = 1 / √2, cos A = 1 / √2Similarly, tan B = 2, so sin B = 2 / √(1 + 2²) = 2 / √5, cos B = 1 / √5tan C = 3, so sin C = 3 / √(1 + 3²) = 3 / √10, cos C = 1 / √10Therefore, the sides a, b, c are proportional to sin A, sin B, sin C, which are 1 / √2, 2 / √5, 3 / √10. Which is the same as before. Then, to simplify the ratio, we can rationalize and scale.Which leads us to the ratio √5 : 2√2 : 3 as before.Alternatively, since the problem says "how do the sides of the triangle relate to each other?", which can be answered by stating the ratio of the sides. So the answer is that the sides are in the ratio √5 : 2√2 : 3.But let me check if there's an alternative way to express this ratio. For example, perhaps squaring the sides gives 5 : 8 : 9, which is a simpler ratio, but the problem asks for the relation of the sides, not their squares. So maybe both answers are related, but the sides themselves are in the ratio √5 : 2√2 : 3.Alternatively, if we consider that 5:8:9 are the squares, then the sides would be in the ratio of the square roots, which is √5 : √8 : √9, but √8 is 2√2, and √9 is 3. So indeed, √5 : 2√2 : 3. Therefore, that confirms the earlier result.Therefore, the sides of the triangle relate to each other in the ratio √5 : 2√2 : 3.Just to recap:1. Let angles be A, B, C with tan A = 1, tan B = 2, tan C = 3.2. Verify angles sum to 180°, which they do (45° + arctan(2) + arctan(3) = 180°).3. Use Law of Sines: sides proportional to sin A, sin B, sin C.4. Compute sines using right triangle definitions from given tangents.5. Simplify the ratio of sines by multiplying by √10 to get rid of denominators, resulting in √5 : 2√2 : 3.6. Confirm via squaring sides to get 5:8:9, which matches the simplified ratio.Hence, the conclusion is that the sides are in the ratio √5 : 2√2 : 3.Final AnswerThe sides of the triangle are in the ratio boxed{sqrt{5} : 2sqrt{2} : 3}.</think>Given that the tangents of the angles of a triangle are in the ratio (1: 2: 3), we denote the angles as (A), (B), and (C) such that (tan A = k), (tan B = 2k), and (tan C = 3k). Since the sum of the angles in a triangle is (180^circ), we use the tangent addition formula and verify that (k = 1). Thus, the angles are (A = 45^circ), (B = arctan(2)), and (C = arctan(3)).Using the Law of Sines, the sides of the triangle are proportional to the sines of the opposite angles. We calculate the sines of these angles:- (sin 45^circ = frac{sqrt{2}}{2})- (sin(arctan(2)) = frac{2}{sqrt{5}})- (sin(arctan(3)) = frac{3}{sqrt{10}})We then simplify the ratio of these sines by multiplying each term by (sqrt{10}) to eliminate the denominators:- (frac{sqrt{2}}{2} cdot sqrt{10} = sqrt{5})- (frac{2}{sqrt{5}} cdot sqrt{10} = 2sqrt{2})- (frac{3}{sqrt{10}} cdot sqrt{10} = 3)Thus, the sides of the triangle are in the ratio (sqrt{5} : 2sqrt{2} : 3).[boxed{sqrt{5} : 2sqrt{2} : 3}],<think>Okay, so I have this problem where the tangents of the angles of a triangle are in the ratio 1:2:3. I need to figure out how the sides of the triangle relate to each other. Hmm, let's start by recalling some basic triangle properties. In any triangle, the sum of the angles is 180 degrees, or π radians. Also, the tangent function relates to angles in a right-angled triangle, but here we have a general triangle. The sides of a triangle are related to the angles through the Law of Sines and the Law of Cosines. Maybe those will come in handy here.Given that the tangents are in the ratio 1:2:3, let's denote the angles as A, B, and C. So, tan A : tan B : tan C = 1:2:3. Let me assign variables to these. Let’s say tan A = k, tan B = 2k, and tan C = 3k for some positive real number k. Since A, B, C are angles of a triangle, each must be between 0 and π radians, and their sum is π. So, A + B + C = π. But how do I relate the tangents of the angles to the sides? Maybe I can use some trigonometric identities or relations. For example, there's a formula that relates the tangent of angles in a triangle. Wait, in any triangle, if I know the tangents of the angles, can I find the angles themselves? If I can find the angles, then I can use the Law of Sines to relate the sides. Let's try that approach.First, let me express A, B, and C in terms of their tangents. Let's denote:tan A = 1ktan B = 2ktan C = 3kSince A + B + C = π, perhaps I can use the tangent addition formula. Let me recall that tan(A + B) = (tan A + tan B)/(1 - tan A tan B). But since A + B = π - C, then tan(A + B) = tan(π - C) = -tan C. So,tan(A + B) = -tan CBut from the formula:tan(A + B) = (tan A + tan B)/(1 - tan A tan B) = (k + 2k)/(1 - k*2k) = 3k/(1 - 2k²)And this equals -tan C = -3kSo, setting them equal:3k/(1 - 2k²) = -3kLet me solve for k. Let's assume k ≠ 0 (since tangent of 0 is 0, which can't be an angle in a triangle), so we can divide both sides by 3k:1/(1 - 2k²) = -1Multiply both sides by (1 - 2k²):1 = - (1 - 2k²)1 = -1 + 2k²Add 1 to both sides:2 = 2k²Divide by 2:1 = k²So, k = ±1. But since tan A, tan B, tan C are all positive (angles in a triangle must be between 0 and π, and their tangents must be positive if the angles are acute or obtuse but in a way that makes sense... Wait, tangent is positive in first and third quadrants, but angles in a triangle are between 0 and π, so first and second quadrants. So tangent is positive for acute angles (first quadrant) and negative for obtuse angles (second quadrant). But the problem states that the tangents are in ratio 1:2:3, which are all positive. Therefore, all angles must have positive tangents, which implies that all angles are acute (since in the second quadrant, tangent is negative). Therefore, all angles must be acute, so the triangle is acute. Therefore, k must be positive, so k=1.Therefore, tan A = 1, tan B = 2, tan C = 3.So now, we can find the angles A, B, C by taking arctangent of 1, 2, 3. Let me compute these in degrees to get a better sense:tan A = 1 ⇒ A = 45°tan B = 2 ⇒ arctan(2) ≈ 63.4349°tan C = 3 ⇒ arctan(3) ≈ 71.5651°Let me check the sum: 45 + 63.4349 + 71.5651 ≈ 180°, which works out. So that's good. So the angles are 45°, arctan(2), arctan(3). Now, how does this relate to the sides?By the Law of Sines, the sides are proportional to the sines of the opposite angles. So, if we denote the sides opposite angles A, B, C as a, b, c respectively, then a/sin A = b/sin B = c/sin C = 2R (where R is the radius of the circumscribed circle). Therefore, the sides are proportional to sin A, sin B, sin C.So, to find the ratio of the sides, we can compute sin A : sin B : sin C.We have A = 45°, so sin A = sin(45°) = √2 / 2 ≈ 0.7071.Angle B is arctan(2). Let's find sin(arctan(2)). Let me consider a right triangle where the opposite side is 2 and adjacent side is 1, so hypotenuse is √(1² + 2²) = √5. Therefore, sin(arctan(2)) = opposite/hypotenuse = 2/√5 ≈ 0.8944.Similarly, angle C is arctan(3). So sin(arctan(3)) can be found by considering a right triangle with opposite side 3 and adjacent side 1, hypotenuse √(1² + 3²) = √10. Thus, sin(arctan(3)) = 3/√10 ≈ 0.9487.Therefore, the ratio of the sides a : b : c is:√2 / 2 : 2/√5 : 3/√10But we need to express this ratio in simplest form, possibly rationalizing denominators or scaling to eliminate radicals.First, let's write all terms with denominator rationalized:√2 / 2 = (√2)/22/√5 = (2√5)/53/√10 = (3√10)/10So, the ratio is (√2)/2 : (2√5)/5 : (3√10)/10To express these ratios more cleanly, we can find a common denominator. The denominators are 2, 5, 10. The least common denominator is 10.Multiply each term by 10 to eliminate denominators:(√2)/2 * 10 = 5√2(2√5)/5 * 10 = 4√5(3√10)/10 * 10 = 3√10Therefore, the ratio becomes 5√2 : 4√5 : 3√10But this still has different radicals. Maybe we can simplify further?Alternatively, perhaps we can write each term in terms of √10 or some common radical. Let me check.Alternatively, maybe we can square each term to eliminate radicals and then find the ratio.So, squaring each term:(5√2)^2 = 25 * 2 = 50(4√5)^2 = 16 * 5 = 80(3√10)^2 = 9 * 10 = 90Therefore, the squares of the sides are in the ratio 50 : 80 : 90, which simplifies to 5 : 8 : 9 (dividing each by 10).Wait, 50:80:90 reduces by dividing each term by 10: 5:8:9. Then, 5,8,9 are in the ratio.But wait, if the squares of the sides are in the ratio 5:8:9, then the sides themselves are in the ratio sqrt(5):sqrt(8):sqrt(9), which is sqrt(5):2sqrt(2):3. But wait, that contradicts the earlier ratio. Hmm, perhaps this approach isn't correct because when we square the terms, the ratio of the squares is not the same as the square of the ratio unless all terms are positive, which they are, but squaring can create different relations.Wait, maybe this suggests that the sides are proportional to sqrt(5), sqrt(8), sqrt(9), but that can't be because the original ratio from the sines gave us 5√2 : 4√5 : 3√10. Let's check:If we square 5√2, we get 25*2=50.Similarly, (4√5)^2=16*5=80.(3√10)^2=9*10=90.So indeed, the squares are 50,80,90 which simplifies to 5:8:9.But 50:80:90 reduces by dividing each term by 10: 5:8:9. So the squares of the sides are in ratio 5:8:9, which might suggest the sides are in ratio sqrt(5):sqrt(8):sqrt(9) or sqrt(5):2sqrt(2):3. However, in the previous approach, we had 5√2 : 4√5 : 3√10. Let's check if these two expressions are proportional.Let me compute (5√2)^2 = 50, (4√5)^2 = 80, (3√10)^2 = 90. So these are the same as the squares. So 5√2, 4√5, 3√10 are sides whose squares are 50,80,90. So if we take the ratio of the sides as 5√2 :4√5 :3√10, that's equivalent to sqrt(50):sqrt(80):sqrt(90). Which simplifies to sqrt(50):sqrt(80):sqrt(90) = 5√2:4√5:3√10.Alternatively, if we factor sqrt(10) out of each term:5√2 = 5√24√5 = 4√53√10 = 3√10But sqrt(10) is common in all? No, wait:5√2 = 5√24√5 = 4√53√10 = 3√2*√5So maybe factor out sqrt(2) and sqrt(5):But perhaps there's a better way to write the ratio. Let me see.Alternatively, note that 5√2 :4√5 :3√10 can be written as 5√2 :4√5 :3√(2*5) = 5√2 :4√5 :3√2√5. So if we divide each term by √5, we get 5√(2/5) :4 :3√2. Not sure if that helps. Alternatively, divide each term by √2: 5 :4√(5/2) :3√5. Hmm, not helpful.Alternatively, maybe express all terms in terms of √10:5√2 = 5*(√10 / √5) = 5√10 / √5 = √10*(5/√5) = √10*(√5) = √50Wait, that seems a convoluted way. Alternatively, 5√2 is equal to √(25*2)=√50, 4√5=√(16*5)=√80, 3√10=√(9*10)=√90. So indeed, the sides are √50, √80, √90, which are in the ratio sqrt(50):sqrt(80):sqrt(90). Which reduces to sqrt(5*10):sqrt(8*10):sqrt(9*10) = sqrt(10)*(sqrt5):sqrt(10)*sqrt8:sqrt(10)*sqrt9. So they all have a factor of sqrt10, so we can factor that out:sqrt10*(sqrt5 : sqrt8 : sqrt9) = sqrt10*(sqrt5 : 2sqrt2 : 3). Therefore, the ratio simplifies to sqrt5 : 2sqrt2 : 3 when we factor out sqrt10. But since ratios are up to a constant multiple, we can write the sides as being proportional to sqrt5, 2sqrt2, and 3. Wait, but when we factor out sqrt10, we get sqrt10 multiplied by (sqrt5 : 2sqrt2 : 3). So the sides are proportional to sqrt5 : 2sqrt2 : 3. That seems a simpler ratio. Let me check if that is consistent with the sine ratios.Wait, initially, we had sin A = √2/2 ≈ 0.7071, sin B = 2/√5 ≈ 0.8944, sin C = 3/√10 ≈ 0.9487. So if we take the ratio of these sines:√2/2 : 2/√5 : 3/√10Multiply numerator and denominator appropriately to rationalize:√2/2 = (√2)/22/√5 = (2√5)/53/√10 = (3√10)/10So the ratio is (√2)/2 : (2√5)/5 : (3√10)/10To find a common denominator, say 10:Multiply each term by 10:(√2)/2 *10 = 5√2(2√5)/5 *10 =4√5(3√10)/10 *10= 3√10So ratio is 5√2 :4√5 :3√10. Which, as before, can be written as sqrt5 :2sqrt2 :3 multiplied by sqrt10. Wait, but 5√2 is 5*√2, 4√5 is 4*√5, 3√10 is 3*√10. If we factor sqrt10 from each term, we get:5√2 = 5√2 = 5√2*(sqrt5/sqrt5) = 5*sqrt2*sqrt5 / sqrt5 = 5*sqrt10 / sqrt5 = (5/sqrt5)*sqrt10 = sqrt5*sqrt10Wait, 5/sqrt5 is sqrt5, so 5√2 = sqrt5*sqrt10. Similarly, 4√5 = 4√5*(sqrt2/sqrt2) = 4*sqrt10 / sqrt2 = (4/sqrt2)*sqrt10 = 2*sqrt2*sqrt10. Similarly, 3√10 is 3*sqrt10. Therefore, the terms can be written as sqrt5*sqrt10, 2sqrt2*sqrt10, 3*sqrt10. So factoring sqrt10 out, we get sqrt10*(sqrt5 : 2sqrt2 : 3). Therefore, the sides are proportional to sqrt5 : 2sqrt2 : 3 times sqrt10. But since the entire ratio is scaled by sqrt10, which is a common factor, we can say the sides are proportional to sqrt5 : 2sqrt2 : 3. But in the problem statement, it's asking "how do the sides of the triangle relate to each other?" So they probably want the ratio in terms of simplest radical form. So sqrt5 : 2sqrt2 :3. Let me check if this ratio is equivalent to the ratio 5√2 :4√5 :3√10. Let's see:If we take sqrt5 : 2sqrt2 :3 and multiply each term by sqrt10, we get sqrt5*sqrt10 :2sqrt2*sqrt10 :3sqrt10. Which is sqrt50 : sqrt40 : sqrt90. Which simplifies to 5√2 : 2√10 : 3√10. Wait, that's not the same as 5√2 :4√5 :3√10. Hmm, so maybe my previous factoring was incorrect.Wait, perhaps there's a miscalculation here. Let me check:sqrt5*sqrt10 = sqrt(5*10)=sqrt50=5√22sqrt2*sqrt10=2sqrt(2*10)=2sqrt20=2*2√5=4√53*sqrt10=3sqrt10Ah, yes! So sqrt5*sqrt10 =5√2, 2sqrt2*sqrt10=4√5, and 3*sqrt10=3√10. Therefore, the ratio sqrt5 :2sqrt2 :3 scaled by sqrt10 gives us 5√2 :4√5 :3√10. So this shows that the original ratio 5√2 :4√5 :3√10 can be written as sqrt10*(sqrt5 :2sqrt2 :3). Therefore, the sides are proportional to sqrt5 :2sqrt2 :3. But since ratios can be scaled by any constant factor, sqrt5 :2sqrt2 :3 is the simplified form. Alternatively, if we want to express the sides in terms of integers multiplied by common radicals, perhaps 1 : sqrt(8/5) : 3/sqrt5. But that seems more complicated.Alternatively, maybe rationalize each term. Let's see:sqrt5 :2sqrt2 :3.But that is already simplified. Alternatively, the problem might prefer the sides expressed as multiples of some common radical, but since they are different radicals, sqrt5, sqrt8, and 3, which is rational, it might be acceptable as sqrt5 :2sqrt2 :3.But let me confirm with the Law of Sines. If angles are A=45°, B=arctan(2), C=arctan(3), then sides are proportional to sin A, sin B, sin C. Which we found as √2/2, 2/√5, 3/√10. Let's compute these numerically:√2/2 ≈0.70712/√5≈0.89443/√10≈0.9487So the ratio is approximately 0.7071 :0.8944 :0.9487. Let's divide each term by the smallest one, which is 0.7071:0.7071/0.7071 :0.8944/0.7071 :0.9487/0.7071 ≈1 :1.2649 :1.3416But sqrt5 ≈2.236, 2sqrt2≈2.828, 3 is 3. So if we take sqrt5 :2sqrt2 :3, which is approximately 2.236 :2.828 :3. Dividing each by 2.236 gives 1 :1.2649 :1.3416, which matches the above. So yes, the ratio sqrt5 :2sqrt2 :3 is equivalent to the original sine ratios. Therefore, the sides of the triangle are in the ratio sqrt5 :2sqrt2 :3.But let me verify once more. Alternatively, perhaps there's a different approach. Let's consider using the Law of Cosines. If we can compute the sides using the angles, we can relate them.Alternatively, since we know the tangents of all angles, and we can find the sines and cosines.Given tan A =1, so A=45°, sin A=√2/2, cos A=√2/2.tan B=2, so we can think of a right triangle with opposite side 2, adjacent side 1, hypotenuse sqrt(1+4)=sqrt5. Therefore, sin B=2/sqrt5, cos B=1/sqrt5.Similarly, tan C=3, so opposite=3, adjacent=1, hypotenuse sqrt(10). Therefore, sin C=3/sqrt10, cos C=1/sqrt10.Using the Law of Sines, the sides a, b, c opposite angles A, B, C are proportional to sin A, sin B, sin C. So:a : b : c = sin A : sin B : sin C = √2/2 : 2/sqrt5 :3/sqrt10.Which is the same as before. Rationalizing denominators:√2/2 = (√2)/22/sqrt5 = (2sqrt5)/53/sqrt10 = (3sqrt10)/10Then, multiplying each term by 10 to eliminate denominators:5√2 :4sqrt5 :3sqrt10.Which simplifies to sqrt5 :2sqrt2 :3 when factoring sqrt10. So both methods lead to the same ratio.Alternatively, since the problem says "how do the sides relate to each other", maybe there's a Pythagorean triplet or some relation. Let me check if 5√2,4√5,3√10 satisfy the Pythagorean theorem. Let's square each:(5√2)^2 = 25*2=50(4√5)^2=16*5=80(3√10)^2=9*10=90Check if 50 +80=130, which is not 90, so not a right triangle. So that approach isn't helpful.Alternatively, maybe the triangle with angles 45°, arctan(2), arctan(3) is a specific type, but it's not a right-angled triangle, as the angles are all acute. So perhaps the answer is that the sides are in the ratio sqrt5 :2sqrt2 :3.Alternatively, maybe the problem expects the sides to be in the ratio 1:2:3, but that seems unlikely because the sides are related to the sines of the angles, not the tangents. Since the angles are not in the ratio 1:2:3, but their tangents are. So the sides can't be in 1:2:3. Therefore, the ratio sqrt5 :2sqrt2 :3 is the correct one.Alternatively, perhaps the problem wants the sides expressed as multiples of some common term. For example, factoring sqrt5 from the first term, sqrt2 from the second, but that seems not useful. Alternatively, note that sqrt5 :2sqrt2 :3 can be written as 1 : (2sqrt2)/sqrt5 : 3/sqrt5. But again, not sure if that's helpful.Alternatively, rationalizing the ratios:sqrt5 :2sqrt2 :3Multiply each term by sqrt5 to eliminate the radical in the first term:sqrt5*sqrt5 :2sqrt2*sqrt5 :3*sqrt5Which is 5 :2sqrt10 :3sqrt5But this complicates the ratio further. So probably not.Therefore, the most simplified form of the ratio is sqrt5 :2sqrt2 :3.Alternatively, perhaps the problem expects the sides to be proportional to 1, sqrt(2), sqrt(3) or something else, but given the process, it's sqrt5 :2sqrt2 :3.But let me verify once again. Let me compute the sides using Law of Sines with specific angles. Suppose we take angle A=45°, angle B=arctan(2)≈63.4349°, angle C≈71.5651°, summing to 180°.Let’s compute the sine of each angle:sin(45°)=√2/2≈0.7071sin(arctan(2))=2/sqrt5≈0.8944sin(arctan(3))=3/sqrt10≈0.9487So the ratio is approximately 0.7071 :0.8944 :0.9487. Let's divide each by 0.7071 to get a ratio relative to the first side:1 : 1.2649 : 1.3416Compare this to sqrt5≈2.236, 2sqrt2≈2.828, 3. So sqrt5 :2sqrt2 :3≈2.236:2.828:3. If we divide each term by sqrt5≈2.236, we get 1 :1.2649 :1.3416, which matches the above. Therefore, yes, the ratio sqrt5 :2sqrt2 :3 is equivalent to the sine ratios. So the sides are in the ratio sqrt5 :2sqrt2 :3.Thus, after careful consideration, the sides relate to each other in the ratio sqrt(5) : 2sqrt(2) : 3.Final AnswerThe sides of the triangle are in the ratio boxed{sqrt{5} : 2sqrt{2} : 3}.</think>Given that the tangents of the angles of a triangle are in the ratio (1: 2: 3), we denote the angles as (A), (B), and (C) such that (tan A = k), (tan B = 2k), and (tan C = 3k). Using the identity for the tangent of a sum of angles, we find that (k = 1), leading to (tan A = 1), (tan B = 2), and (tan C = 3). This implies the angles are (45^circ), (arctan(2)), and (arctan(3)).Using the Law of Sines, the sides opposite these angles are proportional to the sines of the angles. Calculating the sines of these angles:- (sin(45^circ) = frac{sqrt{2}}{2})- (sin(arctan(2)) = frac{2}{sqrt{5}})- (sin(arctan(3)) = frac{3}{sqrt{10}})Rationalizing the denominators and simplifying the ratios, we find the sides are proportional to:[frac{sqrt{2}}{2} : frac{2sqrt{5}}{5} : frac{3sqrt{10}}{10}]Multiplying each term by 10 to clear the denominators, we get:[5sqrt{2} : 4sqrt{5} : 3sqrt{10}]Factoring out (sqrt{10}) from each term, we simplify this to:[sqrt{5} : 2sqrt{2} : 3]Thus, the sides of the triangle are in the ratio (boxed{sqrt{5} : 2sqrt{2} : 3}).]

🔑:I. Solution: Let operatorname{tg} alpha=lambda, operatorname{tg} beta=2 lambda, operatorname{tg} gamma=3 lambda.operatorname{tg} gamma=operatorname{tg}left[180^{circ}-(alpha+beta)right]=-operatorname{tg}(alpha+beta)=frac{operatorname{tg} alpha+operatorname{tg} beta}{1-operatorname{tg} alpha operatorname{tg} beta}thus3 lambda=-frac{lambda+2 lambda}{1-lambda cdot 2 lambda}=-frac{3 lambda}{1-2 lambda^{2}}from which1-2 lambda^{2}=-1hencelambda^{2}=1Since lambda cannot be negative (otherwise all angles of the triangle would be greater than a right angle), therefore lambda=1, and thusoperatorname{tg} alpha=1, quad operatorname{tg} beta=2, quad operatorname{tg} gamma=3Thereforebegin{gathered}a: b: c=sin alpha: sin beta: sin gamma=frac{operatorname{tg} alpha}{sqrt{1+operatorname{tg}^{2} alpha}}: frac{operatorname{tg} beta}{sqrt{1+operatorname{tg}^{2} beta}}: frac{operatorname{tg} gamma}{sqrt{1+operatorname{tg}^{2} gamma}}= =frac{1}{sqrt{2}}: frac{1}{sqrt{5}}: frac{1}{sqrt{10}}=sqrt{5}: 2 sqrt{2}: 3end{gathered}Zawadowski Alfréd (Bp., I. Petőfi g. III. o. t.)II. Solution: Based on the first solution, operatorname{tg} alpha=1, operatorname{tg} beta=2, operatorname{tg} gamma=3.We construct a triangle where the height corresponding to side b is B B_{1}=m_{b}=3 (see the first figure) and the two segments on side b are A B_{1}=b_{1}=3 and B_{1} C=b_{2}=1.In this triangle, indeed operatorname{tg} alpha=1, operatorname{tg} gamma=3 and operatorname{tg} beta=-frac{1+3}{1-3}=2.The sides of this triangle arebegin{aligned}& a=sqrt{m_{b}^{2}+b_{2}^{2}}=sqrt{9+1}=sqrt{10} & b=b_{1}+b_{2}=4end{aligned}andc=sqrt{m_{b}^{2}+b_{1}^{2}}=sqrt{9+9}=sqrt{18}Thereforea: b: c=sqrt{10}: sqrt{16}: sqrt{18}=sqrt{5}: sqrt{8}: sqrt{9}=sqrt{5}: 2 sqrt{2}: 3Lázár László (Mezőkövesd, I. László g. III. o. t.)

🧠:[<think>Okay, let me try to work through this problem step by step. The inequality given is:frac{-log _{2}(105+2 x sqrt{x+19})^{3}+left|log _{2} frac{105+2 x sqrt{x+19}}{left(x^{2}+x+3right)^{4}}right|}{9 log _{5}(76+2 x sqrt{x+19})-4 log _{2}(105+2 x sqrt{x+19})} geqslant 0Wow, that's a mouthful. Let me break it down. First, I need to simplify the numerator and the denominator separately. Let's start by understanding the domain of the function, because with logarithms and square roots, there are some restrictions on x.Starting with the square root: sqrt{x + 19}. The expression inside the square root must be non-negative, so:x + 19 geq 0 implies x geq -19Then, the arguments of the logarithms must be positive. Let's look at the arguments:1. 105 + 2x sqrt{x + 19} must be positive.2. 76 + 2x sqrt{x + 19} must be positive.3. x^2 + x + 3 must be positive (since it's in the denominator inside a logarithm).Let's check each of these.First, x^2 + x + 3. Since this is a quadratic in x, let's compute its discriminant: D = 1 - 12 = -11 < 0. Since the discriminant is negative and the coefficient of x^2 is positive, the quadratic is always positive. So, we don't need to worry about the denominator inside the logarithm; it's always positive.Now, 105 + 2x sqrt{x + 19} > 0 and 76 + 2x sqrt{x + 19} > 0.Let me handle these inequalities. Let's denote y = sqrt{x + 19}. Then, y geq 0, and x = y^2 - 19. Substitute into the expressions:For 105 + 2x y:105 + 2(y^2 - 19)y = 105 + 2y^3 - 38ySimilarly, for 76 + 2x y:76 + 2(y^2 - 19)y = 76 + 2y^3 - 38ySo, both expressions become cubic polynomials in y. Let me check if they are positive. But maybe substituting y is complicating things. Let's see if we can analyze the original expressions directly.Alternatively, perhaps consider that since x >= -19, and sqrt(x + 19) is real. Let's see for x >= -19, when is 105 + 2x sqrt(x + 19) positive?Let me check at x = -19: sqrt(-19 + 19) = sqrt(0) = 0, so 105 + 0 = 105 > 0. So at x = -19, it's positive. What about as x increases? Let's see:If x is very large, then 2x sqrt(x + 19) ~ 2x * x^{1/2} = 2x^{3/2}, which goes to infinity as x increases. So, for x > 0, 2x sqrt(x + 19) is positive, so 105 + positive is positive. For x between -19 and 0, 2x sqrt(x + 19) would be negative, since x is negative. So, in that case, we have 105 + (negative number). So, we need to make sure that 105 + 2x sqrt(x + 19) > 0 when x is negative.Let me solve 105 + 2x sqrt(x + 19) > 0 for x in (-19, 0).Let me set x = -t, where t is between 0 and 19. Then, sqrt(x +19) = sqrt(-t +19) = sqrt(19 - t). So, the expression becomes:105 + 2*(-t)*sqrt(19 - t) = 105 - 2t sqrt(19 - t) > 0So, 105 > 2t sqrt(19 - t)We need to find t in [0,19) such that 2t sqrt(19 - t) < 105.This might be a bit involved. Maybe instead of solving this inequality, we can check if for all x >= -19, the expression 105 + 2x sqrt(x +19) is positive.At x = -19, as we saw, it's 105. At x approaching 0 from the left, sqrt(x +19) approaches sqrt(19), so the expression becomes 105 + 2x sqrt(19). When x approaches 0 from the left, x is small negative, so 2x sqrt(19) is small negative. So, 105 - small positive = still positive. Maybe the minimum occurs somewhere in between?Let me take the derivative of 105 + 2x sqrt(x +19) with respect to x to see if it has a minimum.But since we have x in [-19, ∞), and the function is 105 + 2x sqrt(x +19). Let's compute its derivative:d/dx [2x sqrt(x +19)] = 2 sqrt(x +19) + 2x*(1/(2 sqrt(x +19))) ) = 2 sqrt(x +19) + x / sqrt(x +19) = (2(x +19) + x)/sqrt(x +19) = (3x + 38)/sqrt(x +19)Set derivative to zero: (3x +38)/sqrt(x +19) = 0. The numerator must be zero: 3x +38 =0 => x = -38/3 ≈ -12.6667.So the critical point is at x ≈ -12.6667. Let's check the value there.x = -38/3 ≈ -12.6667. Compute sqrt(x +19):sqrt(-38/3 +19) = sqrt(19 - 38/3) = sqrt( (57 -38)/3 ) = sqrt(19/3) ≈ sqrt(6.333) ≈ 2.5166Compute 2x sqrt(x +19):2*(-38/3)*sqrt(19/3) ≈ (-76/3)*2.5166 ≈ -25.333 *2.5166 ≈ -63.8Therefore, 105 + (-63.8) ≈ 41.2 > 0.So the minimum value is approximately 41.2, so 105 + 2x sqrt(x +19) is always greater than 41.2, which is positive. Therefore, the argument of log2 is always positive. Similarly, check 76 + 2x sqrt(x +19):Again, let's check at x = -19: 76 + 2*(-19)*0 =76 >0. At x approaching 0 from the left: 76 + 2x sqrt(19). Since x is small negative, 2x sqrt(19) is small negative, so 76 - small positive is still positive. Let's check the critical point.The derivative of 76 + 2x sqrt(x +19) is the same as above: (3x +38)/sqrt(x +19). So the critical point is also at x = -38/3 ≈ -12.6667. Let's compute 76 + 2x sqrt(x +19) at x = -38/3:sqrt(x +19) ≈ 2.5166 as before.2x sqrt(x +19) ≈ 2*(-38/3)*2.5166 ≈ (-76/3)*2.5166 ≈ -63.8Therefore, 76 + (-63.8) ≈ 12.2 >0.Hence, 76 + 2x sqrt(x +19) is always positive as well. Therefore, both logarithms in the expression are defined for all x >= -19. So, the domain is x >= -19.Now, moving on to the inequality:Numerator: -log2(105 + 2x sqrt(x +19))^3 + |log2 [ (105 + 2x sqrt(x +19)) / (x² +x +3)^4 ] |Denominator: 9 log5(76 + 2x sqrt(x +19)) -4 log2(105 + 2x sqrt(x +19))We need the entire fraction to be >=0.Let me first simplify the numerator and denominator separately.Starting with the numerator:Let me denote A = 105 + 2x sqrt(x +19), and B = x² + x +3. Then, the numerator is:-3 log2 A + | log2 (A / B^4) | = -3 log2 A + | log2 A - 4 log2 B |.Similarly, the denominator:9 log5 (76 + 2x sqrt(x +19)) -4 log2 A. Let me denote C = 76 + 2x sqrt(x +19). Then denominator is 9 log5 C -4 log2 A.So, numerator: -3 log2 A + | log2 A -4 log2 B |.Denominator: 9 log5 C -4 log2 A.Let me see if I can express log5 C in terms of log2 C or something else. Since log5 C = (log2 C)/(log2 5). That might be helpful.But let's first handle the numerator. Let me consider two cases based on the absolute value.Case 1: log2 (A / B^4) >= 0, which means A / B^4 >=1, so A >= B^4.Case 2: log2 (A / B^4) < 0, which means A / B^4 <1, so A < B^4.Therefore, the absolute value can be split into these two cases.So, in Case 1: |log2 (A / B^4)| = log2 (A / B^4). Then, the numerator becomes:-3 log2 A + log2 (A / B^4) = -3 log2 A + log2 A -4 log2 B = (-2 log2 A) -4 log2 B.In Case 2: |log2 (A / B^4)| = -log2 (A / B^4). Then, the numerator becomes:-3 log2 A - log2 (A / B^4) = -3 log2 A - log2 A +4 log2 B = (-4 log2 A) +4 log2 B.So, numerator is:Case 1: -2 log2 A -4 log2 BCase 2: -4 log2 A +4 log2 BSimilarly, denominator: 9 log5 C -4 log2 A. Let's see if we can write log5 C in terms of log2 C:log5 C = log2 C / log2 5. Let me denote log2 5 as a constant (approximately 2.3219, but we can keep it as log2 5 for exactness). So denominator becomes:9*(log2 C / log2 5) -4 log2 A = (9 / log2 5) log2 C -4 log2 A.Alternatively, if we want to write everything in terms of log2, perhaps factor log2?Alternatively, maybe express log5 C as ln C / ln 5 and log2 A as ln A / ln 2. Then denominator would be:9*(ln C / ln5) -4*(ln A / ln2). Not sure if that helps.Alternatively, let's note that 105 + 2x sqrt(x +19) and 76 + 2x sqrt(x +19) are similar expressions. Let's see:A = 105 + 2x sqrt(x +19)C =76 + 2x sqrt(x +19)Therefore, A = C + 29. So, A is 29 more than C. Not sure if that helps, but perhaps it's a relationship to note.Alternatively, maybe express A in terms of C?Alternatively, note that both A and C have the term 2x sqrt(x +19). Let's denote that term as D = 2x sqrt(x +19). Then, A = D +105, C = D +76.But perhaps that's not particularly helpful. Alternatively, perhaps express in terms of D: C = A -29. So, log5 C = log5 (A -29).But maybe this is not helpful.Alternatively, perhaps try to express the denominator in terms of log2 terms. Let me attempt that.Since log5 C = log2 C / log2 5, so denominator is 9*(log2 C / log2 5) -4 log2 A.Let me write that as (9 / log2 5) * log2 C -4 log2 A. Let me denote k = 9 / log2 5 ≈ 9 / 2.3219 ≈ 3.875. So, approximately 3.875 * log2 C -4 log2 A.But perhaps exact expressions are better. Let's keep it as (9 / log2 5) log2 C -4 log2 A.Alternatively, maybe write everything over a common denominator?Alternatively, perhaps note that log2 C and log2 A are separate terms, so unless there's a relation between C and A, maybe not. Since A = C +29, but unless C can be expressed in terms of A, maybe not directly.Alternatively, let me check if there's a substitution that can be made. Let me see:Let me define t = sqrt(x +19). Then, t >=0, and x = t^2 -19. Then, substitute into A, B, C:Compute A =105 +2x t =105 +2(t^2 -19)t =105 +2t^3 -38t.Similarly, C =76 +2x t =76 +2(t^2 -19)t =76 +2t^3 -38t.And B =x² +x +3 = (t^2 -19)^2 + (t^2 -19) +3.Let me compute B:(t^2 -19)^2 = t^4 -38t^2 +361So B = t^4 -38t^2 +361 + t^2 -19 +3 = t^4 -37t^2 +345Therefore, A =2t^3 -38t +105C =2t^3 -38t +76B =t^4 -37t^2 +345So, substituting t might make the expressions in terms of polynomials. Let me see if this substitution helps.So, numerator and denominator now become:Numerator (Case 1): -2 log2 A -4 log2 BNumerator (Case 2): -4 log2 A +4 log2 BDenominator: 9 log5 C -4 log2 ABut even with t substitution, we still have logs of polynomials. Hmm.Alternatively, perhaps there's a relationship between A, B, C. Let me check.Is there a way that A, B, C relate? Let me plug in t=... Maybe specific values?Wait, perhaps if we can find t such that A = B^4 or A < B^4, but since B is a quartic in t, this might not be straightforward.Alternatively, maybe consider that for x >= -19, t >=0, and analyze the behavior as t increases.Alternatively, note that both numerator and denominator are combinations of logs, so maybe we can combine them.Alternatively, consider the entire inequality:[ -3 log2 A + |log2 (A/B^4)| ] / [9 log5 C -4 log2 A] >=0The sign of the fraction depends on the signs of numerator and denominator. So, the fraction is >=0 when numerator and denominator have the same sign (both positive or both negative).Therefore, perhaps split into two cases:1. Numerator >=0 and Denominator >=02. Numerator <=0 and Denominator <=0But given the complexity of expressions, this might be difficult. Alternatively, perhaps analyze critical points where numerator or denominator is zero, then test intervals.Alternatively, let's first find when the numerator is zero and when the denominator is zero, and use those as critical points.First, solve numerator =0:Case1: If A >= B^4, then numerator = -2 log2 A -4 log2 B =0=> -2 log2 A -4 log2 B =0 => 2 log2 A +4 log2 B =0 => log2 A^2 + log2 B^4 =0 => log2 (A^2 B^4)=0 => A^2 B^4 =1Case2: If A < B^4, then numerator = -4 log2 A +4 log2 B =0=> -4 log2 A +4 log2 B =0 => log2 B - log2 A =0 => log2 (B/A)=0 => B/A =1 => B =ASimilarly, denominator =0:9 log5 C -4 log2 A =0 => 9 log5 C =4 log2 A => 9*(ln C / ln5) =4*(ln A / ln2) => (9 / ln5) ln C = (4 / ln2) ln ABut this seems complicated. Alternatively, express log5 C in terms of log2 C:log5 C = log2 C / log2 5. So,9*(log2 C / log2 5) -4 log2 A =0 => (9 / log2 5) log2 C -4 log2 A =0 => 9 log2 C =4 log2 5 log2 ABut unless there's a relation between C and A, this might not be helpful. Since A and C are related as A = C +29, maybe substitute A = C +29:9 log2 C =4 log2 5 log2 (C +29)But this equation would need to be solved numerically, perhaps. This might be complicated. Similarly, for the numerator equations.Given the complexity, perhaps a different approach is needed. Let's consider substituting specific values for x and see if we can find the smallest solution.Since the problem asks for the smallest solution, which would likely be the leftmost x in the domain where the inequality holds. The domain is x >= -19. Let's check x = -19 first.At x = -19:Compute A =105 +2*(-19)*sqrt(-19 +19)=105 +0=105C=76 +2*(-19)*0=76B= (-19)^2 + (-19) +3=361 -19 +3=345Compute numerator:-3 log2 105 + |log2 (105 / 345^4)|First, compute log2 (105 / 345^4). Since 345^4 is a huge number, so 105 / 345^4 is a very small number, so log2 of that is negative. Therefore, the absolute value becomes -log2 (105 / 345^4) = log2 (345^4 /105).Therefore, numerator:-3 log2 105 + log2 (345^4 /105) = -3 log2 105 +4 log2 345 - log2 105 = (-4 log2 105) +4 log2 345=4 (log2 345 - log2 105) =4 log2 (345/105)=4 log2 (23/7) ≈4*(1.726)≈6.904Denominator:9 log5 76 -4 log2 105Compute log5 76 = ln76 / ln5 ≈4.3301 /1.6094≈2.690So 9*2.690≈24.21log2 105≈6.714So 4*6.714≈26.856Therefore, denominator≈24.21 -26.856≈-2.646So numerator≈6.904, denominator≈-2.646, so the fraction≈6.904 / (-2.646)≈-2.61 <0. So the inequality is not satisfied at x=-19.Wait, but the problem says "Find the smallest solution", so maybe x=-19 is not the solution. Let's check x approaching -19 from the right. But x cannot be less than -19. Wait, x >= -19. So x=-19 is the left endpoint. Since at x=-19, the value is negative, so the inequality is not satisfied here. Therefore, the solution must be somewhere to the right of x=-19.Let me try x=0.At x=0:Compute A=105 +0=105C=76 +0=76B=0 +0 +3=3Numerator:-3 log2 105 + |log2 (105 /3^4)|= -3 log2 105 + |log2 (105 /81)|= -3 log2 105 + log2 (105/81) ≈ -3*6.714 + log2(1.296)≈-20.142 +0.375≈-19.767Denominator:9 log5 76 -4 log2 105 ≈9*2.690 -4*6.714≈24.21 -26.856≈-2.646So the fraction≈-19.767 / -2.646≈7.47>0. Therefore, the inequality holds at x=0.So somewhere between x=-19 and x=0, the fraction changes from negative to positive. So the smallest solution is in that interval. Wait, but at x=-19 it's negative, and at x=0 it's positive, so by Intermediate Value Theorem, there is a crossing point somewhere between x=-19 and x=0.But maybe there are multiple crossing points. Let's check another point. Let's try x=-10.At x=-10:Compute sqrt(x+19)=sqrt(9)=3So A=105 +2*(-10)*3=105 -60=45C=76 +2*(-10)*3=76 -60=16B=(-10)^2 + (-10) +3=100 -10 +3=93Numerator:Case: Check if A >= B^4? A=45, B=93. B^4 is 93^4 which is way larger than 45, so A < B^4. Therefore, Case2 applies.Numerator= -4 log2 A +4 log2 B =4 (log2 B - log2 A)=4 log2 (B/A)=4 log2(93/45)=4 log2(2.0666)≈4*1.047≈4.188Denominator:9 log5 C -4 log2 A=9 log5 16 -4 log2 45log5 16=ln16/ln5≈2.7726/1.6094≈1.722So 9*1.722≈15.5log2 45≈5.491So 4*5.491≈21.964Denominator≈15.5 -21.964≈-6.464Thus, fraction≈4.188 / (-6.464)≈-0.648<0. So at x=-10, the fraction is negative. So the inequality does not hold here.Hmm, so between x=-19 and x=0, at x=-19 it's negative, x=-10 negative, x=0 positive. So the crossing point is between x=-10 and x=0. Let's check x=-5.x=-5:sqrt(x+19)=sqrt(14)≈3.7417A=105 +2*(-5)*3.7417≈105 -37.417≈67.583C=76 +2*(-5)*3.7417≈76 -37.417≈38.583B=(-5)^2 +(-5)+3=25 -5 +3=23Check A vs B^4: A≈67.583, B^4=23^4≈279841. So A < B^4. So Case2.Numerator= -4 log2 A +4 log2 B=4(log2 B - log2 A)=4(log2(23/67.583))≈4(log2(0.340))≈4*(-1.556)≈-6.224Denominator=9 log5 C -4 log2 A≈9*(ln38.583/ln5) -4*(log2 67.583)log5 38.583≈ln38.583/ln5≈3.652/1.609≈2.269So 9*2.269≈20.42log2 67.583≈6.0984*6.098≈24.39Denominator≈20.42 -24.39≈-3.97So fraction≈-6.224 / -3.97≈1.57>0. Therefore, the inequality holds at x=-5.Wait, so at x=-10, the fraction was -0.648, at x=-5 it's 1.57. So between x=-10 and x=-5, it crosses from negative to positive. Let's check x=-7.5.x=-7.5:sqrt(x+19)=sqrt(11.5)≈3.391A=105 +2*(-7.5)*3.391≈105 -50.865≈54.135C=76 +2*(-7.5)*3.391≈76 -50.865≈25.135B=(-7.5)^2 +(-7.5)+3=56.25 -7.5 +3≈51.75Check A vs B^4: A≈54.135, B^4≈51.75^4≈(approx) Let's compute 50^4=6,250,000, 51.75^4 is a bit more, but even 54.135 < 51.75^4 (which is way larger). So A < B^4, Case2.Numerator=4(log2 B - log2 A)=4(log2(51.75/54.135))≈4(log2(0.956))≈4*(-0.065)= -0.26Denominator=9 log5 C -4 log2 A≈9*(ln25.135/ln5) -4*(log2 54.135)log5 25.135≈ln25.135/ln5≈3.224/1.609≈2.0039*2.003≈18.027log2 54.135≈5.7574*5.757≈23.028Denominator≈18.027 -23.028≈-5.001Fraction≈-0.26 / -5.001≈0.052>0. So inequality holds here.Wait, so at x=-7.5, the fraction is ≈0.052, which is barely positive. So the crossing point is between x=-10 and x=-7.5? Wait, at x=-10, fraction≈-0.648, at x=-7.5≈0.052. So crosses from negative to positive between x=-10 and x=-7.5.Wait, but let's check x=-8:x=-8:sqrt(x+19)=sqrt(11)=3.3166A=105 +2*(-8)*3.3166≈105 -53.065≈51.935C=76 +2*(-8)*3.3166≈76 -53.065≈22.935B=(-8)^2 +(-8)+3=64 -8 +3=59Check A vs B^4: A≈51.935, B=59. B^4 is way larger, so Case2.Numerator=4(log2 B - log2 A)=4(log2(59/51.935))≈4(log2(1.136))≈4*0.184≈0.736Denominator=9 log5 C -4 log2 A≈9*(ln22.935/ln5) -4*(log2 51.935)log5 22.935≈ln22.935/ln5≈3.133/1.609≈1.9479*1.947≈17.523log2 51.935≈5.7004*5.700≈22.800Denominator≈17.523 -22.800≈-5.277Fraction≈0.736 / -5.277≈-0.14<0. So at x=-8, the fraction is negative. So between x=-8 and x=-7.5, the fraction crosses zero.Wait, at x=-8: fraction≈-0.14, at x=-7.5≈0.052. So crossing between x=-8 and x=-7.5.To find the exact point where the fraction crosses zero, we can set numerator=0 and solve for x.But numerator is in Case2 (A < B^4), so:Numerator= -4 log2 A +4 log2 B=0 => log2 B - log2 A=0 => B/A=1 => B=A.Therefore, when B=A, the numerator is zero. So solving B=A:x² +x +3 =105 +2x sqrt(x +19)Let me write the equation:x² +x +3 =105 +2x sqrt(x +19)Rearranged:x² +x +3 -105 -2x sqrt(x +19)=0x² +x -102 -2x sqrt(x +19)=0Let me denote sqrt(x +19)=y. Then, x = y² -19. Substitute into the equation:(y² -19)^2 + (y² -19) -102 -2(y² -19)y =0Expand (y² -19)^2:y⁴ -38y² +361So:y⁴ -38y² +361 + y² -19 -102 -2y(y² -19)=0Simplify terms:y⁴ -38y² +361 + y² -19 -102 -2y³ +38y=0Combine like terms:y⁴ -2y³ -37y² +38y +240=0So, we have a quartic equation:y⁴ -2y³ -37y² +38y +240=0Hmm, solving quartic equations is complicated. Let me attempt to factor this.Let me try rational roots. Possible rational roots are factors of 240 divided by factors of 1, so ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±16, ±20, ±24, ±30, etc.Test y=5:5⁴ -2*5³ -37*5² +38*5 +240=625 -250 -925 +190 +240= (625-250)=375; 375 -925= -550; -550 +190= -360; -360 +240= -120 ≠0y=4:256 - 128 - 592 +152 +240= (256-128)=128; 128 -592=-464; -464+152=-312; -312+240=-72≠0y=6:1296 - 432 -1332 +228 +240= (1296-432)=864; 864 -1332=-468; -468 +228=-240; -240 +240=0. Bingo!So y=6 is a root. Therefore, factor (y -6).Use polynomial division or synthetic division.Divide y⁴ -2y³ -37y² +38y +240 by (y -6).Using synthetic division:6 | 1 -2 -37 38 240Multiply 6*1=6, add to -2: 46*4=24, add to -37: -136*(-13)=-78, add to38: -406*(-40)=-240, add to240:0Thus, the quartic factors as (y -6)(y³ +4y² -13y -40)Now, factor the cubic: y³ +4y² -13y -40Try possible roots: ±1, ±2, ±4, ±5, ±8, etc.Test y=5:125 +100 -65 -40=120≠0y=4:64 +64 -52 -40=36≠0y=8:512 +256 -104 -40=624≠0y= -5:-125 +100 +65 -40=0. Wait:y=-5:(-5)^3 +4*(-5)^2 -13*(-5) -40= -125 +100 +65 -40=0. Yes! So y=-5 is a root.Therefore, factor (y +5):Divide y³ +4y² -13y -40 by (y +5):Using synthetic division:-5 | 1 4 -13 -40Bring down 1. Multiply -5*1=-5. Add to 4: -1Multiply -5*(-1)=5. Add to -13: -8Multiply -5*(-8)=40. Add to -40:0Thus, cubic factors as (y +5)(y² -y -8)Thus, quartic equation factors as:(y -6)(y +5)(y² -y -8)=0Solutions y=6, y=-5, and y=(1 ± sqrt(1 +32))/2=(1 ± sqrt33)/2≈(1 ±5.7446)/2≈3.3723 or -2.3723But since y = sqrt(x +19) >=0, so valid roots are y=6, y=(1 +sqrt33)/2≈3.3723. y=-5 and y=(1 -sqrt33)/2≈-2.3723 are negative, discard.Therefore, solutions y=6 and y≈3.3723.Therefore, corresponding x:For y=6: x = y² -19=36 -19=17For y≈3.3723: x≈(3.3723)^2 -19≈11.372 -19≈-7.628So, the equation B=A is satisfied at x≈-7.628 and x=17.But since we were in Case2 (A < B^4), which is true for these x values (since we had B=A here, which is much less than B^4 unless B=1, which it's not). Wait, but when B=A, we are at the boundary between Case1 and Case2. Hmm.But in our previous check at x=-7.5 (~-7.628), we saw the fraction was near zero.So, the points where the numerator is zero are x≈-7.628 and x=17. Let's check x=17.At x=17:sqrt(x +19)=sqrt(36)=6A=105 +2*17*6=105 +204=309C=76 +2*17*6=76 +204=280B=17² +17 +3=289 +17 +3=309So, A=309 and B=309. So B/A=1. Therefore, log2 (B/A)=0. So numerator=0. Denominator:9 log5 280 -4 log2 309.log5 280=ln280/ln5≈5.634/1.609≈3.5019*3.501≈31.509log2 309≈8.2774*8.277≈33.108Denominator≈31.509 -33.108≈-1.599≈-1.6Thus, the fraction is 0 / (-1.6)=0. So at x=17, the fraction is 0, which satisfies the inequality (since >=0). So x=17 is a solution.But we are looking for the smallest solution. So x≈-7.628 and x=17. But need to check which is smaller.x≈-7.628 is less than x=17, so the smallest solution is around x≈-7.628.But need to verify if around that x, the inequality holds.Wait, when we solved B=A, we found x≈-7.628 and x=17. At those points, the numerator is zero. So the inequality is satisfied (since >=0). So those are boundary points.But also, we need to check the denominator at those points. At x≈-7.628, denominator:9 log5 C -4 log2 A. Since A=B= x² +x +3. Let's compute at x≈-7.628.First, compute y=(1 +sqrt(33))/2≈3.3723. Then x= y² -19≈(3.3723)^2 -19≈11.372 -19≈-7.628.Compute A=B= x² +x +3≈(-7.628)^2 +(-7.628)+3≈58.18 -7.628 +3≈53.55C=76 +2x sqrt(x +19)=76 +2*(-7.628)*3.3723≈76 -2*7.628*3.3723≈76 -51.7≈24.3Compute denominator=9 log5 24.3 -4 log2 53.55log5 24.3≈ln24.3/ln5≈3.191/1.609≈1.9829*1.982≈17.838log2 53.55≈5.7414*5.741≈22.964Denominator≈17.838 -22.964≈-5.126Therefore, the fraction at x≈-7.628 is 0 / (-5.126)=0, which satisfies the inequality. So x≈-7.628 is a solution.However, since the problem asks for the smallest solution, which is x≈-7.628. But we need to check if there are solutions smaller than that. Wait, but when we checked x=-10, the fraction was negative, and at x=-7.5, it was positive. Therefore, between x≈-7.628 and x=-7.5, the fraction transitions from 0 to positive. Wait, no. At x≈-7.628, the numerator is zero. To the left of x≈-7.628 (i.e., x < -7.628), since in Case2 (A < B^4), the numerator is -4 log2 A +4 log2 B=4(log2 B - log2 A). Since B=A at x≈-7.628, so for x < -7.628, A < B (since x is smaller, A decreases and B increases? Wait, need to check.Wait, when x decreases from -7.628 to -19, how do A and B behave?A=105 +2x sqrt(x +19). As x decreases, sqrt(x +19) decreases, but x is negative, so 2x sqrt(x +19) becomes more negative. So A decreases as x decreases.B=x² +x +3. As x decreases from -7.628 to -19, x² increases (since squaring a negative number makes it positive and larger as the magnitude increases), x is negative, so B = x² +x +3. Let's see, at x=-19, B=361 -19 +3=345. At x=-7.628, B≈53.55. So as x decreases from -7.628 to -19, B increases.Therefore, for x < -7.628, B > A (since B increases and A decreases). So log2 B - log2 A=log2 (B/A). Since B/A >1, then log2 (B/A) >0, so numerator=4 log2 (B/A) >0.Denominator: For x < -7.628, in that region, denominator=9 log5 C -4 log2 A. Let's see:C=76 +2x sqrt(x +19). As x decreases, sqrt(x +19) decreases, and x is negative, so 2x sqrt(x +19) becomes more negative. Therefore, C decreases as x decreases. So log5 C decreases. Also, A decreases as x decreases, so log2 A decreases (but A is positive). Therefore, 9 log5 C decreases and -4 log2 A increases (since log2 A is decreasing, multiplied by -4). Hmm, complicated.But at x≈-7.628, denominator≈-5.126. To the left of that, x is smaller. Let's take x=-8, which we checked earlier:At x=-8, denominator≈-5.277, and numerator≈-0.14, which was negative. Wait, but according to our previous logic, for x < -7.628, numerator should be positive. Wait, but at x=-8, which is less than -7.628, we computed numerator≈-0.14. That contradicts our previous logic.Wait, this suggests an error in my reasoning. Let me double-check.At x≈-7.628, we have B=A. For x < -7.628, since B increases and A decreases, B >A. So B/A >1. Therefore, log2 (B/A) >0. Therefore, numerator=4 log2 (B/A) >0. But in our calculation at x=-8, the numerator was≈-0.14. That's inconsistent. What's wrong here?Wait, when we calculated numerator at x=-8:A≈51.935, B=59. So log2 (B/A)=log2(59/51.935)=log2≈1.136≈0.184. Therefore, 4*0.184≈0.736. Wait, but in my previous calculation, I thought I got -0.14, but that was a mistake.Wait, wait, no. Wait, x=-8:Numerator is Case2: -4 log2 A +4 log2 B=4(log2 B - log2 A)=4*log2(59/51.935)≈4*0.184≈0.736. Then denominator was≈-5.277. Therefore, fraction≈0.736 / (-5.277)≈-0.14. But according to this, numerator is positive, denominator is negative, so fraction is negative.But according to the previous reasoning, when x < -7.628, B/A >1, so log2 (B/A) >0, so numerator=4 log2 (B/A) >0. But denominator=9 log5 C -4 log2 A. If numerator>0 and denominator<0, then fraction<0, which doesn't satisfy the inequality. However, at x≈-7.628, the numerator is zero, and denominator is negative. To the right of x≈-7.628 (i.e., x >-7.628), B/A <1, so log2 (B/A) <0, so numerator=4 log2 (B/A) <0. But denominator? At x=-7.5, denominator≈-5.001, which is still negative, so numerator≈-0.26, denominator≈-5.001, so fraction≈0.052>0.Wait, but when x increases past -7.628, B/A becomes less than 1. So log2 (B/A) <0, so numerator becomes negative. Denominator is negative. Therefore, negative divided by negative is positive. Therefore, for x > -7.628, numerator is negative and denominator is negative, so fraction positive. For x < -7.628, numerator is positive and denominator is negative, so fraction negative. At x=-7.628, fraction=0. Therefore, the inequality is satisfied for x >=-7.628 where the fraction is >=0. Wait, but when x > -7.628, the fraction is positive, and when x=-7.628, it's zero. So the solution is x >=-7.628. However, we also have x=17 where the fraction is zero. But at x=17, what happens beyond that?Wait, at x=17, we have B=A=309. For x >17, check if A >=B^4 or A < B^4. Let's check x=20:A=105 +2*20*sqrt(39)≈105 +40*6.245≈105 +249.8≈354.8B=400 +20 +3=423A=354.8 < B=423. So A < B, so B/A≈1.192, so log2 (B/A)≈0.255. So numerator=4*0.255≈1.02. Denominator=9 log5 C -4 log2 A.C=76 +2*20*sqrt(39)≈76 +40*6.245≈76 +249.8≈325.8log5 325.8≈ln325.8/ln5≈5.786/1.609≈3.5969*3.596≈32.364log2 354.8≈8.4684*8.468≈33.872Denominator≈32.364 -33.872≈-1.508Fraction≈1.02 / -1.508≈-0.677 <0. So at x=20, the fraction is negative. But at x=17, the fraction is zero. So between x=17 and higher x, the fraction is negative. So the solution set is x in [-7.628,17]. But the problem asks for the smallest solution, which is x≈-7.628.But the problem says "Find the smallest solution", and likely expects an exact value. Wait, we found that x= y² -19, where y=(1 +sqrt(33))/2.Let me compute that exactly.Given y=(1 + sqrt(33))/2, then x= y² -19.Compute y²:y² = [(1 + sqrt(33))/2]^2 = (1 + 2 sqrt(33) +33)/4=(34 + 2 sqrt(33))/4=(17 + sqrt(33))/2Thus, x= (17 + sqrt(33))/2 -19= (17 + sqrt(33))/2 -38/2=(17 -38 + sqrt(33))/2=(-21 + sqrt(33))/2Therefore, x=(sqrt(33) -21)/2. That's the exact value.Check with approximate value:sqrt(33)≈5.7446, so (5.7446 -21)/2≈(-15.2554)/2≈-7.6277, which matches our earlier approximation.Therefore, the smallest solution is x=(sqrt(33) -21)/2.But wait, let's confirm that this is indeed the smallest solution. Since for x < this value, the fraction is negative, and for x >= this value up to x=17, the fraction is non-negative, so the smallest solution is x=(sqrt(33) -21)/2.But the problem might expect the answer in a certain form. Let me check if this is the only solution in the negative x range. From our earlier analysis, yes, the only crossing point from negative to positive is at x=(sqrt(33) -21)/2≈-7.6277, and then it stays positive until x=17, where it becomes zero again but then negative beyond that.Therefore, the smallest solution is x=(sqrt(33) -21)/2.Final AnswerThe smallest solution is boxed{dfrac{sqrt{33} - 21}{2}}.</think>To find the smallest solution of the inequalityfrac{-log _{2}(105+2 x sqrt{x+19})^{3}+left|log _{2} frac{105+2 x sqrt{x+19}}{left(x^{2}+x+3right)^{4}}right|}{9 log _{5}(76+2 x sqrt{x+19})-4 log _{2}(105+2 x sqrt{x+19})} geqslant 0,we start by analyzing the domain and simplifying the expressions. The domain is ( x geq -19 ) due to the square root and logarithm arguments. We denote ( A = 105 + 2x sqrt{x + 19} ), ( B = x^2 + x + 3 ), and ( C = 76 + 2x sqrt{x + 19} ). The inequality simplifies to considering the numerator and denominator separately. The numerator is split into two cases based on the absolute value:- Case 1: ( A geq B^4 ) leads to ( -2 log_2 A - 4 log_2 B )- Case 2: ( A < B^4 ) leads to ( -4 log_2 A + 4 log_2 B )The denominator is ( 9 log_5 C - 4 log_2 A ).By solving for when the numerator is zero, we find the critical points where ( B = A ). This leads to the quadratic equation ( x^2 + x + 3 = 105 + 2x sqrt{x + 19} ). Substituting ( y = sqrt{x + 19} ) and solving, we find the roots ( y = 6 ) and ( y = frac{1 + sqrt{33}}{2} ).The corresponding ( x ) values are ( x = 17 ) and ( x = frac{sqrt{33} - 21}{2} ). Checking the intervals around these points, we determine that the smallest solution is when the numerator transitions from negative to positive, which occurs at ( x = frac{sqrt{33} - 21}{2} ).Thus, the smallest solution is [boxed{dfrac{sqrt{33} - 21}{2}}],<think>Okay, so I need to solve this inequality and find the smallest solution. Let me take a deep breath and try to parse this step by step. The inequality is a big fraction, and it's supposed to be greater than or equal to zero. The numerator and denominator both have logarithmic expressions. Hmm, logarithms can be tricky, especially with different bases and absolute values. Let me write down the inequality again to have it clear:frac{-log _{2}(105+2 x sqrt{x+19})^{3}+left|log _{2} frac{105+2 x sqrt{x+19}}{left(x^{2}+x+3right)^{4}}right|}{9 log _{5}(76+2 x sqrt{x+19})-4 log _{2}(105+2 x sqrt{x+19})} geqslant 0First, I need to figure out the domain of this inequality because there are square roots and logarithms involved. The expressions inside the square roots and the arguments of the logarithms must be positive. Let's start with the domain.Starting with the square roots: we have sqrt{x + 19}. The expression under the square root must be non-negative, so:x + 19 geq 0 implies x geq -19Next, the arguments of the logarithms must be positive. Let's look at each logarithm:1. log_{2}(105 + 2xsqrt{x + 19})^3: The argument here is (105 + 2xsqrt{x + 19})^3. Since any real number cubed is positive if and only if the base is positive. So, we need 105 + 2xsqrt{x + 19} > 0.But since x geq -19, let's check if 2xsqrt{x + 19} could make the whole expression negative. Let's see, if x is negative, but sqrt{x + 19} is real, so x >= -19. Let's check when x is negative. Let's take x = -19: Then sqrt(0) = 0, so 105 + 0 = 105 > 0. For x approaching -19 from above, sqrt(x + 19) approaches 0, so 105 + something approaching 0, so still positive. For x negative but greater than -19, 2x is negative, but sqrt(x + 19) is positive, so 2x*sqrt(x + 19) is negative. So, 105 + negative. We need to make sure that 105 + 2x*sqrt(x + 19) > 0.So, let's solve 105 + 2x*sqrt(x + 19) > 0.Similarly, the denominator has log_{5}(76 + 2xsqrt{x + 19}), so 76 + 2x*sqrt(x + 19) > 0.So, let's find the domain by solving these inequalities.First, solve 105 + 2x*sqrt(x + 19) > 0.Let’s denote y = sqrt(x + 19). Then y >= 0, and x = y^2 - 19.Substituting into the expression:105 + 2(y^2 - 19)y > 0= 105 + 2y^3 - 38y > 0= 2y^3 - 38y + 105 > 0Hmm, this is a cubic equation. Maybe I can find roots?Let’s try y=3: 2*27 - 38*3 +105 = 54 - 114 +105 = 45 >0y=4: 128 - 152 +105 = 81 >0y=5: 250 - 190 +105=165>0y=2: 16 -76 +105=45>0y=1: 2 -38 +105=69>0y=0: 0 -0 +105=105>0How about negative y? But y = sqrt(x +19) >=0, so we don't need to check negative y.So for all y >=0, which means x >= -19, the expression 2y^3 -38y +105 is positive. Wait, but is that true? Let me check maybe a fractional y.Wait, let's take y= sqrt(x + 19). For example, when x approaches -19 from above, y approaches 0. Then 2y^3 -38y +105 ~ -38y +105. For very small y, that's approximately 105 -38y. Since y is approaching 0, this is positive. For y=0, it's 105. For y=1, 69, which is positive. For y=2, 45. y=3, 45. y=4,81, etc. So seems like this expression is always positive for y >=0, hence 105 + 2x*sqrt(x+19) >0 for all x >= -19.Similarly, check 76 + 2x*sqrt(x +19) >0.Again, substitute y = sqrt(x +19), x = y^2 -19.So 76 + 2(y^2 -19)y =76 +2y^3 -38y.Again, let's check for y >=0.y=0:76>0y=1:76 +2 -38=76-36=40>0y=2:76 +16 -76=16>0y=3:76 +54 -114=16>0y=4:76 +128 -152=52>0Similarly, maybe between y=0 and y=1, etc. Let's check y= sqrt(x +19). For y approaching 0, 76 + 0 -0=76>0. For y=0.5:76 + 2*(0.5)^3 -38*(0.5)=76 +0.25 -19=57.25>0. So yes, seems like 76 +2x*sqrt(x +19) >0 for all x >= -19. Hence, the domain is x >= -19.Therefore, the domain of the inequality is x >= -19.Now, proceed to simplify the inequality.First, the numerator:Numerator: -log _{2}(105+2 x sqrt{x+19})^{3}+left|log _{2} frac{105+2 x sqrt{x+19}}{left(x^{2}+x+3right)^{4}}right|Let’s denote A = 105 + 2x sqrt(x +19), and B = x² + x +3. Then the numerator becomes:-3log_2 A + | log_2 (A / B^4) | = -3log_2 A + | log_2 A - 4log_2 B |Similarly, the denominator:Denominator: 9 log _{5}(76+2 x sqrt{x+19})-4 log _{2}(105+2 x sqrt{x+19}) = 9 log_5 C -4 log_2 A where C =76 +2x sqrt(x+19)So, let me write this as:Numerator: -3log_2 A + | log_2 A - 4log_2 B |Denominator: 9 log_5 C -4 log_2 AWe need the entire fraction to be >=0. So, the sign of the fraction depends on the signs of numerator and denominator.Let’s first try to simplify the numerator. Let’s consider the expression inside the absolute value: log_2 A - 4log_2 B = log_2 (A / B^4). Depending on whether this is positive or negative, the absolute value will resolve differently.So, let’s let’s consider two cases:Case 1: log_2 (A / B^4) geq 0 implies A / B^4 geq 1 implies A geq B^4Case 2: log_2 (A / B^4) < 0 implies A / B^4 < 1 implies A < B^4So, in each case, the numerator becomes:Case 1: -3log_2 A + (log_2 A -4log_2 B ) = (-3 +1)log_2 A -4log_2 B = -2log_2 A -4log_2 B = -2(log_2 A + 2log_2 B)Case 2: -3log_2 A + (-(log_2 A -4log_2 B )) = -3log_2 A - log_2 A +4log_2 B = -4log_2 A +4log_2 B = -4(log_2 A - log_2 B ) = -4log_2 (A/B)So, numerator in Case 1: -2(log A + 2 log B) = -2 log (A B²)In Case 2: -4 log (A/B) = -4 log A +4 log BHmm, not sure if that helps yet. Let me note the denominator as well. Let’s see if we can relate C and A. Let’s note that C =76 +2x sqrt(x +19) and A =105 +2x sqrt(x +19). Therefore, C = A -29. So, C = A -29.So, perhaps expressions in terms of A and B? Not sure yet.Alternatively, maybe we can define some variables or expressions.Alternatively, perhaps we can look for possible substitutions.Let me check if A and C are related. Since C = A -29, as above.But maybe not directly helpful. Let’s see.First, let's try to analyze the numerator and denominator expressions.Alternatively, maybe we can write the numerator and denominator in terms of log A and log B, or something else.Let me consider the denominator: 9 log5 C -4 log2 A.Since C = A -29, perhaps if A is known, we can express C in terms of A, but not sure.Alternatively, perhaps convert log5 C to log2 C using change of base formula.Recall that log5 C = log2 C / log2 5. So, 9 log5 C = 9 log2 C / log2 5So denominator: (9 / log2 5) log2 C -4 log2 ASo denominator is in terms of log2 C and log2 A. Maybe that can help.But perhaps not immediately. Let me check the numerator first.So, in Case 1: A >= B^4. Then numerator is -2(log A + 2 log B) = -2 log (A B²). The denominator is 9 log5 C -4 log2 A.So, to determine if the fraction is non-negative, we need to check the sign of numerator and denominator.Similarly, in Case 2: A < B^4. Then numerator is -4 log (A/B) = -4 (log A - log B). Denominator same as before.Alternatively, perhaps we can look for specific values of x that make the numerator zero or the denominator zero, and then test intervals.But this might be complicated. Alternatively, maybe there is a substitution that can make this simpler.Let me note that both A and C are linear in x sqrt(x +19). Let’s see:A =105 +2x sqrt(x +19)C =76 +2x sqrt(x +19) = A -29Similarly, let’s define t = 2x sqrt(x +19)So, A =105 + t, C =76 + tSo t = A -105 = C -76Hence, t = A -105 = C -76 => A = C +29So, with t = 2x sqrt(x +19). Maybe this substitution can help.But perhaps also, t can be expressed in terms of x. Let me compute t =2x sqrt(x +19). Let’s square both sides if we need to solve for x, but maybe not yet.Alternatively, perhaps we can set y = sqrt(x +19). Then, as before, x = y² -19, and sqrt(x +19)=y. Then t =2x y =2(y² -19)y =2y³ -38y.Therefore, A =105 + t =105 +2y³ -38ySimilarly, C=76 + t=76 +2y³ -38yBut we already did that earlier. So maybe in terms of y, the expressions are:Numerator and denominator can be expressed in terms of y. Let's try that.Express A and B in terms of y:A =105 +2y³ -38yB =x² +x +3 = (y² -19)^2 + (y² -19) +3Let me compute B:First, expand (y² -19)^2:= y^4 -38y² + 361Then add (y² -19):= y^4 -38y² +361 + y² -19 = y^4 -37y² +342Then add 3:B = y^4 -37y² +345So B = y^4 -37y² +345Therefore, the numerator in terms of y:Case 1: A >= B^4 (seems very large, but let's check)But A is a cubic in y, and B is a quartic in y. For y >=0, as y increases, B^4 would grow much faster than A. So, probably A >= B^4 only for small y? Let's check when y=3:A =105 +2*27 -38*3=105+54-114=45B=81 -37*9 +345=81-333+345=93So B=93, B^4 is 93^4, which is way larger than 45. So A=45 < B^4. So even at y=3, A < B^4. When y=0: A=105, B=0 -0 +345=345. B^4 is 345^4, which is way larger than 105. So maybe A >= B^4 never happens? That would mean that Case 1 is empty, and we only have Case 2.Wait, but maybe for very small y? Let's check when y is approaching 0. Let y approach 0 from above. Then A approaches 105, B approaches 345. B^4 is a huge number, so A=105 < B^4. So even as y approaches 0, A < B^4.Wait, but maybe when y is negative? But y is sqrt(x +19) >=0, so y cannot be negative. Therefore, for all y >=0, A < B^4. Therefore, Case 1 is impossible. Therefore, we can discard Case 1 and only consider Case 2, where A < B^4, so the absolute value becomes -(log_2 (A / B^4)).Therefore, the numerator simplifies to:-3 log2 A + (-log2 (A / B^4)) = -3 log2 A - log2 A +4 log2 B = -4 log2 A +4 log2 B =4 (log2 B - log2 A) =4 log2 (B/A)So the numerator is 4 log2 (B/A)So the entire fraction becomes [4 log2 (B/A)] / [9 log5 C -4 log2 A]We need this fraction to be >=0.Therefore, the inequality reduces to:4 log2 (B/A) / [9 log5 C -4 log2 A] >=0Since 4 is positive, the sign of the numerator is determined by log2 (B/A). The denominator's sign is determined by 9 log5 C -4 log2 A.Therefore, the inequality is equivalent to:[ log2 (B/A) ] / [9 log5 C -4 log2 A ] >=0Since 4 is positive, cancels out.Therefore, the inequality holds when numerator and denominator have the same sign (both positive or both negative).Therefore, either:1. log2 (B/A) >0 and 9 log5 C -4 log2 A >0OR2. log2 (B/A) <0 and 9 log5 C -4 log2 A <0So, we need to solve these two cases.First, analyze log2 (B/A) >0. This implies B/A >1 => B > ASimilarly, log2 (B/A) <0 implies B < A.So, rewrite the cases:Case 1: B > A and 9 log5 C -4 log2 A >0Case 2: B < A and 9 log5 C -4 log2 A <0Therefore, we need to find x >= -19 where either both B > A and 9 log5 C -4 log2 A >0, or both B < A and 9 log5 C -4 log2 A <0.So, let's first express all terms in terms of y for easier handling. Remember that y = sqrt(x +19), so x = y² -19, and y >=0.Earlier, we have:A =105 +2y³ -38yB = y^4 -37y² +345C =76 +2y³ -38ySo, B > A would be:y^4 -37y² +345 >105 +2y³ -38ySimplify:y^4 -2y³ -37y² +38y +240 >0Similarly, B < A would reverse the inequality.So, need to solve this quartic inequality. Hmm, quartic might be difficult, but maybe factorable.Let me write the quartic polynomial:P(y) = y^4 -2y³ -37y² +38y +240We can attempt to factor this. Let's try possible rational roots using Rational Root Theorem. Possible roots are factors of 240 divided by factors of 1, so ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±16, ±20, ±24, ±30, ±40, ±48, ±60, ±80, ±120, ±240.Test y=5: 625 -250 -925 +190 +240= (625 -250) + (-925 +190 +240)= 375 -495= -120 ≠0y=4: 256 -128 -592 +152 +240= (256 -128) + (-592 +152 +240)=128 -200= -72≠0y=6: 1296 - 432 -1332 +228 +240= (1296 -432)+( -1332 +228 +240)=864 -864=0. Hey, y=6 is a root!So, (y -6) is a factor. Let's perform polynomial division.Divide P(y) by (y -6). Let's use synthetic division:Coefficients: 1, -2, -37, 38, 240Using y=6:6 | 1   -2   -37   38   240Multiply 6*1=6, add to -2: 46*4=24, add to -37: -136*(-13)= -78, add to 38: -406*(-40)= -240, add to 240:0So, the quotient polynomial is y³ +4y² -13y -40Therefore, P(y)=(y -6)(y³ +4y² -13y -40)Now, factor the cubic: y³ +4y² -13y -40Try possible roots: factors of 40: ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40Test y=5:125 +100 -65 -40=125+100=225; 225 -105=120≠0y=4:64 +64 -52 -40=128 -92=36≠0y= -5: -125 +100 +65 -40= (-125+100)+(65-40)=(-25)+25=0. So y=-5 is a root.Therefore, factor as (y +5)(y² - y -8)Check: (y +5)(y² - y -8)= y³ -y² -8y +5y² -5y -40= y³ +4y² -13y -40. Correct.Therefore, P(y)=(y -6)(y +5)(y² - y -8)Therefore, quartic factors as (y -6)(y +5)(y² - y -8). Therefore, roots are y=6, y=-5, and roots of y² - y -8=0, which are y=(1 ± sqrt(33))/2≈ (1 ±5.7446)/2≈3.3723 and -2.3723.But since y >=0 (because y = sqrt(x +19)), the roots in the domain y >=0 are y=6 and y=(1 + sqrt(33))/2≈3.3723. The other roots are negative and discarded.Therefore, the quartic P(y) factors as (y -6)(y +5)(y - (1 + sqrt(33))/2)(y - (1 - sqrt(33))/2)But since we are only considering y >=0, the sign of P(y) is determined by the roots at y≈3.3723 and y=6.Let’s make a sign chart for P(y):For y in [0, (1 + sqrt(33))/2), which is approx [0, 3.3723):Test y=0: P(0)=0 -0 -0 +0 +240=240>0At y=3.3723, P(y)=0For y in (3.3723,6):Test y=4: As before, P(4)= -72 <0At y=6, P(y)=0For y >6:Test y=7: P(7)=7^4 -2*7^3 -37*7^2 +38*7 +240=2401 -686 -1813 +266 +240=2401 -686=1715; 1715 -1813= -98; -98 +266=168; 168 +240=408>0Therefore, the sign of P(y) is:- Positive for y in [0, (1 + sqrt(33))/2)- Negative for y in ((1 + sqrt(33))/2,6)- Positive for y >6Therefore, B > A when y in [0, (1 + sqrt(33))/2) or y >6Similarly, B < A when y in ((1 + sqrt(33))/2,6)Therefore, the original cases:Case 1: B > A (i.e., y in [0,3.3723) or y >6) and 9 log5 C -4 log2 A >0Case 2: B < A (i.e., y in (3.3723,6)) and 9 log5 C -4 log2 A <0So, now we need to analyze the expression 9 log5 C -4 log2 A in each interval.Let’s note that C =76 +2y³ -38y, and A =105 +2y³ -38y. So, C = A -29.So, C = A -29. Therefore, if we can express log5 C in terms of log5 (A -29). But perhaps not helpful.Alternatively, let me denote D = log5 C = log5 (A -29)But maybe not helpful. Alternatively, convert 9 log5 C -4 log2 A into the same base.Recall that log5 C = log2 C / log2 5Therefore, 9 log5 C = 9 * log2 C / log2 5Therefore, 9 log5 C -4 log2 A = (9 / log2 5) log2 C -4 log2 ALet’s compute the constants:log2 5 ≈2.3219So, 9 / log2 5 ≈9 /2.3219≈3.875Therefore, approximately:9 log5 C ≈3.875 log2 CSo, the denominator expression is approximately 3.875 log2 C -4 log2 A = log2 C^{3.875} - log2 A^4 = log2 (C^{3.875}/A^4). But exact expression is better to keep as (9 / log2 5) log2 C -4 log2 ABut this might not help. Alternatively, let me write the denominator expression as:(9 log5 C -4 log2 A) = log2 A [ (9 / log2 5) (log2 C / log2 A) -4 ]But not sure. Alternatively, perhaps write everything in terms of log base 2:Let’s let’s define:Let’s let’s let’s let’s denote:Let’s denote k = log2 ALet’s denote m = log2 CThen, 9 log5 C =9 * (m / log2 5)Therefore, denominator becomes (9 m / log2 5) -4kSo, denominator = (9 / log2 5) m -4kWe need to analyze the sign of this.But perhaps not helpful. Alternatively, let's try to see if for specific y values, we can compute the denominator.Alternatively, let's attempt to graph or test values in the intervals.First, let's note the intervals for y:Case 1: y in [0, (1 + sqrt(33))/2) ≈[0,3.3723) and y >6Case 2: y in (3.3723,6)So, let's pick test values in each interval and compute the denominator and numerator.First, let's pick y=0 (which corresponds to x = y² -19= -19). But y=0 is the lower bound.Compute A, B, C at y=0:A =105 +0 -0=105B=0 -0 +345=345C=76 +0 -0=76Numerator: log2(B/A)=log2(345/105)=log2(3.2857)≈1.72>0Denominator:9 log5 76 -4 log2 105Compute log5 76: ln76/ln5≈4.3301/1.6094≈2.690So 9*2.690≈24.21log2 105≈ln105/ln2≈4.653/0.6931≈6.715So, 4*6.715≈26.86Therefore, denominator≈24.21 -26.86≈-2.65<0Therefore, at y=0: numerator>0, denominator<0. Therefore, the fraction is negative. So, Case 1 at y=0 does not satisfy the inequality.Hmm, so even though B > A here (345 >105), the denominator is negative, so the fraction is negative. Therefore, this interval [0,3.3723) may have parts where denominator is negative or positive. Let's check another point in this interval.Take y=3 (x=3² -19=9 -19=-10)Compute A=105 +2*27 -38*3=105+54-114=45B=81 -37*9 +345=81-333+345=93C=76 +54 -114=16Numerator: log2(93/45)=log2(2.0667)≈1.05>0Denominator:9 log5 16 -4 log2 45Compute log5 16: ln16/ln5≈2.7726/1.6094≈1.7229*1.722≈15.5log2 45≈5.49194*5.4919≈21.967Denominator≈15.5 -21.967≈-6.467<0Therefore, numerator>0, denominator<0. Fraction negative. Not satisfying the inequality.Another test point: y=2 (x=4 -19=-15)Compute A=105 +2*8 -38*2=105+16-76=45Wait, same as y=3?Wait, y=2: A=105 +2*(8) -38*2=105 +16 -76=45B=16 -37*4 +345=16 -148 +345=213C=76 +2*(8) -38*2=76 +16 -76=16Wait, same as y=3? That can't be. Wait, x= y² -19=4 -19=-15Wait, 2x sqrt(x +19)=2*(-15)*sqrt(-15 +19)=2*(-15)*2= -60Wait, A=105 + (-60)=45C=76 + (-60)=16So yes, same as y=3. That's because when y=3, x=9-19=-10, so sqrt(x+19)=sqrt(9)=3, so 2x*3=6x. If x=-10, 6*(-10)=-60. Similarly, when y=2, x=4-19=-15, sqrt(x+19)=2, so 2x*2=4x=-60. So same value. Therefore, A=45, C=16 for both y=2 and y=3? Wait, no. Wait, y=2, x=-15. Then, A=105 +2*(-15)*2=105 -60=45. Similarly, y=3, x=-10: A=105 +2*(-10)*3=105 -60=45. So same A and C.But B is different. For y=2, B=(-15)^2 +(-15)+3=225 -15 +3=213. For y=3, B=(-10)^2 +(-10)+3=100 -10 +3=93. So B decreases as y increases.So, at y=2: log2(213/45)=log2(4.733)≈2.24>0Denominator:9 log5 16 -4 log2 45≈15.5 -21.97≈-6.47<0So, same as previous.Another test point: y=1 (x=1 -19=-18)A=105 +2*1 -38*1=105 +2 -38=69B= (-18)^2 + (-18) +3=324 -18 +3=309C=76 +2*1 -38*1=76 +2 -38=40Numerator: log2(309/69)=log2≈4.478≈2.16>0Denominator:9 log5 40 -4 log2 69log5 40=ln40/ln5≈3.6889/1.6094≈2.2929*2.292≈20.628log2 69≈6.1104*6.110≈24.44Denominator≈20.628 -24.44≈-3.81<0Again, numerator>0, denominator<0. Fraction negative. So in the entire interval y in [0,3.3723), the denominator is negative while the numerator is positive, leading to the fraction being negative. Therefore, Case 1 (B > A) does not contribute to solutions here.Now, check y >6. Let's take y=7 (x=49 -19=30)Compute A=105 +2*343 -38*7=105 +686 -266=525B=30^2 +30 +3=900 +30 +3=933C=76 +2*343 -38*7=76 +686 -266=496Numerator: log2(933/525)=log2(1.777)≈0.83>0Denominator:9 log5 496 -4 log2 525Compute log5 496: ln496/ln5≈6.2066/1.6094≈3.8569*3.856≈34.704log2 525≈ln525/ln2≈6.263/0.693≈9.034*9.03≈36.12Denominator≈34.704 -36.12≈-1.416<0So, numerator>0, denominator<0. Fraction negative. Therefore, even for y>6, the fraction is negative. Therefore, Case 1 (B > A) does not yield solutions.Now, check Case 2: B < A (i.e., y in (3.3723,6)) and denominator <0.Wait, denominator is 9 log5 C -4 log2 A. Let's check in this interval.Take y=4 (x=16 -19=-3)A=105 +2*64 -38*4=105 +128 -152=81B=(-3)^2 + (-3) +3=9 -3 +3=9C=76 +2*64 -38*4=76 +128 -152=52Numerator: log2(9/81)=log2(1/9)= -log2(9)= -3.1699<0Denominator:9 log5 52 -4 log2 81log5 52≈ln52/ln5≈3.9512/1.6094≈2.4559*2.455≈22.095log2 81= log2(3^4)=4*1.58496≈6.33984*6.3398≈25.359Denominator≈22.095 -25.359≈-3.264<0Thus, numerator<0 and denominator<0. Fraction positive. Therefore, this interval satisfies the inequality.Another test point: y=5 (x=25 -19=6)A=105 +2*125 -38*5=105 +250 -190=165B=6^2 +6 +3=36 +6 +3=45C=76 +2*125 -38*5=76 +250 -190=136Numerator: log2(45/165)=log2(0.2727)≈-1.874 <0Denominator:9 log5 136 -4 log2 165log5 136≈ln136/ln5≈4.913/1.609≈3.0539*3.053≈27.477log2 165≈ln165/ln2≈5.105/0.693≈7.374*7.37≈29.48Denominator≈27.477 -29.48≈-2.003<0Thus, numerator<0 and denominator<0: fraction positive. Satisfies the inequality.Another test point: y=3.5 (x=12.25 -19=-6.75)A=105 +2*(42.875) -38*3.5=105 +85.75 -133=57.75B=(-6.75)^2 +(-6.75)+3=45.5625 -6.75 +3=41.8125C=76 +2*(42.875) -38*3.5=76 +85.75 -133=28.75Numerator: log2(41.8125/57.75)=log2≈0.723≈-0.468<0Denominator:9 log5 28.75 -4 log2 57.75log5 28.75≈ln28.75/ln5≈3.358/1.609≈2.0879*2.087≈18.783log2 57.75≈5.8854*5.885≈23.54Denominator≈18.783 -23.54≈-4.757<0Again, numerator<0, denominator<0: fraction positive. So, within the interval (3.3723,6), the fraction is positive. Therefore, the solutions lie in y ∈ (3.3723,6). Converting back to x, since x = y² -19.Therefore, we need to find the corresponding x values.First, the lower bound of y is (1 + sqrt(33))/2≈(1 +5.7446)/2≈3.3723So, x_lower = ( (1 + sqrt(33))/2 )² -19Let me compute this:Let’s compute (1 + sqrt(33))/2 squared:= [1 + 2 sqrt(33) +33]/4 = (34 + 2 sqrt(33))/4 = (17 + sqrt(33))/2Therefore, x_lower = (17 + sqrt(33))/2 -19 = (17 + sqrt(33) -38)/2 = (-21 + sqrt(33))/2Similarly, the upper bound of y is6, so x_upper=6² -19=36 -19=17Thus, the solution in terms of x is x ∈ ( (-21 + sqrt(33))/2 ,17 )But we need to find the smallest solution. The left endpoint is (-21 + sqrt(33))/2. Let's compute its approximate value.sqrt(33)≈5.7446So, (-21 +5.7446)/2≈(-15.2554)/2≈-7.6277Therefore, the solution interval is x ∈ (-7.6277,17). However, we need to check if the inequality is satisfied throughout this interval or if there are any breaks.But from our analysis earlier, in the interval y ∈ (3.3723,6), which corresponds to x ∈ (-7.6277,17), the inequality holds. Therefore, the solutions are all x in (-7.6277,17).But we need the smallest solution, which is the left endpoint x≈-7.6277. But we need to check if this endpoint is included or not. Since the inequality is >=0, and at y=(1 + sqrt(33))/2≈3.3723, the numerator is log2(B/A)=0 (since B=A there), so numerator=0. The denominator at this point is:At y=(1 + sqrt(33))/2, compute denominator:A=105 +2y³ -38yB=y^4 -37y² +345= A (since B=A at this point)Wait, no, at y=(1 + sqrt(33))/2, we have B=A, so log2(B/A)=0. So the numerator is zero, and denominator:At this y, which is a root of B=A. Let's compute denominator:9 log5 C -4 log2 AC=76 +2y³ -38y= A -29 (since C = A -29)But since B=A at this point, and B= y^4 -37y² +345, and A=105 +2y³ -38y.Wait, but this seems complicated. Alternatively, let's plug y=(1 + sqrt(33))/2 into A and C.Given that y=(1 + sqrt(33))/2≈3.3723Compute A=105 +2y³ -38yBut since y=(1 + sqrt(33))/2, let's compute y³:First, note that y satisfies y² -y -8=0 (from the quadratic factor), so y² = y +8Then y³ = y*(y²)= y*(y +8)=y² +8y= (y +8) +8y=9y +8Therefore, A=105 +2*(9y +8) -38y=105 +18y +16 -38y=121 -20ySimilarly, C=76 +2y³ -38y=76 +2*(9y +8) -38y=76 +18y +16 -38y=92 -20yTherefore, log5 C= log5(92 -20y)log2 A=log2(121 -20y)So denominator=9 log5(92 -20y) -4 log2(121 -20y)But y=(1 + sqrt(33))/2≈3.3723, so compute 92 -20y≈92 -20*3.3723≈92 -67.446≈24.554121 -20y≈121 -67.446≈53.554So, log5≈24.554≈log5(24.554)=ln24.554/ln5≈3.200/1.609≈1.988Multiply by 9:≈17.892log2≈53.554≈log2(53.554)=ln53.554/ln2≈3.981/0.693≈5.747Multiply by4:≈22.988Denominator≈17.892 -22.988≈-5.096<0Therefore, at y=(1 + sqrt(33))/2, the numerator is zero and the denominator is negative. Therefore, the entire fraction is 0 / negative=0. But the inequality is >=0, so this point is included. Therefore, x=(-21 + sqrt(33))/2 is included in the solution set.Similarly, check the upper limit y=6 (x=17). At y=6:A=105 +2*216 -38*6=105 +432 -228=309B=6^4 -37*6² +345=1296 -1332 +345=309C=76 +2*216 -38*6=76 +432 -228=280Numerator: log2(309/309)=log2(1)=0Denominator:9 log5 280 -4 log2 309log5 280≈ln280/ln5≈5.635/1.609≈3.5029*3.502≈31.518log2 309≈8.2764*8.276≈33.104Denominator≈31.518 -33.104≈-1.586<0Thus, the fraction is 0 / (-1.586)=0. Therefore, x=17 is also included.Therefore, the solution set is x ∈ [ (-21 + sqrt(33))/2 ,17 ]Thus, the smallest solution is x=(-21 + sqrt(33))/2. Let's compute this exactly:sqrt(33)=sqrt(33), so x=(-21 + sqrt(33))/2.We can rationalize or leave it as is. But let's check if this is the correct minimal solution. Since x must be >=-19, and (-21 + sqrt(33))/2≈(-21 +5.7446)/2≈-15.2554/2≈-7.6277, which is greater than -19, so it's valid.Therefore, the smallest solution is (-21 + sqrt(33))/2. But maybe we can write this in a different form.Alternatively, sqrt(33) is irrational, so this is as simplified as it gets.Final AnswerThe smallest solution is boxed{dfrac{-21 + sqrt{33}}{2}}.</think>To find the smallest solution of the inequalityfrac{-log _{2}(105+2 x sqrt{x+19})^{3}+left|log _{2} frac{105+2 x sqrt{x+19}}{left(x^{2}+x+3right)^{4}}right|}{9 log _{5}(76+2 x sqrt{x+19})-4 log _{2}(105+2 x sqrt{x+19})} geqslant 0we first determine the domain. The expressions inside the square roots and logarithms must be positive, leading to ( x geq -19 ).Next, we simplify the inequality. Let ( A = 105 + 2x sqrt{x + 19} ), ( B = x^2 + x + 3 ), and ( C = 76 + 2x sqrt{x + 19} ). The inequality becomes:frac{-3 log_2 A + left| log_2 frac{A}{B^4} right|}{9 log_5 C - 4 log_2 A} geq 0We consider two cases for the absolute value term:1. ( log_2 frac{A}{B^4} geq 0 ) (i.e., ( A geq B^4 ))2. ( log_2 frac{A}{B^4} < 0 ) (i.e., ( A < B^4 ))Through analysis, we find that ( A geq B^4 ) is not possible for any ( x geq -19 ). Thus, we only consider the second case where ( A < B^4 ). The inequality simplifies to:frac{4 log_2 frac{B}{A}}{9 log_5 C - 4 log_2 A} geq 0This requires the numerator and denominator to have the same sign. We find that ( log_2 frac{B}{A} < 0 ) and ( 9 log_5 C - 4 log_2 A < 0 ) in the interval ( y in (3.3723, 6) ) where ( y = sqrt{x + 19} ). Converting back to ( x ), we find the interval ( x in left( frac{-21 + sqrt{33}}{2}, 17 right) ).The smallest solution is the left endpoint of this interval, which is:boxed{dfrac{-21 + sqrt{33}}{2}}]

🔑:Answer: frac{-21+sqrt{33}}{2} approx-7.63.

🧠:[<think>Okay, so there's this six-digit number that starts with 2. If we move that 2 from the first position to the last, keeping the other digits in order, the new number is three times the original number. Hmm, interesting. Let me try to figure this out step by step.First, let's denote the original number. Since it's a six-digit number starting with 2, we can write it as 2ABCDE, where A, B, C, D, E are the remaining five digits. Each of these can be from 0 to 9, right? So the original number is 200000 + 10000A + 1000B + 100C + 10D + E. Now, when we move the first digit (which is 2) to the end, the new number becomes ABCDE2. This new number is three times the original number. So, ABCDE2 = 3 * (2ABCDE). Let's translate this into an equation.The original number is 2ABCDE. Let me represent this as N = 200000 + X, where X is the number formed by the last five digits, ABCDE. So X can be any number from 00000 to 99999, right? Then, when we move the 2 to the end, the new number becomes 10X + 2. Because we're shifting all the digits to the left and appending 2 at the end. So 10X + 2 is three times the original number, which is 3*(200000 + X). So the equation is: 10X + 2 = 3*(200000 + X)Let me write that out:10X + 2 = 600000 + 3XNow, let's solve for X. Subtract 3X from both sides:7X + 2 = 600000Then subtract 2:7X = 599998Now divide both sides by 7:X = 599998 / 7Let me compute that. 7*85,714 = 599,998. Let me check:7*80,000 = 560,0007*5,714 = 7*(5,000 + 714) = 35,000 + 4,998 = 39,998So 560,000 + 39,998 = 599,998. So yes, 7*85,714 = 599,998. Therefore, X = 85,714.Therefore, the original number is 2 followed by X, which is 285714.Wait, let's verify this. If the original number is 285714, moving the first digit 2 to the end gives us 857142. Let's check if 857142 is three times 285714.Compute 285714 * 3:285714 * 3:4*3=12, carryover 11*3=3 +1=47*3=21, carryover 25*3=15 +2=17, carryover 18*3=24 +1=25, carryover 22*3=6 +2=8So the result is 857142. Which matches the number obtained by moving the first digit to the end. So it works!Therefore, the original number is 285714.Final AnswerThe original number is boxed{285714}.</think>Let the original six-digit number be (2ABCDE), where (A, B, C, D, E) are the remaining five digits. This number can be expressed as (200000 + X), where (X) is the five-digit number (ABCDE).When the digit 2 is moved to the end, the new number becomes (ABCDE2), which can be written as (10X + 2). According to the problem, this new number is three times the original number:[10X + 2 = 3 times (200000 + X)]Expanding and solving for (X):[10X + 2 = 600000 + 3X]Subtract (3X) from both sides:[7X + 2 = 600000]Subtract 2 from both sides:[7X = 599998]Divide by 7:[X = frac{599998}{7} = 85714]Thus, the original number is (200000 + 85714 = 285714). Verifying, moving the first digit 2 to the end gives 857142, which is indeed three times 285714.The original number is (boxed{285714}).,<think>Okay, so I need to find a six-digit number that starts with 2. When we move that first digit (which is 2) to the end, the new number becomes three times the original number. Let me try to figure this out step by step.First, let's denote the original number. Since it's a six-digit number starting with 2, the number can be written as 2ABCDE, where A, B, C, D, E are the remaining five digits. Each of these can be from 0 to 9, right? So the original number is 200000 + 10000A + 1000B + 100C + 10D + E. But maybe there's a better way to represent this.Alternatively, if we think of the original number as N, then moving the first digit (2) to the end would result in a new number. Let's call this new number N'. The problem states that N' = 3 * N.Let me express N mathematically. The original number is 2 followed by five digits, so N = 200000 + x, where x is the number formed by the last five digits. For example, if the number was 212345, then x would be 12345. So N = 200000 + x.When we move the first digit (2) to the end, the new number N' becomes ABCDE2. Wait, ABCDE are the last five digits, so N' is formed by taking those five digits and putting 2 at the end. So how can we express N' in terms of x?If the original number is 2ABCDE, then the new number after moving 2 to the end is ABCDE2. So, ABCDE is x, which is the last five digits of N. Therefore, N' is equal to (x) * 10 + 2. Because we take the five digits x, shift them left by one place (which multiplies by 10) and then add 2 at the end.So, according to the problem, N' = 3 * N.But N is 200000 + x, and N' is 10x + 2. Therefore, the equation is:10x + 2 = 3 * (200000 + x)Let me write that out:10x + 2 = 600000 + 3xNow, let's solve for x. Subtract 3x from both sides:7x + 2 = 600000Then, subtract 2:7x = 599998Divide both sides by 7:x = 599998 / 7Let me calculate that. Hmm, 7 times 85714 is 599,998? Let's check:7 * 80000 = 5600007 * 5000 = 35000 → 560000 + 35000 = 5950007 * 700 = 4900 → 595000 + 4900 = 5999007 * 14 = 98 → 599900 + 98 = 600, 998? Wait, 599900 + 98 = 599,998. Exactly!So 7 * 85,714 = 599,998. Therefore, x = 85,714.So x is 85714. Therefore, the original number N is 200,000 + 85,714 = 285,714.Let me check if this works. If we move the first digit 2 to the end, we get 85,7142. Wait, 857142. Is that equal to 3 times 285,714?Let's compute 285,714 * 3:285,714 * 3:First, 200,000 * 3 = 600,000Then 85,714 * 3: 80,000*3=240,000; 5,714*3=17,142; total is 240,000 +17,142=257,142Add to 600,000: 600,000 +257,142=857,142. Which is exactly the number we get when moving the 2 to the end. So yes, 285,714 *3 =857,142. Therefore, that's correct.Wait, but let me verify that moving the 2 to the end of 285714 gives 857142. Original number: 285714. Moving the first digit (2) to the end: 857142. Yes, that's exactly the result. So 857142 is indeed three times 285714. So that seems correct.Therefore, the original number is 285714. Let me just check if there might be another solution. The problem states that the original number is a six-digit number starting with 2, so x must be a five-digit number. In our case, x was 85714, which is a five-digit number. If x had fewer digits, like four digits, then adding 200,000 would result in a six-digit number starting with 2. But since x is 85,714, which is five digits, that's correct. So I think that's the only solution. Let me confirm that.The equation was 10x + 2 = 3*(200000 +x). Solving gives x = (600000 -2)/7 =599998/7=85,714, which is exact. So there are no other solutions. Therefore, the original number is 285714.Final AnswerThe original number is boxed{285714}.</think>Let the original six-digit number be ( N ), which starts with 2. We can express ( N ) as ( 200000 + x ), where ( x ) is the number formed by the last five digits. When the digit 2 is moved to the end, the new number ( N' ) becomes ( 10x + 2 ). According to the problem, ( N' ) is three times the original number ( N ). Therefore, we have the equation:[10x + 2 = 3(200000 + x)]Expanding and simplifying this equation:[10x + 2 = 600000 + 3x][7x + 2 = 600000][7x = 599998][x = frac{599998}{7} = 85714]Thus, the original number ( N ) is:[N = 200000 + x = 200000 + 85714 = 285714]Verification shows that moving the digit 2 from the start to the end of 285714 results in 857142, which is indeed three times 285714. Therefore, the original number is:[boxed{285714}]]

🔑:Solution.The original six-digit number has the form 2 cdot 10^{5}+x. After moving the digit 2 to the last place, we get the number 10 x+2. According to the condition, 10 x+2=3left(2 cdot 10^{5}+xright), from which x=85714.Answer: 285714.

🧠:[<think>Okay, let's see. So the problem is about Xiaohua distributing apples into baskets. There are two scenarios here. First, if he puts 4 apples in each basket, he has 20 left over. Second, if he puts 8 apples in each basket, there's one basket that doesn't have enough. We need to find out how many apples he originally had, with the options given.Let me start by translating the problem into mathematical equations. Let’s denote the number of baskets as 'b' and the number of apples as 'a'. In the first scenario: When he puts 4 apples in each basket, he has 20 left. So, the total number of apples can be expressed as 4 times the number of baskets plus 20. That gives us the equation:a = 4b + 20Okay, so that's straightforward. The second scenario is a bit trickier. If he puts 8 apples in each basket, there's one basket that doesn't have enough. Hmm. So, this probably means that when he tries to divide the apples into baskets with 8 each, he can fill (b - 1) baskets completely, and the last basket will have fewer than 8 apples. So, the total number of apples here would be 8 times (b - 1) plus some remainder, which is less than 8.So, mathematically, that would be:a = 8(b - 1) + rWhere r is the remainder apples in the last basket, and r < 8.But since the number of apples is the same in both scenarios, we can set these two equations equal to each other. So from the first equation, a = 4b + 20. From the second scenario, a = 8(b - 1) + r. Therefore:4b + 20 = 8(b - 1) + rLet me simplify that equation. Let's expand the right side:4b + 20 = 8b - 8 + rNow, subtract 4b from both sides:20 = 4b - 8 + rThen, add 8 to both sides:28 = 4b + rSo, 4b + r = 28. And we know that r is less than 8, right? Because the last basket doesn't have enough, so it's less than 8. So, r < 8.But we need to find integer values for b and r. Both b and r should be positive integers, with r between 1 and 7 (since it can't be zero because there's at least some apples in the last basket, otherwise there wouldn't be a basket with not enough). Wait, actually, hold on. If he tries to put 8 apples in each basket and there's one basket that doesn't have enough, does that mean that he must have at least one apple in that last basket? Because if there were zero, maybe that basket isn't needed? Hmm. The problem says "there will be one basket that does not have enough apples." So, maybe that basket is still considered, even if it has zero? But if he had exactly 8(b -1) apples, then he could have just used (b -1) baskets. So, maybe the total number of baskets is still the same as in the first scenario? Wait, that might not be the case. Let me think.Wait, in the first scenario, he's putting 4 apples in each basket, and has 20 left. So the number of baskets in that case would be (a - 20)/4. But in the second scenario, he's trying to put 8 apples in each basket, but the number of baskets isn't specified. Hmm. Wait, maybe the number of baskets is the same in both cases? Is that an assumption I need to make?The problem says "If he puts 4 apples in each basket, he will have 20 apples left undistributed." So, the number of baskets here would be some number, let's say b. Then, "If he puts 8 apples in each basket, there will be one basket that does not have enough apples." So, in the second case, is he using the same number of baskets? If he uses the same number of baskets, then trying to put 8 apples in each would leave one basket with not enough. Alternatively, he might be trying to use as many baskets as needed for 8 apples each, but there's ambiguity here. Wait, the problem doesn't specify whether the number of baskets is fixed or not. Hmm. This is a bit confusing. Let's re-examine the problem statement."Xiaohua has a number of apples that he wants to distribute into several baskets. If he puts 4 apples in each basket, he will have 20 apples left undistributed. If he puts 8 apples in each basket, there will be one basket that does not have enough apples. How many apples did Xiaohua originally have?"So, the key here is that he is distributing into "several baskets" in both cases. But does "several baskets" refer to the same number of baskets each time, or different numbers? The problem doesn't specify. However, the way the problem is phrased might imply that the number of baskets is the same in both scenarios. Because otherwise, if he can choose different numbers of baskets, then perhaps he could adjust the number of baskets to avoid having an underfilled basket. But the problem says "there will be one basket that does not have enough," which suggests that regardless of how he distributes them, when trying to use 8 per basket, there's one basket left with not enough. Hmm, actually, maybe not. Maybe the number of baskets is fixed, and he's trying different distributions. Let me assume that the number of baskets is fixed. That is, in both scenarios, he is using the same number of baskets. If he puts 4 apples in each, he has 20 left. If he tries to put 8 apples in each, he can't, because there's one basket that doesn't have enough. Therefore, the number of baskets is fixed. So, with that in mind, let's proceed.So, if the number of baskets is fixed, say, b baskets. Then, when he puts 4 apples in each, the total apples would be 4b + 20. When he tries to put 8 apples in each basket, he can fill (b - 1) baskets with 8 apples each, and the last basket will have a remainder. So, total apples would be 8(b - 1) + r, where r is between 1 and 7 (since it's not enough for 8). Since the total apples are the same, we have:4b + 20 = 8(b - 1) + rWhich simplifies to:4b + 20 = 8b - 8 + rThen:20 + 8 = 8b - 4b + r28 = 4b + rSo, 4b + r = 28, where 1 ≤ r ≤7.Since both b and r are positive integers, let's see possible values of b and r.4b = 28 - rSo, 28 - r must be divisible by 4. Let's see:If r = 0, 28 - 0 = 28, which is divisible by 4, but r must be at least 1.r =1: 28 -1=27, which is not divisible by 4.r=2: 28-2=26, 26/4=6.5, not integer.r=3:25, not divisible by 4.r=4:24, which is divisible by 4. 24/4=6. So, b=6, r=4.r=5:23, not divisible by 4.r=6:22, not divisible by 4.r=7:21, not divisible by 4.So the only possible solution is r=4, which gives b=6.Therefore, the number of apples is:a = 4b + 20 = 4*6 + 20 = 24 +20=44.Wait, 44 is option D. But let's check if that makes sense in the second scenario.If he has 44 apples and tries to put 8 in each basket, how would that go?Number of baskets filled with 8 apples: Let's see, 44 divided by 8 is 5 with a remainder of 4. So, 5 baskets with 8 apples each, and one basket with 4 apples. Since 4 is less than 8, that means there's one basket that doesn't have enough. So that matches the problem statement.But wait, if the number of baskets is 6 (from the first scenario), then in the second scenario, he is using the same 6 baskets. If he tries to put 8 apples in each basket, he can only fill 5 baskets with 8 apples, and the 6th basket has 4 apples. So, that's correct.But let's check the answer options. The options are 38, 40, 42, 44. So, 44 is D. However, let me check with another approach.Alternatively, maybe the number of baskets is different in the two scenarios. For instance, in the first case, when he uses 4 per basket, he has 20 left. So, the number of baskets here would be (a -20)/4. In the second case, if he uses 8 per basket, he needs (a /8) baskets, but there is one basket that doesn't have enough. So, the number of baskets required would be floor(a /8) +1, but the last basket has less than 8. However, the problem states that he wants to distribute into "several baskets," so maybe the number of baskets he uses is the minimum required to hold all apples. But it's unclear.Alternatively, perhaps the number of baskets is not fixed. Let's suppose that in the first scenario, he uses as many baskets as needed to hold 4 apples each, leaving 20. So, the number of baskets here would be (a -20)/4. In the second scenario, he uses as many baskets as needed to hold 8 apples each, but there's one basket that is not full. So, the number of baskets here would be floor(a /8) +1. But the problem says "there will be one basket that does not have enough," implying that only one basket is not full, which would mean that when he divides a apples into baskets of 8, the remainder is non-zero, so there is one basket with the remainder, which is less than 8. So, in that case, the number of baskets used in the second scenario would be (a //8) +1, where // is integer division. But the problem is that the number of baskets in each scenario could be different. But the problem says "distribute into several baskets" in both cases. The problem doesn't specify if the number of baskets is the same or different. So, this is a bit ambiguous.But let's try this approach. Let's let a be the number of apples. Then, in the first case, when he puts 4 apples per basket, he has 20 left. So, the number of baskets he uses is (a -20)/4. Let’s call this number b1. So, b1 = (a -20)/4. Since the number of baskets must be an integer, a -20 must be divisible by 4, so a ≡ 20 mod 4, which simplifies to a ≡ 0 mod 4, since 20 is divisible by 4. Therefore, a must be a multiple of 4. Let's check the options: A:38 (not divisible by 4), B:40 (yes), C:42 (no), D:44 (yes). So the possible answers are B and D.In the second scenario, when he puts 8 apples per basket, there is one basket with not enough. So, the number of baskets needed would be ceil(a /8). But in this case, the number of full baskets is floor(a /8), and the remaining apples are a mod 8. But the problem states that there is one basket without enough, so the number of baskets used is floor(a /8) +1. Therefore, the number of baskets used in the second scenario is b2 = floor(a /8) +1. The problem says "there will be one basket that does not have enough apples," so the remainder is a mod 8, which must be less than 8, which is always true. But it's possible that the remainder is zero, but in that case, there would be no basket with not enough. Therefore, a mod 8 cannot be zero. So, in the second scenario, a is not divisible by 8. Hence, a mod 8 ≠ 0. Therefore, a is not a multiple of 8. Let's check the possible answers: B:40 (40 is divisible by 8, since 40/8=5). D:44 (44 divided by 8 is 5.5, so remainder 4). Therefore, 40 is divisible by 8, which would mean that if he tried to put 8 apples in each basket, he would have exactly 5 baskets with 8 apples each, and no leftover. Therefore, there would be no basket with not enough. But the problem states that there is one basket with not enough. So, 40 cannot be the answer. Therefore, 44 is the only other possibility. So, 44 divided by 8 is 5 baskets with 8 apples each, and 4 apples left, which go into the 6th basket. So, that basket has only 4 apples, which is not enough. Therefore, 44 fits.Wait, but in the first scenario, with 44 apples:He puts 4 apples in each basket, leaving 20. So, the number of baskets is (44 -20)/4 = 24/4 = 6 baskets. Then, in the second scenario, putting 8 apples in each basket would require 5 full baskets (40 apples) and 4 left, so 6 baskets, with the last one having 4. Therefore, same number of baskets. So, that's consistent with the initial assumption that the number of baskets is the same. Therefore, 44 is the answer. So, D.But let me check why the other options don't work.Option B:40 apples.First scenario: 40 apples. If he puts 4 per basket, leaving 20. Then, the number of baskets is (40 -20)/4 = 20/4=5 baskets. So, 5 baskets with 4 apples each, 20 apples left. Then, in the second scenario, putting 8 apples in each basket. If he uses 5 baskets, he can put 8 apples each, needing 40 apples. Which exactly uses up all apples, no leftovers. Therefore, all baskets have 8 apples. Therefore, there are no baskets with not enough. But the problem says there is one basket that does not have enough. Therefore, 40 cannot be the answer.Option A:38 apples.First, check if 38 is a multiple of 4. 38 divided by 4 is 9.5. Not an integer. So, in the first scenario, putting 4 apples per basket would require (38 -20)/4 = 18/4=4.5 baskets. But you can't have half a basket, so this is invalid. Therefore, A is out.Option C:42 apples.42 -20=22. 22 divided by 4 is 5.5. Again, not an integer. So, same problem as A. So, invalid.Therefore, only options B and D are possible, but B fails the second condition, so D is the answer.Alternatively, let's think using inequalities.In the second scenario, when trying to put 8 apples in each basket, there's one basket with not enough. So, if he had enough apples to fill all baskets with 8, then there would be none left. But since there's one basket not full, the number of apples is less than 8 times the number of baskets. But how does this relate to the first scenario?Wait, if the number of baskets in the first scenario is b, then apples a =4b +20.In the second scenario, if he uses b baskets again, trying to put 8 apples each, he needs 8b apples, but he only has a apples. Therefore, a <8b, but there's one basket not full. Wait, but if a <8b, then all baskets would have less than 8 apples. But the problem states "there will be one basket that does not have enough apples," implying that only one basket is underfilled. So, this suggests that (b -1) baskets are filled with 8 apples, and the last one has a -8*(b -1) apples, which is less than 8. So:a -8*(b -1) <8Which simplifies to:a <8*(b -1) +8a <8bBut we also know that in the first scenario, a =4b +20. Therefore:4b +20 <8bSubtract 4b:20 <4bDivide by 4:5 <bSo, b>5. Since b must be an integer (number of baskets), then b≥6.Also, in the second scenario, the number of apples left in the last basket is a -8*(b -1). Which must be at least 1, because if it were zero, there would be no basket with not enough. So:a -8*(b -1) ≥1So:a ≥8*(b -1) +1But a =4b +20, so:4b +20 ≥8b -8 +1Simplify:4b +20 ≥8b -7Subtract 4b:20 ≥4b -7Add 7:27 ≥4bSo:4b ≤27Therefore, b ≤27/4=6.75. Since b is an integer, b≤6.But earlier, we had b≥6. Therefore, b=6.Therefore, the number of baskets is 6. Therefore, apples a=4*6 +20=24 +20=44. So, 44 apples. Which is option D.Therefore, the answer is D:44.Final Answerboxed{D}</think>Xiaohua has a number of apples that he wants to distribute into several baskets. If he puts 4 apples in each basket, he will have 20 apples left undistributed. If he puts 8 apples in each basket, there will be one basket that does not have enough apples. We need to determine the original number of apples.1. First Scenario: When putting 4 apples in each basket, the number of apples can be expressed as: [ a = 4b + 20 ] where (a) is the number of apples and (b) is the number of baskets.2. Second Scenario: When putting 8 apples in each basket, there is one basket with not enough apples. This can be expressed as: [ a = 8(b - 1) + r ] where (r) is the remainder apples in the last basket, and (r < 8).3. Equating the two expressions for (a): [ 4b + 20 = 8(b - 1) + r ] Simplifying this equation: [ 4b + 20 = 8b - 8 + r implies 28 = 4b + r ] Since (r < 8), we test possible values for (r) and find that (r = 4) and (b = 6) satisfy the equation.4. Substituting (b = 6) back into the first equation: [ a = 4 times 6 + 20 = 24 + 20 = 44 ]5. Verification: With 44 apples, distributing 8 apples per basket results in 5 full baskets and one basket with 4 apples, which is less than 8. This satisfies the problem's conditions.Thus, the original number of apples is (boxed{D}).,<think>Okay, let's see. So the problem is about Xiaohua distributing apples into baskets. If he puts 4 apples in each basket, he'll have 20 left over. But if he tries to put 8 apples in each basket, there's one basket that doesn't have enough. The question is asking how many apples he originally had, with options A to D given.Hmm, let me break this down. Let's start by translating the problem into mathematical equations. Let me denote the number of apples as N and the number of baskets as B. First scenario: If he puts 4 apples in each basket, he has 20 left. So that means N = 4B + 20. Because each basket has 4 apples, times the number of baskets, plus the leftover 20.Second scenario: If he puts 8 apples in each basket, there's one basket that doesn't have enough. So, trying to put 8 apples in each basket, but one basket ends up with less. That suggests that the total number of apples isn't a multiple of 8, and when he tries to divide them into baskets with 8 apples each, he can fill some baskets completely, but the last one doesn't have enough. So, in mathematical terms, this would mean that when dividing N by 8, the quotient is (B - 1), and the remainder is less than 8. Wait, but how does the number of baskets relate here?Wait a second, maybe I need to clarify the second scenario. If he uses the same number of baskets as in the first scenario? Or is the number of baskets different? Hmm, the problem doesn't specify whether he's using the same number of baskets or not. That might be a point of confusion.Wait, let's read the problem again: "If he puts 4 apples in each basket, he will have 20 apples left undistributed. If he puts 8 apples in each basket, there will be one basket that does not have enough apples." The problem says "several baskets" both times, but doesn't specify if it's the same number of baskets. Hmm. But maybe we can assume that he is using the same number of baskets in both scenarios? Or is the number of baskets different?Wait, actually, in the first case, he is distributing into several baskets with 4 each, and in the second case, he's also distributing into several baskets, but with 8 each. However, the problem says "there will be one basket that does not have enough apples." So, in the second case, the number of baskets might be the same? Or maybe not. The problem is a bit ambiguous here.Alternatively, maybe the number of baskets is different. Let me think. Let's try to parse the problem again.Original problem: Xiaohua has a number of apples to distribute into several baskets. If he puts 4 apples in each basket, he will have 20 left. If he puts 8 apples in each basket, there will be one basket that does not have enough. So, when he switches from putting 4 apples per basket to 8 apples per basket, the number of baskets might be different? But how?Alternatively, maybe when he tries to use 8 apples per basket, he's using fewer baskets, but since he doesn't have enough apples to fill all baskets with 8, one basket ends up with less. So, perhaps the number of baskets he tries to use when putting 8 apples is such that (number of baskets) times 8 would require more apples than he has, except for one basket. Wait, that's a bit unclear.Alternatively, maybe he tries to use the same number of baskets as before. So in the first case, he has B baskets, puts 4 each, leaving 20. Then in the second case, he tries to put 8 apples into each of those B baskets, but can't because he doesn't have enough apples. So the total apples required would be 8B, but he has N apples, which is 4B + 20. So, if he tries to put 8 in each of B baskets, he would need 8B apples, but he only has 4B + 20. Therefore, the difference is 8B - (4B + 20) = 4B - 20. Since there's one basket that doesn't have enough, that would imply that he can fill (B - 1) baskets with 8 apples each, and the last basket has some remaining apples. Let me formalize this.If he tries to put 8 apples into each of B baskets, he can fill (B - 1) baskets completely, and the last basket has N - 8*(B - 1) apples. But since he doesn't have enough for the last basket to have 8, that means 0 < N - 8*(B - 1) < 8.So, substituting N from the first equation: N = 4B + 20. Therefore, substituting into the inequality:0 < (4B + 20) - 8*(B - 1) < 8Let me compute that:First, expand 8*(B - 1) = 8B - 8So, the expression becomes:0 < 4B + 20 - 8B + 8 < 8Simplify the middle part:(4B - 8B) + (20 + 8) = -4B + 28So:0 < -4B + 28 < 8Now, let's split this into two inequalities:1. 0 < -4B + 282. -4B + 28 < 8Starting with the first inequality:0 < -4B + 28Subtract 28 from both sides:-28 < -4BDivide both sides by -4, remembering to reverse the inequality:7 > BOr, equivalently:B < 7Second inequality:-4B + 28 < 8Subtract 28 from both sides:-4B < -20Divide both sides by -4, reversing the inequality:B > 5So, combining both inequalities:5 < B < 7Since B must be an integer (number of baskets), the possible values for B are 6.So B = 6.Then, substitute back into N = 4B + 20:N = 4*6 + 20 = 24 + 20 = 44.Wait, but 44 is option D. However, let's check if that's correct.Let me verify with B = 6:First scenario: 4 apples per basket, 6 baskets: 4*6 = 24 apples used, leaving 20. So total apples N = 24 + 20 = 44.Second scenario: trying to put 8 apples in each basket. If he uses 6 baskets, he would need 8*6 = 48 apples. But he only has 44. So, 44 - 8*5 = 44 - 40 = 4 apples left for the 6th basket. So the 6th basket has 4 apples, which is less than 8. Therefore, there is one basket (the 6th) that does not have enough. That matches the problem's description.Therefore, N = 44, which is option D. However, looking at the options, D is 44, which seems correct. Wait, but let me check if there are other possible values for B.Wait, in the inequality above, B had to be greater than 5 and less than 7. Since B must be an integer, only 6 is possible. Therefore, N must be 44. So the answer should be D: 44.But hold on, let me check the answer options again. The options are A:38, B:40, C:42, D:44. So D is indeed 44, which is what I got.But let me check another approach to be sure. Let me consider that maybe the number of baskets is different in the two scenarios. The problem says "several baskets" in both cases, but doesn't specify the same number. However, that might complicate things. Let's see.Suppose in the first scenario, he uses B baskets, so N = 4B + 20.In the second scenario, if he uses a different number of baskets, say C baskets. Then putting 8 apples in each basket would require 8*(C -1) apples, and the last basket has N -8*(C -1) apples, which is less than 8. So, N -8*(C -1) < 8, and N -8*(C -1) >0 (since there's at least some apples). Therefore, 0 < N -8C +8 <8.But without knowing the relation between B and C, this might be too vague. So maybe the problem is intended to have the same number of baskets. Therefore, the first approach is correct.Alternatively, perhaps the problem is that when he switches to putting 8 apples per basket, he needs more baskets? Wait, but if he uses more baskets, he could distribute all the apples. But the problem says there's one basket that doesn't have enough. Hmm.Wait, the problem says "there will be one basket that does not have enough apples." So, maybe he tries to use the same number of baskets as before. If he uses the same number of baskets B, then putting 8 apples in each would require 8B apples, but he only has N =4B +20. Therefore, the shortfall is 8B - (4B +20) =4B -20. Therefore, the number of apples he is short by is 4B -20, which would mean he can't fill all B baskets with 8 apples each. Therefore, how many baskets can he fill? Let's see.Alternatively, maybe he fills (B -1) baskets with 8 apples each, and the last basket has N -8*(B -1) apples. Since the last basket has less than 8, that would be the case. So:N -8*(B -1) <8But N =4B +20So:4B +20 -8B +8 <8-4B +28 <8-4B < -20B>5Similarly, the number of apples in the last basket must be positive:4B +20 -8*(B -1) >0Which gives:-4B +28 >04B <28B<7Therefore, same as before: B=6, N=44.Therefore, the answer is D:44.But let me check if N=44 is one of the options. Yes, D is 44.But just to be thorough, let's check the other options to see if they fit.Take option A:38 apples.If he puts 4 apples per basket: number of baskets B= (38 -20)/4=18/4=4.5. But number of baskets must be integer. So this is not possible. So A is invalid.Option B:40 apples.If he puts 4 apples per basket: B=(40-20)/4=20/4=5 baskets. So 5 baskets, 4 each, 20 apples left. Then, when he tries to put 8 apples per basket into 5 baskets, he needs 40 apples. But he has 40 apples. So he can exactly fill 5 baskets with 8 apples each, no leftovers. But the problem says there's one basket that doesn't have enough. But in this case, all baskets are filled exactly. So this doesn't fit the problem statement. Therefore, B is invalid.Option C:42 apples.Putting 4 apples per basket: B=(42 -20)/4=22/4=5.5. Not integer. So invalid.Option D:44 apples.As computed before: B=6 baskets. 4*6=24, 44-24=20 left. Then, trying to put 8 apples in 6 baskets: 8*5=40, 44-40=4 in the last basket. So one basket with only 4 apples, which is less than 8. Fits the problem statement. Therefore, D is correct.Therefore, the answer is D:44.Final Answerboxed{D}</think>Xiaohua has a number of apples ( N ) that he wants to distribute into baskets. If he puts 4 apples in each basket, he will have 20 apples left. This can be expressed as:[ N = 4B + 20 ]where ( B ) is the number of baskets.If he puts 8 apples in each basket, there will be one basket that does not have enough apples. This implies that when distributing the apples into ( B ) baskets, he can fill ( B - 1 ) baskets completely with 8 apples each, and the last basket will have less than 8 apples. This can be expressed as:[ 0 < N - 8(B - 1) < 8 ]Substituting ( N = 4B + 20 ) into the inequality:[ 0 < (4B + 20) - 8(B - 1) < 8 ]Simplifying the middle part:[ 0 < 4B + 20 - 8B + 8 < 8 ][ 0 < -4B + 28 < 8 ]Splitting this into two inequalities:1. ( 0 < -4B + 28 )2. ( -4B + 28 < 8 )Solving the first inequality:[ 0 < -4B + 28 ][ -28 < -4B ][ 7 > B ][ B < 7 ]Solving the second inequality:[ -4B + 28 < 8 ][ -4B < -20 ][ B > 5 ]Combining both inequalities, we get:[ 5 < B < 7 ]Since ( B ) must be an integer, the only possible value is ( B = 6 ). Substituting ( B = 6 ) back into the equation ( N = 4B + 20 ):[ N = 4 times 6 + 20 = 24 + 20 = 44 ]Verification:- Putting 4 apples in each of 6 baskets uses 24 apples, leaving 20, which matches the first scenario.- Putting 8 apples in each of 6 baskets requires 48 apples, but Xiaohua only has 44. He can fill 5 baskets completely (40 apples) and the last basket has 4 apples, which matches the second scenario.Thus, the number of apples Xiaohua originally had is (boxed{D}).]

🔑:Let's denote the number of baskets as x. Then, the total number of apples can be expressed as (4x + 20). Considering the conditions given:1. When distributing 8 apples per basket, one basket is not full, which implies: - 4x + 20 - 8 times (x - 1) > 0 - 4x + 20 - 8 times (x - 1) < 8Solving these inequalities, we find that 5 < x < 7. Since x must be an integer, it follows that x = 6. When x = 6, the total number of apples is 4 times 6 + 20 = 44. Therefore, the correct answer is boxed{text{D}}.

🧠:[<think>Okay, so I need to find the original price of a computer before a 40% discount was applied, given that after the discount, the price is 'a' yuan. Let me think step by step.First, let's recall that when something is reduced by a certain percentage, the new price is the original price minus that percentage of the original price. So if the original price is, say, P yuan, then after a 40% discount, the selling price becomes P minus 40% of P.Mathematically, that would be:Selling Price = Original Price - (40% of Original Price)Or in symbols:a = P - 0.4PHmm, simplifying that, 0.4P subtracted from P would leave 0.6P. So:a = 0.6PTherefore, the original price P must be a divided by 0.6. Let me write that:P = a / 0.6Dividing by 0.6 is the same as multiplying by 10/6, which simplifies to 5/3. Because 10 divided by 6 is 5/3. Let me verify that:0.6 is the same as 3/5, right? Because 0.6 = 3/5. So if a = (3/5)P, then P = a divided by (3/5) which is a multiplied by 5/3. So that's the same as (5/3)a.Looking at the options given, option D is 5/3 a yuan. So that must be the correct answer. Let me just check my reasoning again.Original price P is reduced by 40%, so the new price is 60% of P, which is 0.6P. That's equal to a. Therefore, P = a / 0.6 = (10/6)a = (5/3)a. Yep, that seems right.Alternatively, I can plug the options back into the equation to see which one works. Let's test each option:Option A: 0.6a yuan. If the original price was 0.6a, then a 40% discount would be 0.4 * 0.6a = 0.24a. Subtracting that from 0.6a gives 0.36a, which is not equal to a. So A is wrong.Option B: 0.4a yuan. 40% of 0.4a is 0.16a. Subtracting that gives 0.24a, which is also not a. So B is wrong.Option C: (5/2)a yuan. Let's compute 40% of (5/2)a: 0.4*(5/2)a = (2/5)*(5/2)a = a. Wait, that would mean subtracting a from (5/2)a, which gives (5/2)a - a = (3/2)a. But the selling price is supposed to be a, not (3/2)a. So that's not correct. Wait, maybe I did that wrong. Let me recalculate.Original price is (5/2)a. 40% of that is 0.4*(5/2)a = (2/5)*(5/2)a = (10/10)a = a. So the discount is a yuan. Therefore, the selling price would be (5/2)a - a = (5/2 - 2/2)a = (3/2)a. But the problem states the selling price is a, so that's a discrepancy. Therefore, option C is incorrect.Option D: (5/3)a yuan. Let's check. 40% of (5/3)a is 0.4*(5/3)a = (2/5)*(5/3)a = (10/15)a = (2/3)a. Subtracting that from the original price: (5/3)a - (2/3)a = (3/3)a = a. Perfect, that matches the given selling price. So D is correct.Therefore, the answer must be D.Final Answerboxed{D}</think>Given that the selling price of a computer after a 40% discount is (a) yuan, we need to find the original price.1. Let the original price be (P) yuan.2. After a 40% discount, the selling price is (P - 0.4P = 0.6P).3. We know the selling price is (a) yuan, so we set up the equation: [ 0.6P = a ]4. Solving for (P), we get: [ P = frac{a}{0.6} = frac{a}{frac{3}{5}} = frac{5}{3}a ]To verify, if the original price is (frac{5}{3}a) yuan, a 40% discount is calculated as:[0.4 times frac{5}{3}a = frac{2}{5} times frac{5}{3}a = frac{2}{3}a]Subtracting this discount from the original price:[frac{5}{3}a - frac{2}{3}a = frac{3}{3}a = a]Thus, the original price of each unit is (frac{5}{3}a) yuan, which corresponds to option D.[boxed{D}],<think>Okay, let's see. The problem is about finding the original price of a computer after a 40% discount. The current selling price after the discount is 'a' yuan. I need to figure out what the original price was before the discount. The options are given, so maybe I can check each one, but first, let me try to solve it step by step.Alright, so if the price was reduced by 40%, that means the current selling price is 60% of the original price. Because 100% minus 40% is 60%. So if I let the original price be, let's say, 'x' yuan, then 60% of x is equal to 'a' yuan. So mathematically, that would be 0.6 * x = a. Then, to find x, I need to divide both sides by 0.6. So x = a / 0.6. Hmm, let me compute that. Dividing by 0.6 is the same as multiplying by 10/6, which simplifies to 5/3. So x = (5/3)a. Let me verify that.If the original price is (5/3)a, then reducing it by 40% would be 40% of (5/3)a. Let's calculate that. 40% is 0.4, so 0.4 * (5/3)a = (2/5)*(5/3)a = (2/3)a. So the discount is (2/3)a. Therefore, the new price is original price minus discount: (5/3)a - (2/3)a = (3/3)a = a. That checks out. So the original price is indeed (5/3)a, which is option D.Wait, let me check the options again. Option D is 5/3 a yuan, which matches my calculation. Let me just make sure that I didn't make any mistakes in my reasoning. So, the key idea is that after a 40% discount, the price is 60% of the original. Therefore, original price * 0.6 = a, so original price = a / 0.6 = (10/6)a = (5/3)a. Yeah, that's right. Alternatively, maybe I can use fractions instead of decimals to avoid confusion. 40% is equivalent to 2/5. So the discount is 2/5 of the original price. Therefore, the selling price is original price minus 2/5 of original price, which is 3/5 of the original price. So 3/5 * original price = a. Hence, original price = a divided by 3/5, which is a multiplied by 5/3, so 5/3 a. Yep, that's the same answer.So, the original price is 5/3 a yuan. Looking at the options, D is 5/3 a, so that should be the correct answer. Let me just check the other options to make sure none of them could be correct. Option A is 0.6a, which is 60% of a, but since a is the discounted price, that would actually make the original price lower than the discounted price, which doesn't make sense. Option B is 0.4a, which is even less, so that's definitely wrong. Option C is 5/2 a, which is 2.5a. If I check that, 40% of 2.5a is 1a, so subtracting 1a gives 1.5a, which is not equal to a. So that's wrong. Option D is 5/3 a, which as we saw, gives the correct result. So, yeah, D is the answer.Final Answerboxed{D}</think>After a 40% discount, the selling price of the computer is (a) yuan. To find the original price, we note that the selling price is 60% of the original price. Let the original price be (x) yuan. Then we have:[0.6x = a]Solving for (x), we divide both sides by 0.6:[x = frac{a}{0.6} = frac{a}{frac{3}{5}} = frac{5}{3}a]We can verify this by checking that 40% of the original price (frac{5}{3}a) subtracted from the original price gives (a):[text{Discount} = 0.4 times frac{5}{3}a = frac{2}{5} times frac{5}{3}a = frac{2}{3}a][text{Selling Price} = frac{5}{3}a - frac{2}{3}a = frac{3}{3}a = a]Thus, the original price is (frac{5}{3}a) yuan, which corresponds to option D.[boxed{D}]]

🔑:To solve the problem, let's denote the original price of each unit of this brand of computer as x yuan. According to the given information, the price is reduced by 40%, which means the selling price becomes 60% of the original price. This can be mathematically represented as:[x times (1 - 40%) = x times 60% = a]Solving for x, we get:[x times 60% = a][x = frac{a}{60%}][x = frac{a}{0.6}][x = frac{5}{3}a]Therefore, the original price of each unit of this brand of computer is frac{5}{3}a yuan. Hence, the correct answer is:[boxed{D}]