🧠:[<think>Alright, let me try to figure out this problem. So, we have a rectangular field with an area of 864 square steps. The width is 12 steps less than the length. We need to find the perimeter. Hmm, okay, let's break this down step by step.First, I remember that the area of a rectangle is calculated by multiplying its length and width. The formula is Area = length × width. Given that the area is 864, and the width is 12 steps less than the length, we can set up an equation. Let me define the variables first. Let's let the length be 'L' steps. Then, the width would be 'L - 12' steps since it's 12 less than the length.So, plugging into the area formula: L × (L - 12) = 864. That should give me a quadratic equation. Let me expand that: L² - 12L = 864. Then, bringing 864 to the left side, the equation becomes L² - 12L - 864 = 0. Now, I need to solve this quadratic equation for L. Quadratic equations can be solved using the quadratic formula: L = [12 ± √(144 + 3456)] / 2. Wait, let me check that. The quadratic formula is (-b ± √(b² - 4ac)) / 2a. In this equation, a = 1, b = -12, and c = -864. So, substituting those values in:L = [12 ± √( (-12)² - 4*1*(-864) )] / (2*1)L = [12 ± √(144 + 3456)] / 2L = [12 ± √3600] / 2√3600 is 60, right? Because 60*60=3600. So then,L = [12 ± 60] / 2This gives two possible solutions:1. [12 + 60] / 2 = 72 / 2 = 362. [12 - 60] / 2 = -48 / 2 = -24But since length can't be negative, we discard -24. Therefore, the length is 36 steps. Then the width is 36 - 12 = 24 steps. Now, to find the perimeter of a rectangle, the formula is 2*(length + width). So, plugging in the values:Perimeter = 2*(36 + 24) = 2*60 = 120 steps.Wait, let me verify that again. If the length is 36 and the width is 24, then the area should be 36*24. Let me compute that: 36*24. 30*24=720, 6*24=144, so 720+144=864. Yep, that matches the given area. So the calculations seem correct.Therefore, the perimeter is 120 steps.Final AnswerThe perimeter of the rectangular field is boxed{120} steps.</think>Given the area of a rectangular field is 864 square steps, and the width is 12 steps less than the length, we need to find the perimeter.1. Let the length be ( L ) steps. Then, the width is ( L - 12 ) steps.2. The area formula gives us the equation: [ L times (L - 12) = 864 ]3. Expanding and rearranging the equation: [ L^2 - 12L - 864 = 0 ]4. Solving the quadratic equation using the quadratic formula ( L = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1 ), ( b = -12 ), and ( c = -864 ): [ L = frac{12 pm sqrt{144 + 3456}}{2} = frac{12 pm sqrt{3600}}{2} = frac{12 pm 60}{2} ]5. This gives two solutions: [ L = frac{12 + 60}{2} = 36 quad text{and} quad L = frac{12 - 60}{2} = -24 ] Since length cannot be negative, we discard -24. Thus, ( L = 36 ) steps.6. The width is ( 36 - 12 = 24 ) steps.7. The perimeter of the rectangle is calculated as: [ 2 times (36 + 24) = 2 times 60 = 120 ]The perimeter of the rectangular field is (boxed{120}) steps.,<think>Okay, so I need to solve this problem about the rectangular field. Let me read it again to make sure I understand. The area is 864 square steps, and the width is 12 steps less than the length. I need to find the perimeter. Hmm, right. Let's see.First, let's recall that the area of a rectangle is calculated by multiplying its length and width. The formula is Area = length × width. They also mention that the width is 12 steps less than the length. So, if I let the length be a variable, say L, then the width would be L minus 12. Let me write that down:Let length = L stepsThen width = L - 12 stepsGiven that the area is 864 square steps, so according to the formula:L × (L - 12) = 864So that equation should help me find the value of L. Once I know L, I can find the width, and then compute the perimeter. The perimeter of a rectangle is 2 times length plus 2 times width, right? So perimeter P = 2L + 2W. Since W is L - 12, substituting that in, P = 2L + 2(L - 12) = 2L + 2L - 24 = 4L - 24. So once I know L, I can calculate the perimeter.But first, solving the equation L(L - 12) = 864. Let's expand that left side:L² - 12L = 864Then, subtract 864 from both sides to set the equation to zero:L² - 12L - 864 = 0So now we have a quadratic equation. Let's try to factor it. Hmm, quadratic equations can sometimes be factored, but the coefficients here might make it tricky. Let's see. The quadratic is L² -12L -864 = 0.To factor this, we need two numbers that multiply to -864 and add up to -12. Let me think. Since the product is negative, one number is positive and the other is negative. Let's find pairs of factors of 864 and see which pair differs by 12, with the larger one being negative? Wait, maybe I need to adjust my approach.Alternatively, I can use the quadratic formula. The quadratic formula is L = [12 ± sqrt( (-12)^2 - 4*1*(-864) ) ] / (2*1)Calculating the discriminant first: (-12)^2 = 144, and 4*1*864 = 3456. So the discriminant is 144 + 3456 = 3600. The square root of 3600 is 60. So:L = [12 ± 60]/2So that gives two solutions:L = (12 + 60)/2 = 72/2 = 36Or L = (12 - 60)/2 = (-48)/2 = -24But since length can't be negative, we discard L = -24. Therefore, L = 36 steps.So the length is 36 steps, and the width is 36 - 12 = 24 steps.Now, let's compute the perimeter. As I mentioned earlier:Perimeter P = 2L + 2W = 2*36 + 2*24 = 72 + 48 = 120 steps.Wait, let me double-check that. 36 times 2 is 72, and 24 times 2 is 48. Adding them together gives 72 + 48 = 120. Yeah, that seems right.But let me verify the area to make sure I didn't make a mistake. If length is 36 and width is 24, then area is 36*24. Let's compute that: 36*20=720, 36*4=144, so 720+144=864. Exactly the area given. Perfect, that checks out.So all steps seem to be correct. I defined variables correctly, formed the equation, solved the quadratic, checked for valid solutions, computed the perimeter, and verified the area. All steps lead to the perimeter being 120 steps.I think that's the answer. But just to recap the steps again:1. Let length be L, width is L - 12.2. Area = L*(L - 12) = 8643. Quadratic equation: L² -12L -864 = 04. Solving using quadratic formula gives L = 36 (discarding negative solution)5. Width = 36 -12 =246. Perimeter: 2*(36 +24) = 2*60 =120Yep, that all makes sense. I don't see any errors here. The key was setting up the correct equation from the problem statement and then solving it properly. Initially, factoring might have been a bit hard because of the large numbers, but using the quadratic formula was straightforward. Then verifying the answer by plugging back into the area formula is a good practice to ensure no mistakes were made in the algebra.Another way to approach the quadratic equation could be to complete the square, but with these numbers, quadratic formula was faster. Let me try completing the square just to see:Starting from L² -12L =864To complete the square, take coefficient of L, which is -12, divide by 2 to get -6, square it to get 36. Add 36 to both sides:L² -12L +36 =864 +36(L -6)^2 =900Take square roots:L -6 = ±30Therefore, L =6 ±30Which gives L=36 or L=-24. Same solutions. So that's another way, and it's helpful to see that both methods give the same result. So that reinforces that the length is indeed 36.So again, perimeter is 2*(36+24)=120. Confirming once more.Therefore, the perimeter of the rectangular field is 120 steps.Final AnswerThe perimeter of the rectangular field is boxed{120} steps.</think>Let the length of the rectangular field be ( L ) steps. Then the width is ( L - 12 ) steps. The area of the rectangle is given as 864 square steps, so we have the equation:[L times (L - 12) = 864]Expanding and rearranging terms, we get the quadratic equation:[L^2 - 12L - 864 = 0]Using the quadratic formula ( L = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1 ), ( b = -12 ), and ( c = -864 ):[L = frac{12 pm sqrt{144 + 3456}}{2} = frac{12 pm 60}{2}]This gives two solutions:[L = frac{72}{2} = 36 quad text{or} quad L = frac{-48}{2} = -24]Since the length cannot be negative, we discard ( L = -24 ). Thus, the length is ( L = 36 ) steps, and the width is ( 36 - 12 = 24 ) steps.The perimeter ( P ) of the rectangular field is calculated as:[P = 2 times (text{length} + text{width}) = 2 times (36 + 24) = 2 times 60 = 120]Thus, the perimeter of the rectangular field is (boxed{120}) steps.]

🔑:Given the problem, we start by setting up the equation based on the information provided:- Let the width of the rectangle be x steps. Since the width is 12 steps less than the length, we can express the length as (x + 12) steps.- The area of the rectangle is given as 864 square steps. The area of a rectangle is calculated by multiplying its length by its width, so we have the equation x(x + 12) = 864.Now, we solve the equation for x:[begin{align*}x(x + 12) &= 864 x^2 + 12x - 864 &= 0end{align*}]Solving this quadratic equation, we find the values of x:[begin{align*}x_1 &= 24, quad x_2 = -36end{align*}]Since a negative width does not make sense in this context, we discard x_2 = -36 and keep x_1 = 24 as the width of the rectangle.To find the length, we substitute x = 24 into the expression for the length:[begin{align*}x + 12 &= 24 + 12 &= 36end{align*}]Therefore, the length of the rectangle is 36 steps.Finally, to find the perimeter of the rectangle, we use the formula for the perimeter of a rectangle, which is 2(text{length} + text{width}):[begin{align*}text{Perimeter} &= 2(24 + 36) &= 2(60) &= 120 text{ steps}end{align*}]Thus, the perimeter of the rectangular field is boxed{120} steps.

🧠:[<think>Okay, so I need to find the limit as x approaches 1 of [(2 - x)/x] raised to the power of 1 over ln(2 - x). Hmm, let me think about how to approach this. First, when x approaches 1, what happens to the base (2 - x)/x? Let's plug in x = 1. Then the base becomes (2 - 1)/1 = 1/1 = 1. And the exponent is 1 divided by ln(2 - 1) which is ln(1) = 0. So, the expression becomes 1 raised to infinity, which is an indeterminate form. That means I can't just plug in the value directly; I need to manipulate it somehow.I remember that when dealing with limits of the form 1^∞, it's useful to use the exponential function and natural logarithm properties. The standard trick is to take the natural logarithm of the function, evaluate the limit, and then exponentiate the result. So, let me try that.Let L be the original limit. Then,ln(L) = lim_{x→1} [ (1/ln(2 - x)) * ln((2 - x)/x) ]So, if I can compute this limit, then L will be e raised to that result.Let me write that out:ln(L) = lim_{x→1} [ ln((2 - x)/x) / ln(2 - x) ]Simplify the numerator and denominator. Let's see. The numerator is ln((2 - x)/x) which can be written as ln(2 - x) - ln(x). So,ln(L) = lim_{x→1} [ (ln(2 - x) - ln(x)) / ln(2 - x) ]I can split this fraction into two terms:ln(L) = lim_{x→1} [ ln(2 - x)/ln(2 - x) - ln(x)/ln(2 - x) ] = lim_{x→1} [1 - (ln(x)/ln(2 - x))]So, now the expression simplifies to 1 minus the limit of [ln(x)/ln(2 - x)] as x approaches 1. Let's focus on that remaining limit.Let me compute lim_{x→1} [ln(x)/ln(2 - x)].As x approaches 1, ln(x) approaches ln(1) = 0, and ln(2 - x) approaches ln(1) = 0. So, this is a 0/0 indeterminate form, which suggests that I can apply L'Hospital's Rule here. Applying L'Hospital's Rule, we take the derivatives of the numerator and denominator:Derivative of numerator (ln(x)) is 1/x.Derivative of denominator (ln(2 - x)) is (1/(2 - x)) * (-1) = -1/(2 - x).So, the limit becomes:lim_{x→1} [ (1/x) / (-1/(2 - x)) ) ] = lim_{x→1} [ (1/x) * (-(2 - x)/1) ] = lim_{x→1} [ -(2 - x)/x ]Now, substituting x = 1 gives -(2 - 1)/1 = -1/1 = -1.Therefore, the limit of [ln(x)/ln(2 - x)] as x approaches 1 is -1. Plugging this back into our earlier expression for ln(L):ln(L) = 1 - (-1) = 1 + 1 = 2.Therefore, ln(L) = 2, so L = e^2.Wait, hold on. Let me double-check that. If ln(L) = 2, then L = e^2. Hmm, but let me verify each step to make sure I didn't make a mistake.First, when I split ln((2 - x)/x) into ln(2 - x) - ln(x), that's correct. Then, dividing by ln(2 - x) gives 1 - [ln(x)/ln(2 - x)]. That seems right. Then, taking the limit of that as x approaches 1, and handling the 0/0 case with L'Hospital's Rule. The derivatives are correct: derivative of ln(x) is 1/x, derivative of ln(2 - x) is -1/(2 - x). Then, when we take the ratio, the negatives cancel? Wait, let's see:Original ratio after derivatives:(1/x) / (-1/(2 - x)) = (1/x) * (- (2 - x)/1) = - (2 - x)/x. Correct. Then substituting x=1 gives - (2 - 1)/1 = -1. So the limit is -1. Therefore, ln(L) = 1 - (-1) = 2. So L = e^2. That seems okay.But wait, let me check again the original expression. Let me plug in values approaching 1 from both sides to see if it makes sense.For example, take x approaching 1 from the left, say x = 0.9. Then (2 - 0.9)/0.9 = 1.1 / 0.9 ≈ 1.2222. ln(2 - 0.9) = ln(1.1) ≈ 0.09531. So the exponent is 1 / 0.09531 ≈ 10.49. So 1.2222^10.49 ≈ ?Let me compute that. 1.2222^10. Let's see: 1.2^10 is about 6.1917. Since 1.2222 is a bit higher, maybe around 8? Let's see:1.2222^2 ≈ 1.4938^4 ≈ (1.4938)^2 ≈ 2.2314^8 ≈ (2.2314)^2 ≈ 4.979Then 1.2222^10 ≈ 4.979 * 1.4938 ≈ 7.43. But the exponent is 10.49, so maybe around 8.5?But e^2 is approximately 7.389. So, it's in the ballpark. If x approaches 1 from the right, 2 - x would be less than 1, so (2 - x)/x would be less than 1, and ln(2 - x) would be negative. Wait, but if x approaches 1 from the right, then 2 - x approaches 1 from below, so 2 - x is approaching 1 from the left? Wait, no. If x approaches 1 from the right, x is slightly larger than 1, so 2 - x is slightly less than 1. For example, x = 1.1, then 2 - x = 0.9, so (2 - x)/x = 0.9/1.1 ≈ 0.818, and ln(2 - x) = ln(0.9) ≈ -0.10536. So exponent is 1 / (-0.10536) ≈ -9.49. Then, 0.818^-9.49 ≈ 1 / 0.818^9.49. Let's compute 0.818^10 ≈ (0.818^2)^5 ≈ (0.669)^5 ≈ 0.669*0.669=0.447, *0.669=0.3, *0.669≈0.2, *0.669≈0.134. So 1/0.134 ≈ 7.46. Again, close to e^2 ≈ 7.389. So the numerical estimates from both sides give values around 7.4, which is e squared. So that seems to confirm that the limit is indeed e squared.Therefore, I think my answer is correct. But let me check once more the steps for any possible errors.So, original limit: [(2 - x)/x]^{1/ln(2 - x)}. Taking natural log gives [ln((2 - x)/x)] / ln(2 - x). Split into [ln(2 - x) - ln(x)] / ln(2 - x) = 1 - [ln(x)/ln(2 - x)]. Then, the limit of [ln(x)/ln(2 - x)] as x approaches 1 is calculated using L'Hospital's Rule, derivative of numerator is 1/x, derivative of denominator is -1/(2 - x). Their ratio is -(2 - x)/x, which approaches -1 as x approaches 1. Therefore, ln(L) = 1 - (-1) = 2, so L = e^2. All steps seem correct. Okay, I think that's the right answer.Final Answerboxed{e^2}</think>To find the limit of the function:lim _{x rightarrow 1}left(frac{2-x}{x}right)^{frac{1}{ln (2-x)}}we start by recognizing that the expression is of the form (1^infty), an indeterminate form. We use the natural logarithm to simplify the evaluation. Let (L) be the limit, then:ln(L) = lim_{x to 1} left( frac{lnleft(frac{2-x}{x}right)}{ln(2-x)} right)Simplifying the logarithm in the numerator:lnleft(frac{2-x}{x}right) = ln(2-x) - ln(x)Thus, the expression becomes:ln(L) = lim_{x to 1} left( frac{ln(2-x) - ln(x)}{ln(2-x)} right) = lim_{x to 1} left( 1 - frac{ln(x)}{ln(2-x)} right)Next, we evaluate the limit:lim_{x to 1} frac{ln(x)}{ln(2-x)}This is a (0/0) indeterminate form, so we apply L'Hospital's Rule. Taking the derivatives of the numerator and denominator:frac{d}{dx} ln(x) = frac{1}{x} quad text{and} quad frac{d}{dx} ln(2-x) = frac{-1}{2-x}Applying these derivatives:lim_{x to 1} frac{frac{1}{x}}{frac{-1}{2-x}} = lim_{x to 1} left( -frac{2-x}{x} right) = -frac{2-1}{1} = -1Substituting back, we get:ln(L) = 1 - (-1) = 2Therefore, exponentiating both sides gives:L = e^2The final answer is:boxed{e^2}```,<think>Okay, let's see. I need to find the limit as x approaches 1 of [(2 - x)/x] raised to the power of 1 over ln(2 - x). Hmm, that looks a bit complicated, but maybe I can simplify it somehow. First, let me write down the expression again to make sure I have it right:limₓ→1 [(2 - x)/x]^{1/ln(2 - x)}Alright, so when x approaches 1, let's check what happens to the base and the exponent. The base is (2 - x)/x. If x approaches 1, then 2 - x approaches 2 - 1 = 1, and the denominator x approaches 1 as well. So the base becomes 1/1 = 1. Now the exponent is 1 over ln(2 - x). As x approaches 1, 2 - x approaches 1, so ln(2 - x) approaches ln(1) which is 0. Therefore, the exponent is 1 divided by something approaching 0. If the denominator approaches 0 from the positive side, it would be +infty, but if it approaches 0 from the negative side, it would be -infty. Wait, but when x approaches 1 from the left (x → 1⁻), 2 - x is slightly greater than 1, so ln(2 - x) is approaching 0 from the positive side. If x approaches 1 from the right (x → 1⁺), then 2 - x is slightly less than 1, so ln(2 - x) is negative, approaching 0 from the negative side. Therefore, the exponent 1/ln(2 - x) would approach +infty when approaching from the left and -infty when approaching from the right. But the base is approaching 1. So we have 1 raised to ±infty. That's an indeterminate form, right? Because 1^infty is indeterminate. So I need to use a technique to evaluate this limit. Maybe take the natural logarithm, solve for the limit of the logarithm, and then exponentiate the result. Let me set L = limₓ→1 [(2 - x)/x]^{1/ln(2 - x)}. Then, ln L = limₓ→1 [ln((2 - x)/x) / ln(2 - x)]. So that's the exponent times the logarithm of the base. So ln L = limₓ→1 [ln((2 - x)/x)] / ln(2 - x). Let me simplify the numerator and the denominator. First, ln((2 - x)/x) can be written as ln(2 - x) - ln x. So the expression becomes:[ln(2 - x) - ln x] / ln(2 - x) = 1 - [ln x / ln(2 - x)]So, ln L = limₓ→1 [1 - (ln x / ln(2 - x))]Now, let's evaluate this limit. Let me split it into two terms:limₓ→1 1 - limₓ→1 [ln x / ln(2 - x)]The first limit is just 1. The second limit is the limit of ln x divided by ln(2 - x) as x approaches 1. Let's compute that. As x approaches 1, ln x approaches ln 1 = 0, and ln(2 - x) approaches ln 1 = 0. So we have a 0/0 indeterminate form here. That suggests using L'Hospital's Rule. Applying L'Hospital's Rule to limₓ→1 [ln x / ln(2 - x)]:Take the derivative of the numerator and the denominator. The derivative of ln x is 1/x. The derivative of ln(2 - x) is (1/(2 - x)) * (-1) = -1/(2 - x).So by L'Hospital's Rule, the limit becomes:limₓ→1 [ (1/x) / (-1/(2 - x)) ] = limₓ→1 [ (1/x) * (- (2 - x)/1 ) ] = limₓ→1 [ - (2 - x)/x ]Now, plug in x = 1:- (2 - 1)/1 = -1/1 = -1So the limit of [ln x / ln(2 - x)] as x approaches 1 is -1. Therefore, going back to ln L:ln L = 1 - (-1) = 1 + 1 = 2Hence, L = e^2. Wait, hold on. Let me check this again. So the original limit is e^{2}? Let's verify step by step.Starting from ln L = limₓ→1 [1 - (ln x / ln(2 - x))]We found that limₓ→1 (ln x / ln(2 - x)) = -1, so 1 - (-1) = 2. Therefore, ln L = 2 => L = e². Hmm, but let me check if I applied L'Hospital's Rule correctly. When we had [ln x]' / [ln(2 - x)]' = (1/x) / (-1/(2 - x)) = - (2 - x)/x. Then plugging in x =1 gives -1. So that's correct. Therefore, the final answer should be e². Let me think again if there's another approach or if I made any mistake in the steps.Alternatively, maybe substitute t = 2 - x. When x approaches 1, t approaches 1 as well. Let me try that substitution. Let t = 2 - x. Then x = 2 - t. As x →1, t →1. Then the expression becomes:[(t)/(2 - t)]^{1/ln t}So the limit is as t approaches 1. So that's [t/(2 - t)]^{1/ln t}. Let's check the base: t/(2 - t) as t →1 is 1/(2 -1) = 1. The exponent is 1/ln t, which as t approaches 1, ln t approaches 0, so exponent approaches infinity. So same as before, 1^infty indeterminate form. Taking natural log again:ln L = lim_{t→1} [ln(t/(2 - t)) / ln t] = lim_{t→1} [ln t - ln(2 - t)] / ln t = lim_{t→1} 1 - [ln(2 - t)/ln t]Wait, this seems similar but different from before. Wait, in this substitution, maybe I made a different split. Let's see. Wait, ln(t/(2 - t)) = ln t - ln(2 - t). So numerator is ln t - ln(2 - t), denominator is ln t. Therefore, the expression is 1 - [ln(2 - t)/ln t]. So ln L = 1 - lim_{t→1} [ln(2 - t)/ln t]. Let's compute this limit.As t approaches 1, ln(2 - t) approaches ln(1) = 0, and ln t approaches 0. So again 0/0. Apply L'Hospital's Rule. Derivative of numerator: derivative of ln(2 - t) is (-1)/(2 - t). Derivative of denominator: derivative of ln t is 1/t. So the limit becomes lim_{t→1} [ (-1/(2 - t)) / (1/t) ] = lim_{t→1} [ -t / (2 - t) ]Plug in t =1: -1/(2 -1) = -1. Therefore, lim_{t→1} [ln(2 - t)/ln t] = -1. Hence, ln L = 1 - (-1) = 2 => L = e². Same result. So whether I substitute or not, the answer is e squared. Alternatively, maybe another substitution. Let me set h = x -1. Then as x →1, h→0. So x =1 + h. Let's rewrite the original expression in terms of h. Original expression: [(2 - (1 + h))/(1 + h)]^{1 / ln(2 - (1 + h))} = [(1 - h)/(1 + h)]^{1 / ln(1 - h)}. So the limit as h approaches 0 of [(1 - h)/(1 + h)]^{1 / ln(1 - h)}. Hmm, the base is (1 - h)/(1 + h). Let's approximate when h is small. (1 - h)/(1 + h) ≈ (1 - h)(1 - h) ≈ 1 - 2h (using first order approximation). Wait, or more accurately, (1 - h)/(1 + h) ≈ 1 - 2h when h is small. Alternatively, take natural log of the base:ln[(1 - h)/(1 + h)] = ln(1 - h) - ln(1 + h) ≈ (-h - h²/2) - (h - h²/2) = -2h. So the base is approximately e^{-2h} for small h. Therefore, the expression becomes approximately [e^{-2h}]^{1 / ln(1 - h)}. But ln(1 - h) ≈ -h - h²/2 ≈ -h. So 1 / ln(1 - h) ≈ -1/h. Therefore, the exponent is -2h * (-1/h) = 2. So the entire expression approaches e². This gives the same result, which is reassuring. Therefore, regardless of the substitution or method, the limit seems to be e squared. Another way: Let’s consider variable substitution y = 2 - x. As x approaches 1, y approaches 1. Then the expression becomes (y / (2 - y))^{1 / ln y}. Taking natural logarithm, as before, gives the same steps. Alternatively, perhaps rewrite the original expression in exponential form. Let me think. Let’s denote the original function as f(x) = [(2 - x)/x]^{1 / ln(2 - x)}. Then, f(x) = e^{ [ln((2 - x)/x)] / ln(2 - x) }, which is the same as e^{ [ln(2 - x) - ln x] / ln(2 - x) } = e^{1 - [ln x / ln(2 - x)]}, as we did before. So again, that exponent simplifies to 1 - [ln x / ln(2 - x)]. Then, evaluating the limit as x approaches 1, we get 1 - (-1) = 2, so e². Therefore, multiple approaches all lead to the conclusion that the limit is e squared. Let me check if there is any alternative perspective. Suppose I let u = 2 - x. Then x = 2 - u, and as x approaches 1, u approaches 1. So the expression becomes (u / (2 - u))^{1 / ln u}. If I let u approach 1, this is similar to the previous substitution. Alternatively, maybe use Taylor series expansions around x =1. Let’s set x =1 + h, where h approaches 0. Then 2 - x =1 - h. Let’s expand the numerator and denominator. The base (2 - x)/x = (1 - h)/(1 + h). As h approaches 0, we can expand ln((1 - h)/(1 + h)) as ln(1 - h) - ln(1 + h) ≈ (-h - h²/2 - ...) - (h - h²/2 + ...) = (-h - h²/2) - h + h²/2 = -2h. Therefore, the exponent in the logarithm is -2h. Then, the exponent of the entire expression is 1 / ln(2 - x) = 1 / ln(1 - h) ≈ 1 / (-h - h²/2) ≈ -1/h (for small h). Therefore, the exponent of e is (-2h) * (-1/h) = 2. So again, the limit is e². All these methods are converging to the same result. Therefore, I feel confident that the answer is e squared. Just to recap, the key steps were recognizing the indeterminate form 1^infty, taking the natural logarithm to convert the exponent into a product, simplifying the expression, applying L'Hospital's Rule where necessary, and confirming the result through substitutions and series expansions. Therefore, the final answer should be e squared, or e².Final Answerboxed{e^2}</think>To calculate the limit of the function:lim _{x rightarrow 1}left(frac{2-x}{x}right)^{frac{1}{ln (2-x)}}we start by recognizing that as ( x ) approaches 1, the base ( frac{2-x}{x} ) approaches 1, and the exponent ( frac{1}{ln(2-x)} ) approaches infinity, leading to an indeterminate form ( 1^infty ). To resolve this, we take the natural logarithm of the function and evaluate the limit.Let ( L ) be the limit we need to find. Taking the natural logarithm, we have:ln L = lim_{x rightarrow 1} frac{lnleft(frac{2-x}{x}right)}{ln(2-x)}Simplifying the logarithm in the numerator:lnleft(frac{2-x}{x}right) = ln(2-x) - ln(x)Thus, the expression becomes:ln L = lim_{x rightarrow 1} left( frac{ln(2-x) - ln(x)}{ln(2-x)} right) = lim_{x rightarrow 1} left( 1 - frac{ln(x)}{ln(2-x)} right)Next, we evaluate the limit of ( frac{ln(x)}{ln(2-x)} ) as ( x ) approaches 1. This is a ( frac{0}{0} ) indeterminate form, so we apply L'Hospital's Rule:lim_{x rightarrow 1} frac{ln(x)}{ln(2-x)} = lim_{x rightarrow 1} frac{frac{1}{x}}{frac{-1}{2-x}} = lim_{x rightarrow 1} left( -frac{2-x}{x} right) = -1Substituting this result back into the expression for ( ln L ):ln L = 1 - (-1) = 2Exponentiating both sides to solve for ( L ):L = e^2Thus, the limit is:boxed{e^2}]

🔑:## Solutionbegin{aligned}& lim _{x rightarrow 1}left(frac{2-x}{x}right)^{frac{1}{ln (2-x)}}=lim _{x rightarrow 1}left(e^{ln left(frac{2-x}{x}right)}right)^{frac{1}{ln (2-x)}}= & =lim _{x rightarrow 1} e^{frac{1}{ln (2-x)} cdot ln left(frac{2-x}{x}right)}=exp left{lim _{x rightarrow 1} frac{1}{ln (2-x)} cdot ln left(frac{2-x}{x}right)right}=end{aligned}Substitution:x=y+1 Rightarrow y=x-1x rightarrow 1 Rightarrow y rightarrow 0We get:=exp left{lim _{y rightarrow 0} frac{1}{ln (2-(y+1))} cdot ln left(frac{2-(y+1)}{y+1}right)right}==exp left{lim _{y rightarrow 0} frac{1}{ln (1-y)} cdot ln left(frac{1-y}{y+1}right)right}==exp left{lim _{y rightarrow 0} frac{1}{ln (1-y)} cdot ln left(1-frac{2 y}{y+1}right)right}=Using the substitution of equivalent infinitesimals:ln left(1-frac{2 y}{y+1}right) sim-frac{2 y}{y+1} text{ as } y rightarrow 0left(-frac{2 y}{y+1} rightarrow 0right)ln (1-y) sim-y text{ as } y rightarrow 0(-y rightarrow 0)We get:=exp left{lim _{y rightarrow 0} frac{1}{-y} cdot frac{-2 y}{y+1}right}=exp left{lim _{y rightarrow 0} frac{2}{y+1}right}==exp left{frac{2}{0+1}right}=exp {2}=e^{2}## Problem Kuznetsov Limits 19-23

🧠:[<think>Okay, so I need to figure out how many liters of oil are in a 20-liter canister where the mixing ratio of gasoline to oil is 33.33:1. Hmm, let me start by understanding what this ratio means. Ratios can sometimes be confusing, but I think in this case, it's saying that for every 33.33 parts of gasoline, there's 1 part of oil. So the total parts would be 33.33 + 1 = 34.33 parts. Wait, but maybe I should check if that's the right way to interpret it.Yes, typically when a ratio is given like A:B, it means A parts of one substance to B parts of another. So gasoline to oil being 33.33:1 means 33.33 parts gasoline per 1 part oil. Therefore, the total mixture is 33.33 + 1 = 34.33 parts. The whole canister is 20 liters, which equals those 34.33 parts. So to find the amount of oil, which is 1 part, I need to divide 20 liters by 34.33 to find the volume of one part. Then multiply that by 1 to get the oil.Let me write that down step by step:1. Total parts = gasoline parts + oil parts = 33.33 + 1 = 34.33 parts2. Total volume = 20 liters = 34.33 parts3. Therefore, 1 part = 20 liters / 34.334. Oil volume = 1 part = 20 / 34.33 ≈ ?Calculating that division: 20 divided by 34.33. Let me do that on paper. 34.33 goes into 20 how many times? Well, 34.33 times 0.5 is 17.165. That's less than 20. 34.33 times 0.6 is 20.598. Oh, so 0.6 times 34.33 is about 20.598, which is slightly more than 20. So maybe approximately 0.58 liters? Wait, no, wait, that can't be. Wait, no, the oil volume is 20 divided by 34.33. Let me use a calculator here.But since I don't have a calculator, maybe I can approximate. Let's see, 34.33 is approximately 34 and 1/3. So 34.33 is approximately 34.333... which is 103/3. Wait, 103 divided by 3 is 34.333... So 34.33 is roughly 103/3. Therefore, 20 divided by (103/3) is 20 * (3/103) = 60/103 ≈ 0.5825 liters. So approximately 0.5825 liters of oil. That's about 0.583 liters. Let me check that again.Alternatively, perhaps we can convert the ratio to a fraction. The ratio of oil to the total mixture is 1/(33.33 + 1) = 1/34.33. So oil volume is 20 * (1/34.33). Which is the same as 20 divided by 34.33. So yes, that gives the same result. So 20 divided by 34.33. Let me compute this more accurately.34.33 multiplied by 0.5 is 17.165. 34.33 multiplied by 0.6 is 20.598, as before. So 20 is between 0.5 and 0.6 times 34.33. The difference between 20 and 17.165 is 2.835. The difference between 20.598 and 17.165 is 3.433. So 2.835 divided by 3.433 is approximately 0.826. So 0.5 + 0.826*(0.1) ≈ 0.5 + 0.0826 = 0.5826. So that's about 0.5826, which is roughly 0.583 liters. So approximately 0.583 liters of oil.But let me check the exact calculation. Let's write 34.33 as 34.33. So 20 divided by 34.33. Let me compute this division step by step.34.33 × 0.5 = 17.16534.33 × 0.58 = 34.33 × 0.5 + 34.33 × 0.08 = 17.165 + 2.7464 = 19.911434.33 × 0.58 = 19.9114So 0.58 times 34.33 is 19.9114. The difference between 20 and 19.9114 is 0.0886.So now, how much more do we need? So 34.33 × x = 0.0886, where x is the decimal to add to 0.58.x = 0.0886 / 34.33 ≈ 0.00258.Therefore, total is approximately 0.58 + 0.00258 ≈ 0.58258 liters. So approximately 0.5826 liters. Rounding to three decimal places, that's 0.583 liters. If we want to convert this to a fraction, maybe?Alternatively, since the original ratio was given as 33.33:1, which is likely an approximation. 33.33 is approximately 1/3 of 100, so 33.33 is 100/3. So if the ratio is 100/3 : 1, then total parts are (100/3 + 1) = 103/3. Therefore, oil is 1/(103/3) = 3/103 of the total mixture. Therefore, oil volume is 20 * (3/103) = 60/103 ≈ 0.5825 liters. Which is consistent with the previous calculation.So 60/103 liters is the exact fraction, which is approximately 0.5825 liters. If we want to write that as a decimal, maybe round to two decimal places: 0.58 liters. But perhaps the problem expects an exact fractional answer? Let me check the problem again.The problem says "mixing ratio of gasoline to oil = 33.33:1". The use of 33.33 suggests that it's a decimal approximation, possibly of 100/3, since 100 divided by 3 is approximately 33.333... So perhaps the ratio is intended to be 100:3, but written as 33.33:1? Wait, 100:3 is equivalent to 33.333...:1. So perhaps the ratio is 100 parts gasoline to 3 parts oil, which simplifies to approximately 33.33:1. So if that's the case, maybe the original ratio was 100:3, but presented as 33.33:1 for simplicity. Let me check.If gasoline to oil is 100:3, then the ratio is 100/3 :1, which is 33.333...:1. So perhaps the problem is using a ratio of 100:3, approximated as 33.33:1. So if we take that, then total parts would be 100 + 3 = 103 parts, where 3 parts are oil. Therefore, oil is 3/103 of the total mixture. Therefore, 20 liters * (3/103) = 60/103 liters. Which is approximately 0.5825 liters. So in that case, the exact answer is 60/103 liters. But since the problem presented the ratio as 33.33:1, which is a decimal approximation, perhaps the answer is expecting a decimal approximation. Let me check the problem again.The problem states: "mixing ratio of gasoline to oil =33.33: 1". So the ratio is given as 33.33:1. Therefore, in this case, gasoline is 33.33 units, oil is 1 unit. So total units are 34.33, oil is 1/34.33. So 20*(1/34.33) is approximately 0.5825 liters. So depending on how precise the answer needs to be. The problem is from a math puzzle, so maybe an exact fractional answer is expected. Let's see. If we take 33.33 as 33 and 1/3, which is 100/3, then 33.33:1 is equivalent to 100/3 :1. Therefore, total parts is 100/3 +1 = 103/3. Therefore, oil is 1/(103/3) = 3/103. Therefore, 20 *3/103 = 60/103 ≈0.5825 liters. Alternatively, 33.33 is a decimal, so maybe 33.33 +1 =34.33, so oil is 1/34.33, which is approximately 0.02915, and 20 *0.02915≈0.583 liters. So depending on whether the ratio is exact or an approximation.But the problem is presented with 33.33, which is likely an exact value here. Wait, but in reality, mixing ratios for fuel are often given as ratios like 50:1 or 40:1, which are exact. So 33.33:1 seems like an unusual ratio. Maybe in some contexts, but perhaps here it's a simplified version. Alternatively, maybe the ratio is 33.33:1 by volume, so we can take it as given.Therefore, total parts = 33.33 +1 =34.33. Oil is 1/34.33 of 20 liters. So 20 /34.33.Wait, 34.33 times 0.58 is approximately 20, as we calculated. So the answer is approximately 0.583 liters, which is roughly 0.58 liters. But let me check the exact decimal calculation.If I compute 20 divided by 34.33:34.33 ) 20.00000034.33 goes into 200.000 how many times? Wait, let's adjust the decimal. Let me write 20.0000 divided by 34.33.Since 34.33 is larger than 20, we write it as 0.xxxxMultiply numerator and denominator by 100 to eliminate decimals:2000.00 divided by 3433.3433 goes into 2000.0 0 times. So 0.3433 goes into 20000 5 times (5*3433=17165), subtract 20000-17165=2835. Bring down a zero: 28350.3433 goes into 28350 8 times (8*3433=27464), subtract 28350-27464=886. Bring down a zero: 8860.3433 goes into 8860 2 times (2*3433=6866), subtract 8860-6866=1994. Bring down a zero: 19940.3433 goes into 19940 5 times (5*3433=17165), subtract 19940-17165=2775. Bring down a zero: 27750.3433 goes into 27750 8 times (8*3433=27464), subtract 27750-27464=286. Bring down a zero: 2860.3433 goes into 2860 0 times. So next digit is 0. Bring down a zero: 28600.3433 goes into 28600 8 times (8*3433=27464), subtract 28600-27464=1136. Bring down a zero: 11360.3433 goes into 11360 3 times (3*3433=10299), subtract 11360-10299=1061. Bring down a zero: 10610.3433 goes into 10610 3 times (3*3433=10299), subtract 10610-10299=311. Bring down a zero: 3110.3433 goes into 3110 0 times. So next digit is 0. Bring down a zero: 31100.3433 goes into 31100 9 times (9*3433=30897), subtract 31100-30897=203. So the pattern continues.So putting it all together, the division 2000 / 3433 is approximately 0.5825... So 0.5825 liters. So approximately 0.5825 liters. If we round to two decimal places, that's 0.58 liters. If we round to three decimal places, 0.583 liters. However, depending on the required precision, the answer may vary.But let me check if the problem is expecting an exact fraction. If the ratio was 33.33:1, which is 33.33 recurring, meaning 33.3333333..., which is 100/3. So 100/3:1 is the ratio. Then total parts is 100/3 +1 =103/3. Therefore, oil is 1 part out of 103/3 total parts, which is (1)/(103/3) =3/103. Therefore, oil is 3/103 of 20 liters. 3/103 *20=60/103 liters. Let's compute that as a decimal.103 divided into 60.000...103 goes into 600 5 times (5*103=515), subtract 600-515=85. Bring down 0: 850.103 goes into 850 8 times (8*103=824), subtract 850-824=26. Bring down 0:260.103 goes into 260 2 times (2*103=206), subtract 260-206=54. Bring down 0:540.103 goes into 540 5 times (5*103=515), subtract 540-515=25. Bring down 0:250.103 goes into 250 2 times (2*103=206), subtract 250-206=44. Bring down 0:440.103 goes into 440 4 times (4*103=412), subtract 440-412=28. Bring down 0:280.103 goes into 280 2 times (2*103=206), subtract 280-206=74. Bring down 0:740.103 goes into 740 7 times (7*103=721), subtract 740-721=19. Bring down 0:190.103 goes into 190 1 time (1*103=103), subtract 190-103=87. Bring down 0:870.103 goes into 870 8 times (8*103=824), subtract 870-824=46. Bring down 0:460.103 goes into 460 4 times (4*103=412), subtract 460-412=48. Bring down 0:480.103 goes into 480 4 times (4*103=412), subtract 480-412=68. Bring down 0:680.103 goes into 680 6 times (6*103=618), subtract 680-618=62. Bring down 0:620.103 goes into 620 6 times (6*103=618), subtract 620-618=2. Bring down 0:20.At this point, we see that the decimal repeats. So 60/103 ≈0.582524... which is approximately 0.5825 liters, as before.So if the problem expects an exact answer, 60/103 liters is the exact value, approximately 0.5825 liters. Depending on the context, this could be rounded to 0.58 liters or 0.583 liters. However, since it's a math puzzle, maybe the answer is expected in fractional form. Let me check units. The canister is 20 liters, so the answer should be in liters. The problem says "how many liters of oil are in the canister?"Given that, if the ratio is 33.33:1, which is 100/3:1, then the exact answer is 60/103 liters. But 60/103 is approximately 0.5825 liters, which is about 0.583 liters. Alternatively, if the problem uses 33.33 as a terminating decimal, which would be 33.33 exactly, then the total parts are 33.33 +1=34.33, and oil is 1/34.33 of 20 liters. Let's compute 1/34.33 exactly.1 divided by 34.33. Let's compute that:34.33 )1.000000...Multiply numerator and denominator by 100 to remove decimals:100 /3433 ≈0.02915...So 1/34.33≈0.02915. Therefore, 20*0.02915≈0.583 liters. So same as before. So regardless of the method, the approximate answer is 0.583 liters.But the problem might be expecting an exact fraction. If we take 33.33 as exact, then 33.33 is 3333/100. So the ratio is 3333/100 :1. Therefore, total parts is 3333/100 +1 =3333/100 +100/100=3433/100. Therefore, oil is 1 part, which is (1)/(3433/100)=100/3433. Therefore, oil volume is 20*(100/3433)=2000/3433 liters. Let's see if that simplifies. 2000 and 3433. Let's check if they have any common factors. 3433 divided by 2000, remainder 1433. Then 2000 divided by 1433, remainder 567. 1433 divided by 567, remainder 299. 567 divided by 299, remainder 268. 299 divided by 268, remainder 31. 268 divided by 31, remainder 20. 31 divided by 20, remainder 11. 20 divided by 11, remainder 9. 11 divided by 9, remainder 2. 9 divided by 2, remainder 1. 2 divided by 1, remainder 0. So the GCD is 1. Therefore, 2000/3433 is already in simplest terms. So 2000/3433≈0.5825 liters. Which is the same as before.But since the problem says 33.33:1, which is likely a rounded figure, the exact answer depends on whether 33.33 is exact or an approximation. Given that 33.33 is commonly used as an approximation for 1/3 of 100 (i.e., 33.333...), the problem might actually be 100:3 ratio, leading to 60/103 liters. However, if we take 33.33 as exact, then it's 2000/3433 liters. But 2000/3433 cannot be simplified further, as established. However, 2000/3433 is approximately 0.5825, which is the same as 60/103≈0.5825. Wait, is that a coincidence?Wait, 60/103 is approximately 0.5825, and 2000/3433 is also approximately 0.5825. Are these fractions equivalent?Let me check: 60/103 vs. 2000/3433.Cross-multiplying: 60*3433 = 205,9802000*103 = 206,000These are not equal, so the fractions are not equivalent. Therefore, depending on whether the ratio is 33.33 (exact) or 33.33... (recurring), the exact fraction differs. However, since the problem states 33.33, not 33.33..., it's likely that they expect the answer based on 33.33 as a precise decimal, leading to 2000/3433 liters. But 2000/3433 simplifies to approximately 0.5825 liters. However, 60/103 is approximately 0.5825 as well. Given the closeness, perhaps the problem expects the answer as 60/103 liters. Alternatively, maybe rounded to two decimal places, 0.58 liters.But given that it's a math puzzle, and the numbers given (20 liters, ratio 33.33:1), I think the answer is 0.58 liters if rounded to two decimal places, or 0.583 liters to three. Alternatively, perhaps the answer is expected to be in milliliters, but the problem asks for liters. Let me check once again.Wait, 0.5825 liters is 582.5 milliliters. But the problem asks for liters. So unless specified, the answer is in liters. So depending on the required precision, but since the original ratio was given to two decimal places (33.33), maybe the answer should also be to two decimal places: 0.58 liters.Alternatively, if considering that 33.33 is a repeating decimal, then the exact answer is 60/103 liters, which is approximately 0.5825 liters, and the problem might expect this fractional form. But I need to check how the problem is presented. The problem is titled "Math Puzzle 1 / 77", and the ratio is given as 33.33:1. If this is a competition-style problem, they might expect an exact answer. If 33.33 is an exact value, then the answer is 2000/3433 liters. But this fraction is not simplifying, so maybe they expect a decimal. Alternatively, perhaps there's a different approach.Wait, maybe the problem is simpler. If the ratio is gasoline to oil 33.33:1, which is roughly 33.33 parts gasoline to 1 part oil. Then for every liter of oil, there's 33.33 liters of gasoline. So total mixture would be 33.33 +1 =34.33 liters. Therefore, in 34.33 liters of mixture, there's 1 liter of oil. Therefore, in 20 liters, the amount of oil is (1 /34.33)*20≈0.5825 liters. So as before. Therefore, the answer is approximately 0.58 liters.Alternatively, perhaps the problem expects an answer of 0.6 liters, but that seems too rough. Alternatively, maybe they consider 33.33 as 1/3, so 33.33:1 is equivalent to 100:3. Then total parts 103, oil is 3/103, so 20*3/103≈0.5825 liters. So whether you take 33.33 as exact decimal or as 100/3, the approximate answer is the same. Therefore, maybe 0.583 liters. But perhaps the problem expects an exact fraction, so 60/103 or 2000/3433. However, 60/103 is simpler, and is close to the decimal. Alternatively, the problem might expect it written as a fraction. Let me check if 60/103 can be simplified. 60 and 103: factors of 60 are 2, 3, 5; 103 is a prime number (since it's not divisible by 2, 3, 5: 103/2=51.5, 103/3≈34.33, 103/5=20.6). So 103 is a prime, so 60/103 is in simplest terms.Alternatively, 20 liters divided by (33.33 +1) =34.33, so 20 /34.33. If we write 34.33 as 34 +0.33, then 34.33=34 +1/3. So 34 +1/3=103/3. Therefore, 20 divided by (103/3)=20*(3/103)=60/103≈0.5825. Therefore, exact answer is 60/103 liters. Since 0.5825 is approximately 0.583, but as a fraction, 60/103 is the exact value. Therefore, maybe the problem expects the answer as 60/103 liters, which is approximately 0.583 liters. Since math puzzles often prefer exact fractions unless specified otherwise, I think 60/103 liters is the correct exact answer. Therefore, boxed, it would be boxed{dfrac{60}{103}} liters. Alternatively, if they want a decimal approximation, boxed{0.58} or boxed{0.583}. But given the problem's presentation, and the fact that 33.33 is likely an approximation for 100/3, the exact answer is 60/103.Therefore, after considering all possibilities, the answer is boxed{dfrac{60}{103}} liters.Final AnswerThe amount of oil in the canister is boxed{dfrac{60}{103}} liters.</think>To determine the amount of oil in a 20-liter canister with a gasoline to oil mixing ratio of 33.33:1, we start by understanding the ratio. This ratio means there are 33.33 parts of gasoline for every 1 part of oil. 1. Calculate the total parts of the mixture: [ text{Total parts} = 33.33 + 1 = 34.33 ]2. Determine the volume of one part: [ text{Volume of one part} = frac{20 text{ liters}}{34.33} ]3. Calculate the volume of oil (which is 1 part): [ text{Oil volume} = 1 times left( frac{20}{34.33} right) ]Recognizing that 33.33 is an approximation of (frac{100}{3}), we convert the ratio to a fraction: [ text{Gasoline to oil ratio} = frac{100}{3} : 1 ] [ text{Total parts} = frac{100}{3} + 1 = frac{103}{3} ] [ text{Oil volume} = frac{1}{frac{103}{3}} times 20 = frac{3}{103} times 20 = frac{60}{103} ]Thus, the exact amount of oil in the canister is:[boxed{dfrac{60}{103}}] liters.,<think>Okay, so I need to figure out how many liters of oil are in a 20-liter canister where the mixing ratio of gasoline to oil is 33.33:1. Hmm, ratios can sometimes be tricky, but let me break it down step by step.First, the ratio given is 33.33 parts gasoline to 1 part oil. That means for every 33.33 liters of gasoline, there's 1 liter of oil. But the total mixture in the canister is 20 liters. So, the total parts of the mixture would be the sum of the gasoline parts and the oil parts.Let me write that down. If gasoline is 33.33 parts and oil is 1 part, then total parts = 33.33 + 1 = 34.33 parts. Wait, but 33.33 is a repeating decimal, right? 33.33 is probably 33 and 1/3, which is 100/3. Let me check that: 100 divided by 3 is approximately 33.333... So maybe the ratio is actually 100:3, which would be exact. Because 100/3 is 33.333..., which is the same as 33.33 when rounded to two decimal places. Maybe the original ratio was 100:3, but written as 33.33:1 for simplicity. Hmm, this might be important because using exact fractions might be better for precise calculation.So, if the ratio is 100:3, that would mean gasoline to oil is 100:3. Wait, but the problem states gasoline to oil is 33.33:1. Let me confirm. If 100:3 is the ratio, then dividing both by 3 gives approximately 33.33:1. So, either way, the ratio is equivalent. So, 100 parts gasoline to 3 parts oil is the same as 33.33 parts gasoline to 1 part oil. So, either way, the total parts would be 100 + 3 = 103 parts, or 33.33 + 1 = 34.33 parts. But 100:3 is 103 total parts, which when divided by 3 gives approximately 34.33 parts. So, both approaches should lead to the same answer.But maybe the problem wants me to use the given ratio as 33.33:1. Let's stick with that first. So, total parts are 33.33 + 1 = 34.33 parts. The total mixture is 20 liters. Therefore, each part is equal to 20 liters divided by 34.33 parts. Let me compute that.First, 20 divided by 34.33. Let me convert 34.33 into a fraction. Wait, 34.33 is approximately 34 and 1/3, because 0.33 is roughly 1/3. Let me check: 1/3 is approximately 0.3333. So 34.33 is close to 34 and 1/3, which is 103/3. Because 34*3=102, 102 +1=103, so 103/3 is 34.333... So, 34.33 is approximately 103/3. So, 20 divided by (103/3) is 20 * (3/103) = 60/103 liters per part. Therefore, since oil is 1 part, the amount of oil is 60/103 liters. Let me compute that as a decimal. 60 divided by 103. Let me do that division.103 goes into 60 zero times. Add a decimal point. 103 into 600 is 5 times (5*103=515). 600-515=85. Bring down a zero: 850. 103*8=824. 850-824=26. Bring down a zero: 260. 103*2=206. 260-206=54. Bring down a zero: 540. 103*5=515. 540-515=25. Bring down a zero: 250. 103*2=206. 250-206=44. Bring down a zero: 440. 103*4=412. 440-412=28. Bring down a zero: 280. 103*2=206. 280-206=74. Bring down a zero: 740. 103*7=721. 740-721=19. So, so far, I have 0.5825242718... So approximately 0.5825 liters. So, approximately 0.5825 liters of oil in the canister. But this is using the approximate 34.33 parts.Alternatively, if the ratio is 100:3, then total parts are 103. Then each part is 20/103 liters. So oil is 3 parts? Wait, no. Wait, if gasoline is 100 parts and oil is 3 parts, total parts 103, then oil is 3 parts. Wait, but in the problem statement, the ratio is gasoline to oil = 33.33:1, which is equivalent to 100:3. So, gasoline is 100 parts, oil is 3 parts, total 103. Therefore, oil is 3 parts. Therefore, oil = (3/103)*20 = 60/103 liters, which is the same as before. So, that's 60/103 ≈ 0.5825 liters.But wait, the problem says the mixing ratio of gasoline to oil is 33.33:1, which is the same as 100:3. So, maybe they expect the answer as a fraction, like 60/103, which is approximately 0.5825 liters, or maybe they want it rounded to a certain decimal place. But the problem doesn't specify, so perhaps the exact fraction is better. However, 60/103 cannot be simplified further, so that's the exact value. Alternatively, if we use 33.33 as exactly 33.33, then total parts would be 33.33 + 1 = 34.33, and oil is 1 part. Therefore, oil = (1 / 34.33) * 20. Let's compute that exactly.20 divided by 34.33. Let's convert 34.33 to a decimal. 34.33 is 34 + 0.33. So, 20 / 34.33 = 20 / (34 + 0.33) = 20 / 34.33. Let's compute this division.34.33 goes into 20 zero times. 34.33 goes into 200 approximately 5 times (5*34.33=171.65). 200 - 171.65 = 28.35. Bring down a zero: 283.5. 34.33 goes into 283.5 approximately 8 times (8*34.33=274.64). 283.5 - 274.64=8.86. Bring down a zero: 88.6. 34.33 goes into 88.6 approximately 2 times (2*34.33=68.66). 88.6 - 68.66=19.94. Bring down a zero: 199.4. 34.33 goes into 199.4 approximately 5 times (5*34.33=171.65). 199.4 -171.65=27.75. Bring down a zero: 277.5. 34.33 goes into 277.5 approximately 8 times (8*34.33=274.64). 277.5 -274.64=2.86. So, putting it all together: 0.58252... So approximately 0.5825 liters, which is the same as before.So whether we use 33.33:1 as exact decimal or consider that 33.33 is 100/3, leading to 100:3 ratio, we end up with the same decimal approximation of approximately 0.5825 liters. However, if the problem is expecting an exact answer, 60/103 liters is exact, but 60/103 is approximately 0.5825 liters. Wait, but 60/103 is actually approximately 0.58252427, which rounds to 0.5825 or 0.583 depending on the required precision.But let me check the original problem again. It says "mixing ratio of gasoline to oil =33.33:1". The number 33.33 is likely a rounded figure, as 33.33 is a common decimal approximation for 1/3. So, if the original ratio was 100:3, which is exact, then gasoline is 100 parts, oil is 3 parts, total 103 parts. Therefore, oil is 3/103 of the total mixture. Therefore, oil = 3/103 * 20 liters = 60/103 liters ≈ 0.5825 liters. Alternatively, if the ratio is taken as exactly 33.33:1, then total parts 34.33, oil is 1/34.33, which gives the same decimal approximation. However, 60/103 is an exact fraction, so perhaps that's the better answer.But let me confirm with another approach. Suppose the ratio is Gasoline:Oil = 33.33:1, so for every 33.33 liters of gasoline, there is 1 liter of oil. Therefore, the total volume would be 33.33 + 1 = 34.33 liters for each "unit" of the mixture. Therefore, in 34.33 liters of mixture, there is 1 liter of oil. So, in 20 liters of mixture, the amount of oil would be (1 / 34.33) * 20 liters. Which is the same as 20 / 34.33 ≈ 0.5825 liters. So, that's consistent.Alternatively, using fractions, if 33.33 is 100/3, then the ratio is (100/3):1, which can be written as 100/3 : 1. To make this a whole number ratio, multiply both parts by 3: 100:3. Therefore, Gasoline:Oil = 100:3. Therefore, in 103 parts of mixture, 3 parts are oil. Therefore, oil is 3/103 of the total. Therefore, oil = 3/103 * 20 = 60/103 liters. 60 divided by 103 is approximately 0.5825 liters.Therefore, both methods give the same result. So, the answer is 60/103 liters, which is approximately 0.5825 liters. Depending on what the problem expects, either could be acceptable. Since the problem is presented as a math puzzle, perhaps an exact fractional answer is expected, especially since 60/103 is the precise result when considering the ratio as 100:3. However, 60/103 is approximately 0.5825, which is roughly 0.583 liters. But maybe we need to check if 60/103 simplifies, but 60 and 103 have no common factors (103 is a prime number, right? 103 divided by 2, 3, 5, 7, 11—none of these divide 103. 11*9=99, 11*10=110, so no. So 103 is prime. Therefore, 60/103 is in simplest terms.Alternatively, if we use the 33.33:1 ratio as exact, then the answer is 20 / 34.33 liters of oil, which is approximately 0.5825 liters. But this is a decimal approximation. Since the problem might be expecting an exact answer, 60/103 liters is exact. But maybe the problem is presented in a way that expects a decimal answer. Let me check the problem statement again.The problem says: "In a 20-liter canister, there is a fuel mixture with a mixing ratio of gasoline to oil =33.33:1. How many liters of oil are in the canister?" So, it's given as 33.33:1. Since 33.33 is a decimal, maybe they expect a decimal answer. However, 33.33 is a repeating decimal, which is 33.333333... So, perhaps the problem is designed such that the answer is 0.6 liters? Wait, 33.33:1 is 100/3 :1, so 100/3 parts gasoline to 1 part oil. Total parts 100/3 +1 = 103/3. Therefore, oil is 1 / (103/3) of the total. So, oil = (3/103)*20 = 60/103 ≈0.5825 liters. But if I use 33.333... as exact, then it's 60/103. However, perhaps the problem is expecting an approximate answer? Wait, 60 divided by 103 is approximately 0.5825, which is about 0.58 liters. Alternatively, if the problem uses 33.33 as exact, then 20 / (33.33 +1) = 20 /34.33 ≈0.5825 liters.But maybe the problem is designed to use a simpler ratio. Let me think. Wait, 33.33 is approximately 1/3. So, if the ratio is 1/3:1, but that would be gasoline to oil as 1/3:1, which is equivalent to 1:3. But that's not the case here. Wait, no. Wait, gasoline to oil is 33.33:1, which is a higher ratio. Gasoline is 33.33 times more than oil. So, in other words, for every 1 liter of oil, there's 33.33 liters of gasoline. Therefore, the total mixture is 33.33 +1 =34.33 liters. So, oil is 1/34.33 of the total. Therefore, 20 liters total mixture would have 20 * (1/34.33) liters of oil. Which is approximately 0.5825 liters. So, if I round to the nearest hundredth, that would be 0.58 liters. But maybe the problem expects a fractional answer. Wait, 0.5825 is close to 7/12, which is approximately 0.5833. But 7/12 is 0.5833, which is very close. Alternatively, 60/103 is approximately 0.5825. But 60/103 is the exact answer if we consider the ratio as 100:3.Alternatively, maybe the problem is expecting a simpler calculation, like 20 liters divided by (33.33 +1) =34.33, but 20/34.33 is approximately 0.5825. But since 33.33 is a repeating decimal, 33.333..., then 34.333... is 103/3. So, 20 divided by (103/3) is 20 * 3/103 =60/103≈0.5825.Alternatively, maybe the problem is from a source that uses a standard ratio. For example, in two-stroke engines, a common fuel-oil ratio is 50:1 or 40:1, but 33.33:1 is also a possible ratio. However, the answer here is specific to the given ratio.Given that, I think the exact answer is 60/103 liters, which is approximately 0.5825 liters. Depending on the required precision, this could be rounded to 0.58 liters, 0.583 liters, or kept as a fraction. However, in mathematical problems, unless specified, fractions are preferred. However, since the question asks for liters and doesn't specify the format, both could be acceptable. But given that 60/103 is an exact value, I think that's the better answer.But wait, let me check if 33.33:1 is indeed 100:3. 100 divided by 3 is approximately 33.333..., so yes. Therefore, 33.33:1 is equivalent to 100:3. Therefore, gasoline is 100 parts, oil is 3 parts. Total parts 103. Therefore, oil is 3/103 of the total mixture. Therefore, in 20 liters, oil is (3/103)*20 =60/103 liters. 60 divided by 103. Since 103 is a prime number, this fraction cannot be simplified further. Therefore, the exact answer is 60/103 liters, approximately 0.5825 liters.Alternatively, if I use exact decimals, 33.33 +1 =34.33. 20 divided by34.33. Let me compute this more precisely. 34.33 *0.5 =17.165. 34.33*0.58=34.33*0.5 +34.33*0.08=17.165 +2.7464=19.9114. So, 34.33*0.58≈19.9114. Therefore, 20 liters would be 0.58 + (20-19.9114)/34.33≈0.58 +0.0886/34.33≈0.58 +0.00258≈0.58258 liters. So, approximately 0.5826 liters, which matches our previous calculation. Therefore, approximately 0.5826 liters, which rounds to 0.583 liters if we go to three decimal places.But again, the problem doesn't specify how to present the answer. If it's a competition or exam problem, sometimes they expect fractions, sometimes decimals. Given that the ratio was given as 33.33:1, which is a decimal approximation, maybe they expect a decimal answer. However, since 33.33 is a repeating decimal, which is 33.overline{3}, the exact answer would be 60/103. But 60/103 is approximately 0.5825, and maybe the problem expects rounding to two decimal places, which would be 0.58 liters. But 0.5825 is closer to 0.58 than 0.58. Hmm, actually, 0.5825 is 0.583 when rounded to three decimal places, and 0.58 when rounded to two. But without specific instructions, it's hard to tell.Alternatively, maybe the problem expects us to recognize that 33.33:1 is 100:3 and give the answer as 60/103. Given that it's a math puzzle, perhaps the exact fractional form is expected. So, writing the answer as boxed{dfrac{60}{103}} liters. Alternatively, if they want a decimal, approximately 0.58 liters. But given that 60/103 is exact, and math puzzles often prefer exact answers, I think 60/103 is the way to go.But to confirm, let me do a quick check. If I have 60/103 liters of oil, then gasoline is 20 -60/103 = (2060 -60)/103 =2000/103 ≈19.4175 liters. Then the ratio of gasoline to oil is (2000/103)/(60/103)=2000/60=100/3≈33.333..., which matches the given ratio of 33.33:1. Therefore, the exact answer is indeed 60/103 liters. So, that must be the correct answer.Final AnswerThe canister contains boxed{dfrac{60}{103}} liters of oil.</think>Given a 20-liter canister with a fuel mixture having a gasoline to oil ratio of 33.33:1, we need to determine the amount of oil in the canister.1. Understanding the Ratio: - The ratio of gasoline to oil is given as 33.33:1. This can be interpreted as 100 parts gasoline to 3 parts oil (since 33.33 is approximately ( frac{100}{3} )). - This means for every 100 parts of gasoline, there are 3 parts of oil, leading to a total of ( 100 + 3 = 103 ) parts.2. Calculating the Parts: - The total mixture in the canister is 20 liters. - Each part of the mixture is ( frac{20}{103} ) liters.3. Determining the Amount of Oil: - Since oil constitutes 3 parts of the total 103 parts, the amount of oil is ( 3 times frac{20}{103} ).4. Simplifying the Calculation: - The calculation simplifies to ( frac{60}{103} ) liters.5. Verification: - The ratio of gasoline to oil is verified by checking the exactness of the fraction. The ratio ( frac{100}{3}:1 ) confirms that the exact amount of oil is indeed ( frac{60}{103} ) liters.Thus, the canister contains boxed{dfrac{60}{103}} liters of oil.]

🔑:For a fuel mixture ratio of 33.33:1, 1 liter of oil is added to 33.33 liters of gasoline - this results in 34.33 liters of mixture. If 34.33 liters of fuel contain 1 liter of oil, then in a 20-liter canister,frac{1}{34.33} cdot 20 mathrm{l}=0.582 mathrm{l} text { of oil }

🧠:[<think>Alright, let's see. I need to figure out which of the options A to E is closest to the value of this fraction: (17 × 0.3 × 20.16) divided by 999. Hmm, okay. Let me break this down step by step. First, maybe I should calculate the numerator first before dealing with the denominator. The numerator is 17 multiplied by 0.3 multiplied by 20.16. Let me write that out: 17 * 0.3 * 20.16. Let me compute each part one by one. Starting with 17 times 0.3. Hmm, 17 times 0.3 is the same as 17 multiplied by 3 tenths. So 17 * 0.3. Let me compute that. 17 times 0.3. Well, 10 times 0.3 is 3, 7 times 0.3 is 2.1, so 3 + 2.1 is 5.1. So 17 * 0.3 is 5.1. Now, the next step is multiplying that result by 20.16. So 5.1 * 20.16. Hmm, okay. Let me think. Maybe break down 20.16 into 20 + 0.16. So 5.1 times 20 plus 5.1 times 0.16. Let's compute each part. First part: 5.1 * 20. Well, 5 * 20 is 100, and 0.1 * 20 is 2, so total is 102. Then, 5.1 * 0.16. Let's calculate that. 5 * 0.16 is 0.8, and 0.1 * 0.16 is 0.016. So adding those gives 0.8 + 0.016 = 0.816. So then, adding the two parts together: 102 + 0.816 = 102.816. So the numerator is 102.816. Then the denominator is 999. So the entire expression is 102.816 divided by 999. Now, I need to compute that. Let me see. 102.816 ÷ 999. Hmm. Alternatively, maybe approximate the numbers to make the division easier? Let's see. 102.816 divided by 999. Hmm. Well, 999 is very close to 1000. Maybe I can approximate 999 as 1000. Then, 102.816 divided by 1000 is 0.102816. So approximately 0.1028. But since 999 is slightly less than 1000, the actual result would be slightly higher. Because dividing by a smaller number gives a larger result. For example, 100 / 1000 is 0.1, but 100 / 999 is roughly 0.1001. So the difference here would be minimal. Therefore, approximating 999 as 1000 gives us roughly 0.1028, and the actual value is a bit higher. So maybe around 0.103. Let me check that with actual division. Alternatively, let's compute 102.816 ÷ 999 exactly. Let's do the division step by step. First, 999 goes into 102.816 how many times? Let's write it as 102.816 ÷ 999. Alternatively, we can write both numerator and denominator multiplied by 1000 to eliminate decimals. So that would be 102816 ÷ 999000. But that seems more complicated. Maybe we can do the division directly. Let me convert the division into a fraction: 102.816 / 999. To make this easier, perhaps multiply numerator and denominator by 1000 to eliminate the decimal in the numerator. So that gives (102.816 * 1000) / (999 * 1000) = 102816 / 999000. Then, simplifying this fraction. Let's see if 102816 and 999000 have a common divisor. First, let's check if both numbers are divisible by 3. 1+0+2+8+1+6 = 18, which is divisible by 3, so 102816 is divisible by 3. 9+9+9+0+0+0 = 27, which is divisible by 3, so 999000 is also divisible by 3. Let's divide numerator and denominator by 3. 102816 ÷ 3 = let's compute that. 3 into 10 is 3, remainder 1; 3 into 12 is 4, remainder 0; 3 into 2 is 0, remainder 2; 3 into 8 is 2, remainder 2; 3 into 1 is 0, remainder 1; 3 into 6 is 2. So that would be 34272. Wait, maybe my mental division is off. Let me do 102816 ÷ 3: 3 into 1 is 0, 3 into 10 is 3, remainder 1. 3 into 12 is 4, remainder 0. 3 into 2 is 0, remainder 2. 3 into 8 is 2, remainder 2. 3 into 1 is 0, remainder 1. 3 into 6 is 2. So the result is 34,272. Similarly, denominator: 999000 ÷ 3 = 333,000. So now the fraction is 34,272 / 333,000. Let's check if we can simplify further. Both numbers divisible by 3 again? 3+4+2+7+2 = 18, divisible by 3. 3+3+3+0+0+0=9, divisible by 3. Divide numerator and denominator by 3 again. 34,272 ÷ 3 = 11,424. 333,000 ÷ 3 = 111,000. So now it's 11,424 / 111,000. Check again for divisibility by 3: 1+1+4+2+4=12, divisible by 3. 1+1+1+0+0+0=3, divisible by 3. Divide by 3 again: 11,424 ÷ 3 = 3,808. 111,000 ÷ 3 = 37,000. So now we have 3,808 / 37,000. Check if they can be simplified further. Let's see if 3,808 and 37,000 share a common divisor. Let's check 4: 3,808 ÷ 4 = 952, 37,000 ÷ 4 = 9,250. So yes, divide by 4: 952 / 9,250. Check if 952 and 9,250 have a common divisor. 952 ÷ 2 = 476, 9,250 ÷ 2 = 4,625. So divide by 2: 476 / 4,625. Check again, 476 and 4,625. Let's see: 476 is 4*119, 4,625 ÷ 119. Let me check if 119 divides 4,625. 119*39=4,641, which is more than 4,625. So probably no. So 476 and 4,625 have no common divisors? Let's check GCD(476,4625). Find GCD of 476 and 4625. Using Euclidean algorithm: GCD(4625,476). 4625 ÷ 476 = 9 times (9*476=4284), remainder 4625-4284=341. Then GCD(476,341). 476 ÷ 341 = 1 time, remainder 135. GCD(341,135). 341 ÷ 135 = 2 times, remainder 71. GCD(135,71). 135 ÷ 71 = 1 time, remainder 64. GCD(71,64). 71 ÷ 64 =1, remainder 7. GCD(64,7). 64 ÷7=9, remainder 1. GCD(7,1). Which is 1. So GCD is 1. Therefore, the fraction reduces to 476/4625, which cannot be simplified further. Therefore, the simplified fraction is 476/4625. Let's convert that to decimal. 476 divided by 4625. Let me compute this division. Well, 4625 goes into 476.0 how many times? Since 4625 is larger than 476, it goes 0 times. So we write 0. and proceed. 4625 goes into 4760 once (since 4625*1=4625). Subtract 4625 from 4760: 135. Bring down a 0: 1350. 4625 goes into 1350 0 times. Bring down another 0: 13500. 4625 goes into 13500 two times (2*4625=9250). Subtract from 13500: 4250. Bring down a 0: 42500. 4625 goes into 42500 nine times (9*4625=41625). Subtract: 42500 - 41625 = 875. Bring down a 0: 8750. 4625 goes into 8750 once (1*4625=4625). Subtract: 8750 - 4625 = 4125. Bring down a 0: 41250. 4625 goes into 41250 eight times (8*4625=37000). Subtract: 41250 - 37000 = 4250. Bring down a 0: 42500. Wait, we already had 42500 before. So the decimal repeats. So putting this all together: 0.1027... approximately. Let's see the decimal expansion step by step:0.1 (after first division)0.10 (second digit)0.102 (after third division)0.1029 (wait, let me check again). Wait, maybe my previous steps were off. Let's recount:First step: 476 ÷ 4625. Since 4625 > 476, we start with 0. Multiply numerator by 10: 4760 ÷ 4625 = 1, remainder 135. So first decimal digit is 1, remainder 135. Next, bring down a 0: 1350. 1350 ÷ 4625 = 0. So next digit is 0, remainder 1350. Bring down another 0: 13500. 13500 ÷ 4625 = 2 (since 2*4625=9250). Subtract: 13500 - 9250 = 4250. So next digit is 2, remainder 4250. Bring down a 0: 42500. 42500 ÷ 4625. Let's compute 4625*9=41625. 42500 - 41625 = 875. So digit is 9, remainder 875. Bring down a 0: 8750. 8750 ÷ 4625 = 1 (1*4625=4625). Subtract: 8750 - 4625 = 4125. Digit is 1, remainder 4125. Bring down a 0: 41250. 41250 ÷ 4625. Let's see, 4625*8=37000. 41250 - 37000 = 4250. Digit is 8, remainder 4250. Bring down a 0: 42500. Wait, we saw 42500 earlier. So the cycle repeats: 9,1,8,... So the decimal is 0.102918... and then repeats. So approximately, 0.102918... So approximately 0.103. Therefore, the value of the original expression is approximately 0.103. Now, looking back at the options: A is 0.01, B is 0.1, C is 1, D is 10, E is 100. So 0.103 is closest to 0.1 (option B) since the difference between 0.103 and 0.1 is 0.003, whereas the difference between 0.103 and 0.01 is 0.093, which is much larger. Similarly, comparing to 1, the difference is 0.897, which is way bigger. So the closest is B, 0.1. But wait, let me double-check my approximations. Because maybe my initial estimation was too rough. Let me see. Alternatively, maybe there's a smarter way to approximate the original calculation without going through all the decimal division. Let's see. Original problem: (17 * 0.3 * 20.16) / 999. Let's approximate each number to a simpler one. 17 is approximately 17. 0.3 is 0.3. 20.16 is approximately 20. So numerator approx: 17 * 0.3 * 20 = 17 * 0.3 = 5.1; 5.1 * 20 = 102. So numerator approx 102. Denominator 999 approx 1000. So 102 / 1000 = 0.102. So approximately 0.102, which is about 0.1. But in reality, the numerator was 102.816, which is slightly higher, and denominator 999, which is slightly less than 1000. So 102.816 / 999 is a bit higher than 0.102816. Since denominator is 999 instead of 1000, the value is (102.816 / 999) = 102.816 / (1000 - 1) ≈ (102.816 / 1000) * (1 + 1/1000) ≈ 0.102816 * 1.001 ≈ 0.1029. So approximately 0.1029. Therefore, 0.1029 is approximately 0.103, which is indeed closest to 0.1 (option B). Alternatively, let's consider the actual calculation again. If we compute 17 * 0.3 * 20.16 / 999. Wait, 17 * 0.3 is 5.1. Then 5.1 * 20.16. Let me compute that more accurately. 5.1 * 20.16. Let's compute 5 * 20.16 = 100.8, and 0.1 * 20.16 = 2.016. Adding them together: 100.8 + 2.016 = 102.816. So the numerator is indeed 102.816. Divided by 999. So 102.816 / 999. Let me try to compute this division exactly. Let's set it up as 102.816 ÷ 999. First, 999 goes into 1028 (from 102.816) once, since 999*1=999. Subtract 999 from 1028: 29. Bring down the 1: 291. 999 goes into 291 0 times. Bring down the next digit 6: 2916. Wait, maybe I should adjust the decimal. Let's write 102.816 as 102816 divided by 1000. So 102816 / 1000 ÷ 999 = 102816 / (1000*999) = 102816 / 999000. Which simplifies as before. But maybe computing it as a decimal. Alternatively, use a calculator approach. Let me think. Since 999 is 1000 - 1, we can write 102.816 / 999 = 102.816 / (1000 - 1) = [102.816 / 1000] / (1 - 0.001) ≈ (0.102816) * (1 + 0.001 + 0.001² + ...) using the expansion 1/(1 - x) ≈ 1 + x for small x. So since 0.001 is small, this approximation holds. Therefore, approximately 0.102816 * 1.001 ≈ 0.102816 + 0.000102816 ≈ 0.102918816. So approximately 0.1029, which matches our previous calculation. So 0.1029 is the approximate value. Comparing this to the options: A is 0.01 (0.01), B is 0.1 (0.1), C is 1 (1), D is 10 (10), E is 100 (100). The distances from 0.1029 to each option:- Distance to A (0.01): |0.1029 - 0.01| = 0.0929- Distance to B (0.1): |0.1029 - 0.1| = 0.0029- Distance to C (1): |1 - 0.1029| = 0.8971- Distance to D (10): 9.8971- Distance to E (100): 99.8971Clearly, the closest is B, 0.1, with a difference of only 0.0029. But let me just check once more if there's any error in calculation. If we compute 17 * 0.3 is 5.1, correct. 5.1 * 20.16: let's verify. 20.16 * 5 = 100.8. 20.16 * 0.1 = 2.016. So 100.8 + 2.016 = 102.816, yes. Divided by 999. So 102.816 ÷ 999. Let's do a rough check: 999 * 0.1 = 99.9. 999 * 0.102 = 99.9 + (999 * 0.002) = 99.9 + 1.998 = 101.898. 999 * 0.103 = 101.898 + 0.999 = 102.897. Which is very close to 102.816. So 999 * 0.103 ≈ 102.897. The actual numerator is 102.816, which is 102.897 - 102.816 = 0.081 less. So 0.103 - (0.081 / 999). 0.081 / 999 ≈ 0.000081. So approximately 0.103 - 0.000081 ≈ 0.102919. Which again confirms our prior calculation. Therefore, the value is approximately 0.1029, which is 0.103. So the closest option is B, 0.1. Alternatively, maybe check if 0.1029 is closer to 0.1 or 0.11? Wait, but the options don't have 0.11. The options are 0.01, 0.1, 1, 10, 100. So among these, 0.1 is the closest. Alternatively, if there was an option for 0.11, but since there isn't, B is 0.1. So yes, B is the correct answer. Wait, but let me check once more if perhaps there was a miscalculation in the initial steps, because sometimes when approximating, errors can compound. Let's recalculate the numerator precisely. 17 * 0.3 = 5.1. Correct. 5.1 * 20.16. Let's compute this multiplication more carefully. First, 5 * 20.16 = 100.8. Then 0.1 * 20.16 = 2.016. So 100.8 + 2.016 = 102.816. Yes, that's correct. Denominator is 999. So 102.816 divided by 999. Let's compute 999 into 102.816. How many times does 999 go into 1028 (the first four digits of 102.816)? 999 * 1 = 999, which is less than 1028. So 1 time. Subtract 999 from 1028: 1028 - 999 = 29. Bring down the 1: making it 291. 999 goes into 291 zero times. Bring down the 6: 2916. 999 goes into 2916 two times (999*2=1998). Subtract 1998 from 2916: 2916 - 1998 = 918. Bring down the next digit (0, since we're dealing with 102.816 which is 102.8160...). So now it's 9180. 999 goes into 9180 nine times (999*9=8991). Subtract: 9180 - 8991 = 189. Bring down a 0: 1890. 999 goes into 1890 once (999*1=999). Subtract: 1890 - 999 = 891. Bring down a 0: 8910. 999 goes into 8910 eight times (999*8=7992). Subtract: 8910 - 7992 = 918. Bring down a 0: 9180. Wait, we've seen 9180 before. So the decimal repeats here. Putting this all together: 102.816 ÷ 999 = 0.102918102918..., so 0.102918 repeating. So up to four decimal places, it's approximately 0.1029. Therefore, 0.1029 is indeed closest to 0.1 (option B). Thus, the answer should be B.Final Answerboxed{B}</think>To determine which of the given options is closest to the value of (frac{17 times 0.3 times 20.16}{999}), we start by calculating the numerator:1. Calculate (17 times 0.3): [ 17 times 0.3 = 5.1 ]2. Multiply the result by (20.16): [ 5.1 times 20.16 = 5.1 times (20 + 0.16) = 5.1 times 20 + 5.1 times 0.16 = 102 + 0.816 = 102.816 ]3. Now, divide the result by 999: [ frac{102.816}{999} ]Approximating the division by recognizing that 999 is close to 1000: [ frac{102.816}{1000} approx 0.102816 ]Since 999 is slightly less than 1000, the actual value will be slightly higher. Performing the exact division (or using a more precise approximation) shows that the value is approximately 0.1029.Comparing this result to the given options:- The distance from 0.1029 to 0.01 (option A) is 0.0929.- The distance from 0.1029 to 0.1 (option B) is 0.0029.- The distances to the other options (1, 10, 100) are significantly larger.Thus, the closest option is B (0.1).[boxed{B}],<think>Okay, let's see. I need to figure out which of the options A to E is closest to the value of this fraction: (17 × 0.3 × 20.16) divided by 999. Hmm. The options are 0.01, 0.1, 1, 10, or 100. Alright, so it's a multiple-choice question. Let's break it down step by step.First, let me compute the numerator: 17 multiplied by 0.3 multiplied by 20.16. Then divide that result by 999. I need to approximate this without a calculator, right? So maybe there's a way to simplify the calculation or round the numbers to make it easier.Starting with the numerator. Let's do 17 × 0.3 first. 0.3 is the same as 3/10, so 17 × 3 is 51, and then divide by 10 gives 5.1. So 17 × 0.3 = 5.1. Got that part. Now, multiply that by 20.16. Hmm, 5.1 × 20.16. Let's see. Maybe approximate 20.16 as 20 to make it easier? Then 5.1 × 20 = 102. But since it's 20.16, which is a bit more than 20, the actual result should be a bit more than 102. Let me check. The difference between 20.16 and 20 is 0.16, so 5.1 × 0.16. What's 5 × 0.16? That's 0.8, and 0.1 × 0.16 is 0.016. So total is 0.8 + 0.016 = 0.816. So adding that to the 102 gives 102 + 0.816 = 102.816. So approximately, the numerator is 102.816. Let me confirm that calculation again to be sure.Wait, 17 × 0.3 is indeed 5.1. Then 5.1 × 20.16. Breaking down 20.16 into 20 + 0.16. So 5.1 × 20 = 102, 5.1 × 0.16 = 0.816. Adding them gives 102.816. Yep, that seems right. So numerator is approximately 102.816.Now, divide that by 999. So 102.816 ÷ 999. Let's see. 999 is very close to 1000, so maybe approximate it as 1000? Then 102.816 divided by 1000 is 0.102816. But since 999 is slightly less than 1000, the actual value would be slightly higher. Because dividing by a smaller number gives a larger result. So 0.102816 multiplied by (1000/999). Since 1000/999 is approximately 1.001001..., so the actual value is approximately 0.102816 × 1.001001 ≈ 0.103. So roughly 0.103.Looking at the options: A is 0.01, B is 0.1, C is 1, D is 10, E is 100. So 0.103 is closest to 0.1, which is option B. But wait, let me double-check in case my approximations went wrong.Alternatively, maybe compute it more accurately. Let's do the exact calculation.First, numerator: 17 × 0.3 = 5.1. Then 5.1 × 20.16. Let's compute 5.1 × 20 = 102, 5.1 × 0.16. As before, 5 × 0.16 = 0.8, 0.1 × 0.16 = 0.016, so total 0.816. So 102 + 0.816 = 102.816. So numerator is exactly 102.816.Denominator: 999.So 102.816 / 999. Let's compute this division. Let me write it as 102.816 ÷ 999.Alternatively, write both numerator and denominator multiplied by 1000 to eliminate decimals: 102816 / 999000. Then simplify.But maybe that's complicated. Alternatively, let's perform the division step by step.102.816 ÷ 999. Let's see. 999 goes into 102.816 how many times?Since 999 × 0.1 = 99.9. So 0.1 times 999 is 99.9. Subtract that from 102.816: 102.816 - 99.9 = 2.916.Now, bring down a zero (treating it as 2.9160). 999 goes into 2916 how many times? 999 × 2 = 1998, 999 × 3 = 2997, which is too big. So 2 times. 999 × 2 = 1998. Subtract from 2916: 2916 - 1998 = 918.Bring down another zero, making it 9180. 999 goes into 9180 how many times? 999 × 9 = 8991. Subtract: 9180 - 8991 = 189.Bring down another zero, making it 1890. 999 goes into 1890 once (999 × 1 = 999). Subtract: 1890 - 999 = 891.Bring down another zero, making it 8910. 999 goes into 8910 eight times (999 × 8 = 7992). Subtract: 8910 - 7992 = 918.Wait, we've seen this remainder before (918). So the pattern will start repeating.So putting it all together: 0.1 (from the first division), then 2, then 9, then 1, then 8, then the 918 repeats. So the decimal is approximately 0.102918189... So 0.102918 repeating.So approximately 0.1029, which is about 0.103. So 0.103 is the approximate value. Comparing to the options: A is 0.01 (which is 0.01), B is 0.1 (0.1), C is 1, D is 10, E is 100. The value 0.103 is closer to 0.1 (B) than to 0.01 (A) or 1 (C). Because 0.103 is 0.003 away from 0.1 and 0.093 away from 0.11, but the next option up is 1, which is 0.897 away. So definitely closer to 0.1.But wait, let me check if maybe I made a mistake in the initial calculation. Let me verify numerator and denominator again.17 × 0.3 is 5.1. Then 5.1 × 20.16. Let's do that multiplication more carefully.20.16 can be written as 20 + 0.16. So:5.1 × 20 = 102.5.1 × 0.16: Let's compute 5 × 0.16 = 0.8, and 0.1 × 0.16 = 0.016. So adding those gives 0.8 + 0.016 = 0.816. So total is 102 + 0.816 = 102.816. Correct.So numerator is 102.816. Then divide by 999. As above, approximately 0.103. So 0.1 is the closest. Hence, option B.But wait, hold on. Let me check if the problem says "closest to the value". So 0.103 is indeed closer to 0.1 (difference of 0.003) than to 0.01 (difference 0.093) or 1 (difference 0.897). So yes, B is the answer.Alternatively, maybe I can cross-cancel some terms before multiplying. Let me see.Original expression: (17 × 0.3 × 20.16) / 999.Perhaps write 0.3 as 3/10, so:(17 × 3/10 × 20.16) / 999.Then, 17 × 3 = 51, so:(51/10 × 20.16) / 999.Then, 51/10 × 20.16 = (51 × 20.16)/10.51 × 20.16. Let me compute that. 51 × 20 = 1020, 51 × 0.16 = 8.16. So total is 1020 + 8.16 = 1028.16. Then divide by 10: 102.816. Then divide by 999: same as before. So 102.816 / 999 ≈ 0.103. Same result.Alternatively, maybe we can simplify 999 as 1000 - 1. Let's see:102.816 / (1000 - 1) ≈ 102.816 / 1000 × (1 + 1/1000) [using the approximation 1/(1 - x) ≈ 1 + x for small x]. Here, x is 1/1000, so:≈ (0.102816) × (1 + 0.001) = 0.102816 + 0.000102816 ≈ 0.102918, which matches the earlier division result. So again, 0.103.Therefore, the answer is B, 0.1.But let me check once again, maybe approximate all numbers to one significant figure for a quick estimation.Original problem: (17 × 0.3 × 20.16)/999.Approximate 17 ≈ 20, 0.3 stays 0.3, 20.16 ≈ 20, and 999 ≈ 1000.So (20 × 0.3 × 20)/1000 = (12 × 20)/1000 = 240 / 1000 = 0.24. That's an approximate value. Hmm, 0.24 is closer to 0.1 or 0.1 is 0.14 away, and 0.24 is 0.14 away from 0.1, and 0.24 is 0.76 away from 1. Wait, but my first approximation gave 0.24, but the actual value is around 0.103. So this rough approximation is not very accurate. So maybe my initial approach of calculating it step by step is better.Alternatively, maybe approximate 20.16 as 20.2, which is 20.2. Then 17 × 0.3 is still 5.1. Then 5.1 × 20.2. 5 × 20.2 = 101, 0.1 × 20.2 = 2.02. So total is 101 + 2.02 = 103.02. Then divide by 999 ≈ 0.103. So similar result. So the answer is 0.103, which is closest to 0.1.Alternatively, perhaps using fractions.17 is prime, 0.3 is 3/10, 20.16 is 2016/100 = 504/25. So numerator: 17 × 3/10 × 504/25 = (17 × 3 × 504) / (10 × 25) = (17 × 3 × 504) / 250.Denominator: 999. So overall: (17 × 3 × 504) / (250 × 999).Simplify fractions. Let's see if 999 and 504 have common factors. 999 = 27 × 37, 504 = 16 × 32 = 16 × 9 × 7. So 9 is a common factor. So 504 ÷ 9 = 56, 999 ÷ 9 = 111. So now, we have (17 × 3 × 56) / (250 × 111).3 and 111: 111 ÷ 3 = 37. So (17 × 56) / (250 × 37).17 × 56 = 952. 250 × 37 = 9250. So 952 / 9250. Let's divide numerator and denominator by 2: 476 / 4625. Hmm, not sure if they have common factors. 476 is 4 × 119 = 4 × 7 × 17. 4625 is 25 × 185 = 25 × 5 × 37. No common factors. So 476/4625 ≈ 0.1029. Which is the same as before. So 0.1029, which is approximately 0.103. So yes, 0.1 is the closest. Therefore, option B.I think that's thorough. Different methods lead to the same conclusion. So the answer should be B, 0.1.Final Answerboxed{B}</think>To determine which of the given options is closest to the value of (frac{17 times 0.3 times 20.16}{999}), we start by calculating the numerator and then dividing by 999.1. Calculate the numerator: - First, compute (17 times 0.3): [ 17 times 0.3 = 5.1 ] - Next, multiply the result by 20.16. Breaking it down: [ 5.1 times 20 = 102 quad text{and} quad 5.1 times 0.16 = 0.816 ] Adding these results: [ 102 + 0.816 = 102.816 ] - Therefore, the numerator is (102.816).2. Divide the numerator by 999: - Approximating 999 as 1000 for simplicity: [ frac{102.816}{1000} approx 0.102816 ] - Since 999 is slightly less than 1000, the actual value will be slightly higher. Adjusting for this: [ 0.102816 times frac{1000}{999} approx 0.1029 ]3. Compare the result to the given options: - The value (0.1029) is closest to (0.1) (option B), as the difference is (0.0029), which is smaller than the differences to other options.Thus, the closest number to the given value is (boxed{B}).]

🔑:1. B The calculation can be approximated as follows:frac{17 times 0.3 times 20.16}{999} approx frac{17 times 3 times 2}{1000}=frac{51 times 2}{1000} approx frac{100}{1000}=0.1

🧠:[<think>Okay, so I have this problem here about two circles that touch each other externally. A tangent is drawn to the first circle, passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of this circle. I need to find how many times greater the circumference of the first circle is compared to the second one. Hmm, circumference is related to radius, right? So if I can find the ratio of their radii, then since circumference is 2πr, the ratio of the circumferences will be the same as the ratio of the radii. So my main goal is to find the radius of the first circle divided by the radius of the second circle.Let me visualize the problem. There are two circles touching externally. That means they touch at one point, and their centers are separated by a distance equal to the sum of their radii. Let me denote the radius of the first circle as R and the second one as r. So the distance between their centers is R + r.Now, there's a tangent to the first circle that passes through the center of the second circle. The point where this tangent touches the first circle is such that the distance from this point to the center of the second circle is three times the radius of the second circle. So the distance from the point of tangency (on the first circle) to the center of the second circle is 3r.Let me try to draw a diagram mentally. The first circle has center O1 and radius R. The second circle has center O2 and radius r. They touch externally, so the distance between O1 and O2 is R + r. There's a tangent line to the first circle at point T. This tangent line passes through O2. The distance from T to O2 is given as 3r.Since the tangent line passes through O2 and is tangent to the first circle at T, the line O1T is perpendicular to the tangent line. That means triangle O1T O2 is a right triangle, with right angle at T. The legs of this triangle are O1T (which is R) and the tangent segment from T to O2 (which is 3r). The hypotenuse is O1O2, which is R + r.Wait, but in a right triangle, the sum of the squares of the legs equals the square of the hypotenuse. So we can write the equation:R² + (3r)² = (R + r)²Expanding the right side: (R + r)² = R² + 2Rr + r²So substituting back:R² + 9r² = R² + 2Rr + r²Subtract R² from both sides:9r² = 2Rr + r²Subtract r² and 2Rr from both sides:9r² - r² - 2Rr = 08r² - 2Rr = 0Factor out 2r:2r(4r - R) = 0So either 2r = 0 or 4r - R = 0. Since r can't be zero (as it's a circle), we have 4r - R = 0 => R = 4rSo the radius of the first circle is 4 times that of the second. Therefore, the circumference of the first circle is 2πR = 2π(4r) = 8πr, and the circumference of the second is 2πr. Therefore, the first circle's circumference is 4 times greater than the second's. So the answer should be 4.Wait, let me double-check. The tangent line at T is tangent to the first circle, so O1T is perpendicular to the tangent. The tangent passes through O2, so the distance from T to O2 is 3r, which is along the tangent line. So the triangle O1-T-O2 is a right triangle with legs R and 3r, hypotenuse R + r. Then the Pythagorean theorem gives R² + (3r)^2 = (R + r)^2. That leads to R² + 9r² = R² + 2Rr + r². Then 9r² = 2Rr + r², so 8r² = 2Rr, divide both sides by 2r: 4r = R. So yes, R = 4r. Therefore, the circumference ratio is R/r = 4. So the length (circumference) of the first circle is 4 times greater than the second. That seems correct.Final AnswerThe length of the first circle is boxed{4} times greater than that of the second circle.</think>Given two circles that touch each other externally, a tangent is drawn to the first circle passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of the second circle. We need to determine how many times greater the circumference of the first circle is compared to the second circle.1. Let the radius of the first circle be ( R ) and the radius of the second circle be ( r ).2. The distance between the centers of the two circles, which touch externally, is ( R + r ).3. The tangent line at the point of tangency ( T ) on the first circle passes through the center of the second circle ( O_2 ). The distance from ( T ) to ( O_2 ) is ( 3r ).4. Since the tangent line is perpendicular to the radius at the point of tangency, triangle ( O_1TO_2 ) is a right triangle with legs ( R ) (radius of the first circle) and ( 3r ) (distance from ( T ) to ( O_2 )), and hypotenuse ( R + r ).Using the Pythagorean theorem:[R^2 + (3r)^2 = (R + r)^2]Expanding and simplifying:[R^2 + 9r^2 = R^2 + 2Rr + r^2][9r^2 = 2Rr + r^2][8r^2 = 2Rr][4r = R]Thus, the radius of the first circle ( R ) is 4 times the radius of the second circle ( r ). The circumference of a circle is proportional to its radius, so the circumference of the first circle is 4 times that of the second circle.[boxed{4}],<think>Okay, so I have this geometry problem here. Let me read it again carefully."Two circles touch each other externally. A tangent is drawn to the first circle, passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of this circle. How many times greater is the length of the first circle compared to the length of the second circle?"Hmm, let's break this down. First, there are two circles that touch each other externally. That means they're separate and only touching at one point. So, the distance between their centers is equal to the sum of their radii. Let me note that.Let me denote the first circle as Circle A with radius r, and the second circle as Circle B with radius R. Wait, the problem is asking for how many times greater the length (circumference) of the first circle is compared to the second. So, circumference is 2πr for the first and 2πR for the second. Therefore, the ratio would be (2πr)/(2πR) = r/R. So, the answer should be r/R. So, we need to find r in terms of R, or R in terms of r, depending on which circle is which.But wait, the problem says "the length of the first circle compared to the length of the second". So, if the first circle has radius r, then its circumference is 2πr, and the second is 2πR. So, the ratio is (2πr)/(2πR) = r/R. So, we need to find r/R. But let me check again the problem statement.Wait, maybe I misread. Wait, the tangent is drawn to the first circle, passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of this circle. So, "this circle" refers to the second circle, which is Circle B. So, the distance from the point of tangency (on Circle A) to the center of Circle B is 3R, where R is the radius of Circle B.Therefore, if we can model this, maybe with a diagram. Let me visualize.We have two circles touching externally. Let me draw Circle A on the left and Circle B on the right, touching at a single point. The tangent to Circle A passes through the center of Circle B. Let's denote the point of tangency on Circle A as point T. Then, the distance from T to the center of Circle B (let's call it O_B) is given as 3R, where R is the radius of Circle B.We need to relate this to the radii of the two circles. Let's denote the radius of Circle A as r and the radius of Circle B as R. Since the circles touch externally, the distance between their centers (O_A to O_B) is equal to r + R.We also know that the tangent from O_B to Circle A touches at point T. The length of the tangent from a point to a circle is given by the formula sqrt(d^2 - r^2), where d is the distance from the point to the center of the circle, and r is the radius of the circle.In this case, the length of the tangent from O_B to Circle A is the distance from O_B to T, which is given as 3R. Wait, but according to the tangent length formula, that should also be equal to sqrt(d^2 - r^2), where d is the distance between O_A and O_B, which is r + R.Therefore, according to the problem statement, 3R = sqrt((r + R)^2 - r^2)Let me verify. So, if O_B is at distance (r + R) from O_A, and the tangent from O_B to Circle A has length sqrt((r + R)^2 - r^2). But in the problem, this tangent length is given as 3R. Therefore:3R = sqrt((r + R)^2 - r^2)Let me compute the right-hand side:sqrt((r + R)^2 - r^2) = sqrt(r^2 + 2Rr + R^2 - r^2) = sqrt(2Rr + R^2) = sqrt(R^2 + 2Rr)So, setting equal to 3R:sqrt(R^2 + 2Rr) = 3RSquare both sides:R^2 + 2Rr = 9R^2Subtract R^2 from both sides:2Rr = 8R^2Divide both sides by R (assuming R ≠ 0):2r = 8RDivide both sides by 2:r = 4RTherefore, the radius of Circle A is 4 times the radius of Circle B. Therefore, the circumference of Circle A is 2πr = 2π(4R) = 8πR, and the circumference of Circle B is 2πR. Therefore, the ratio is 8πR / 2πR = 4. So, the first circle's circumference is 4 times greater than the second circle's.Wait, but let me check again. Because the problem says "how many times greater is the length of the first circle compared to the length of the second". So, if the first is 4 times the second, then it's 4 times as long, or 3 times greater? Wait, sometimes "times greater" can be ambiguous. But in standard terms, "times greater" usually means the multiplier. If A is 4 times greater than B, then A = 4B. But sometimes people might interpret it as (A - B)/B = 3, which is 3 times greater. But in the problem statement, since they say "how many times greater", I think it's expecting the multiplier. So, since circumference of first is 4 times the second, so 4 times greater.But to confirm, let's see. Let's suppose the radius of the second circle is R, then the first is 4R. The circumference of the first is 2π*4R = 8πR, and the second is 2πR. So, 8πR is 4 times 2πR, so yes, 4 times greater. So, the answer is 4.But let me just recheck the steps.Starting with the two circles touching externally, so centers are (r + R) apart. Then, the tangent from O_B to Circle A has length 3R. The tangent length formula gives sqrt((r + R)^2 - r^2) = sqrt(2Rr + R^2). Setting equal to 3R, square both sides:2Rr + R^2 = 9R^2Subtract R^2:2Rr = 8R^2Divide by R:2r = 8R => r = 4R. Correct.Therefore, circumference ratio is 2πr / 2πR = r/R = 4. So, the first circle is 4 times longer. So, the answer is 4.But just to make sure, let's consider if "times greater" is interpreted as "how many times more than", which is the same as the ratio. If something is 4 times greater, it's 4 times as much. So, if B is 1, A is 4, so 4 times greater. So, yes, 4.Alternatively, sometimes in everyday language, people might say "300% greater" to mean 4 times as much. But in math problems, unless specified, "times greater" is usually taken as the multiplier. So, 4 times greater is 4 times as much.Therefore, the answer should be 4.Final AnswerThe length of the first circle is boxed{4} times greater than that of the second circle.</think>Given two circles that touch each other externally, we need to determine how many times greater the circumference of the first circle is compared to the second. Let the radius of the first circle (Circle A) be ( r ) and the radius of the second circle (Circle B) be ( R ). The distance between their centers is ( r + R ) since they touch externally. A tangent is drawn from the center of Circle B (O_B) to Circle A, touching it at point T. The distance from T to O_B is given as ( 3R ).Using the tangent length formula, the length of the tangent from O_B to Circle A is ( sqrt{(r + R)^2 - r^2} ). This is equal to ( 3R ):[3R = sqrt{(r + R)^2 - r^2}]Simplifying the right-hand side:[sqrt{(r + R)^2 - r^2} = sqrt{r^2 + 2Rr + R^2 - r^2} = sqrt{R^2 + 2Rr}]Equating and squaring both sides:[3R = sqrt{R^2 + 2Rr}][(3R)^2 = R^2 + 2Rr][9R^2 = R^2 + 2Rr][8R^2 = 2Rr][4R = r]Thus, the radius of Circle A is ( 4R ). The circumference of Circle A is ( 2pi cdot 4R = 8pi R ), and the circumference of Circle B is ( 2pi R ). The ratio of their circumferences is:[frac{8pi R}{2pi R} = 4]Therefore, the length of the first circle is boxed{4} times greater than that of the second circle.]

🔑:1.9. Let O_{1} and O_{2} be the centers of the circles, and A be the point of tangency (Fig. 1.8). Then O_{1} A = R_{1}, O_{1} O_{2} = R_{1} + R_{2}, O_{2} A = 3 R_{2} (by the condition). We need to find the ratio 2 pi R_{1} : 2 pi R_{2} = R_{1} : R_{2}. In the right triangle O_{1} A O_{2} left(angle A = 90^{circ}right), we haveFig. 1.8O_{1} O_{2}^{2} = O_{1} A^{2} + O_{2} A^{2}, text{ or } left(R_{1} + R_{2}right)^{2} = R_{1}^{2} + left(3 R_{2}right)^{2}Simplifying this equation, we get R_{1} = 4 R_{2}, hence R_{1} : R_{2} = 4.Answer: 4 times.

🧠:[<think>Okay, so I need to find the smallest positive number λ such that the sum of 1 over the square root of (1 + 2x_i) from i=1 to n is always less than or equal to λ, given that all x_i are positive real numbers and their product is 1. Hmm, let's start by understanding the problem better.First, the constraints: each x_i is a positive real number, and their product is 1. The expression we need to bound is the sum of reciprocals of square roots of (1 + 2x_i). We need to find the minimal λ that works for all possible such x_i. So, essentially, we need to maximize this sum under the given constraint and then set λ equal to that maximum.Since we're dealing with products and sums, maybe some inequality like AM ≥ GM or Hölder's inequality could be useful here. Let me think.First, let's consider the case when all x_i are equal. Since the product is 1, each x_i would be 1. Then the sum would be n / sqrt(1 + 2*1) = n / sqrt(3). So in this case, the sum is n over sqrt(3). But is this the maximum? Maybe not. Because if we let some x_i be very large and others very small, keeping the product 1, perhaps the corresponding terms in the sum could be larger.Wait, let's test that. Suppose n=2. Let's take n=2 for simplicity. Then x1*x2=1. Let's take x1 approaching 0, then x2 approaches infinity. Then 1/sqrt(1 + 2x1) approaches 1/sqrt(1 + 0) = 1, and 1/sqrt(1 + 2x2) approaches 1/sqrt(infinity) which is 0. So the sum approaches 1. If we take x1 = x2 =1, the sum is 2/sqrt(3) ≈1.1547, which is larger than 1. So in this case, the maximum occurs when all variables are equal. So maybe for n=2, the maximum is 2/sqrt(3). Then λ would be 2/sqrt(3).But wait, let's check another case. Suppose n=1. Wait, n is a given positive integer. If n=1, then x1=1, so the sum is 1/sqrt(3). But the problem says "three" in the beginning, maybe that's a typo. Anyway, moving on.Wait, let's check for n=2. If x1 and x2 are equal, then the sum is 2/sqrt(3). If x1 approaches 0, then the sum approaches 1. So the maximum is indeed 2/sqrt(3). Therefore, for n=2, λ=2/sqrt(3).Similarly, let's check for n=3. Suppose n=3. If all x_i=1, sum is 3/sqrt(3)=sqrt(3)≈1.732. If we set two variables to t and one to 1/t^2, so that the product is t * t *1/t^2=1. Then as t approaches 0, the first two terms approach 1/sqrt(1 + 0)=1 each, and the third term approaches 1/sqrt(1 + 2*(1/t^2)) which approaches 0. So the sum approaches 2. Which is larger than sqrt(3). Therefore, in this case, the maximum is 2. Wait, so for n=3, the maximum is 2?Wait, that's interesting. So when n=3, if we take two variables to be very small and one very large, the sum approaches 2, which is higher than when all variables are 1. So perhaps for n=3, the minimal λ is 2?Wait, let's check more carefully. Let's take n=3. Let x1 = x2 = t, x3 = 1/t^2. Then each term:1/sqrt(1 + 2t) + 1/sqrt(1 + 2t) + 1/sqrt(1 + 2*(1/t^2)).As t approaches 0, first two terms approach 1, third term approaches 0. So sum approaches 2. If t is something else, say t=1, sum is 3/sqrt(3) ≈1.732. If we take t=1/4, so x1=x2=1/4, x3=16. Then first two terms: 1/sqrt(1 + 2*(1/4)) = 1/sqrt(1 + 0.5) = 1/sqrt(1.5) ≈0.816. So two terms sum to ~1.632, and third term is 1/sqrt(1 + 2*16) =1/sqrt(33)≈0.174. Total sum≈1.632 +0.174≈1.806, which is still less than 2.Wait, but as t approaches 0, it approaches 2. So maybe the supremum is 2. Is there a configuration where the sum exceeds 2? Let's see. Suppose we set two variables to t, one to 1/t^2, but make t approach 0. Then the first two terms approach 1 each, and the third approaches 0. So the sum approaches 2. So for n=3, the maximum is 2. Hence, λ=2.But wait, can we get more than 2? Suppose we set three variables such that two are approaching 0 and the third is approaching infinity. Wait, but the product has to remain 1. If two variables approach 0, say x1=x2=t, then x3=1/t^2. As t approaches 0, x3 approaches infinity. Then, as before, sum approaches 2. If we set only one variable to approach 0 and the other two to approach infinity such that the product is 1. Wait, how? If x1 approaches 0, then x2*x3 =1/x1. If x1 approaches 0, then x2 and x3 would need to approach infinity such that their product approaches infinity. But how do we distribute that? Let’s set x2 = x3 = sqrt(1/x1). Then x1*x2*x3 = x1*(1/x1) =1. Then as x1 approaches 0, x2 and x3 approach infinity. Then the terms:1/sqrt(1 + 2x1) approaches 1.1/sqrt(1 + 2x2) =1/sqrt(1 + 2*sqrt(1/x1)). As x1 approaches 0, sqrt(1/x1) approaches infinity, so this term approaches 0. Same for the third term. So sum approaches 1. Which is less than 2. So that case isn't better. Hence, the maximum for n=3 is indeed 2.Similarly, maybe for general n, the maximum occurs when k variables approach 0 and n -k variables approach infinity, but keeping the product 1. Wait, but how?Wait, let's suppose that for some k, we set k variables to t and the remaining n -k variables to t^{-k/(n -k)} so that the product is t^k * t^{-k/(n -k)*(n -k)}= t^k * t^{-k} =1. So this way, the product remains 1. Then, as t approaches 0, the first k terms in the sum will be 1/sqrt(1 + 2t) which approaches 1 each, and the remaining n -k terms will be 1/sqrt(1 + 2t^{-k/(n -k)}). As t approaches 0, t^{-k/(n -k)} approaches infinity, so those terms approach 0. Therefore, the sum will approach k. So if we choose k as large as possible, such that n -k variables can be set to infinity while keeping the product 1. But k can be any integer from 1 to n-1. Therefore, the maximum possible k is n-1, but if we set k =n-1 variables to t and the last variable to t^{-(n-1)/1} = t^{-(n-1)}. Then, as t approaches 0, the sum of the first n-1 terms approaches n-1, and the last term approaches 0. Hence, the total sum approaches n-1. But is this the maximum?Wait, but for n=2, we saw that when k=1, the sum approaches 1, but when variables are equal, the sum is 2/sqrt(3)≈1.1547, which is higher. Therefore, the maximum might not be achieved when variables approach 0 or infinity, but somewhere in between. Wait, but in n=3, we saw that when two variables approach 0 and one approaches infinity, the sum approaches 2, which is higher than when all variables are equal (≈1.732). Therefore, in that case, the maximum is achieved when two variables go to 0. So maybe for even n, the maximum is n/ sqrt(3), and for odd n, it's something else? Wait, but in n=3, the maximum is 2, which is less than 3/sqrt(3)=sqrt(3)≈1.732? Wait, no, 2 is larger than sqrt(3). Wait, sqrt(3)≈1.732, and 2 is larger. So for n=3, the maximum is actually higher when two variables go to zero. So in that case, the maximum is 2.Wait, but how does that generalize? For each n, the maximum possible sum seems to be floor(n/2) *1 + something? Wait, no. Let me think again.Wait, when we set k variables to approach zero and n -k variables to approach infinity, keeping the product 1, then the sum approaches k. So if we take k as large as possible, but we need the product of the remaining variables to approach infinity. Wait, but actually, the product of all variables is 1. If k variables are t, and n -k variables are t^{-k/(n -k)}, then the product is t^k * (t^{-k/(n -k)})^{n -k} }= t^k * t^{-k} =1. So regardless of k, the product is 1. So for any k from 1 to n-1, if we set k variables to t and n -k variables to t^{-k/(n -k)}, then as t approaches 0, the sum approaches k. Therefore, the maximum possible sum in this case would be the maximum k, which is n-1. So as t approaches 0, the sum approaches n-1. But in the case of n=3, we saw that the sum approaches 2, which is n-1=2. For n=2, the sum approaches 1, which is n-1=1, but in that case, when variables are equal, the sum is higher. So the question is: is the maximum of the sum the maximum between n / sqrt(3) and n-1? For n=2, n / sqrt(3) ≈1.1547 >1, so maximum is n / sqrt(3). For n=3, n-1=2 > 3/sqrt(3)=sqrt(3)≈1.732, so maximum is 2. Similarly, for n=4, if we set 3 variables to t and one to t^{-3}, then the sum approaches 3. But when variables are all 1, the sum is 4/sqrt(3)≈2.309, which is less than 3. Therefore, for n=4, the maximum would be 3? Wait, but 3 is n-1.But wait, let's test n=4. Let me take n=4. Suppose we set three variables to t and the fourth to t^{-3}. Then as t approaches 0, the three terms 1/sqrt(1 + 2t) approach 1 each, and the fourth term 1/sqrt(1 + 2t^{-3}) approaches 0. So total sum approaches 3. If we set two variables to t, and two variables to t^{-1}, then as t approaches 0, the sum becomes 2*1 + 2*0=2. Which is less than 3. If we set all variables equal to 1, sum is 4/sqrt(3)≈2.309. If we set one variable to t, and three variables to t^{-1/3}. Then as t approaches 0, the sum approaches 1 + 3*0=1. So the maximum in this case is 3.But wait, 3 is n-1=3. So perhaps for n ≥3, the maximum is n-1? Wait, but for n=3, 2 is n-1. For n=4, 3 is n-1. For n=2, the maximum is n/sqrt(3)≈1.1547, which is greater than n-1=1. So perhaps, for n=2, the maximum is n/sqrt(3), for n ≥3, the maximum is n-1. Therefore, the minimal λ is n/sqrt(3) for n=1, 2 and n-1 for n ≥3? Wait, but n=1 is trivial, since x1=1, so sum is 1/sqrt(3). But the problem states n is a given positive integer, so maybe the answer depends on n.Wait, but let's check for n=3 more carefully. Suppose instead of taking two variables to t and one to t^{-2}, we take three variables such that two are t and one is t^{-2}. Then, as t approaches 0, sum approaches 2. Alternatively, can we set variables in a different way to get a higher sum? Suppose we set one variable to t, another variable to t, and the third variable to t^{-2}. Then, sum is 2/sqrt(1 + 2t) + 1/sqrt(1 + 2t^{-2}) → 2*1 +0=2 as t→0. Alternatively, set all three variables differently. Suppose we set x1 = t, x2 = t, x3 = 1/t^2. Then, same as before. Alternatively, set x1 = t, x2 = s, x3 =1/(ts). Then, as t and s approach 0, x3 approaches infinity. Then, sum is 1/sqrt(1 + 2t) +1/sqrt(1 + 2s) +1/sqrt(1 + 2/(ts)). As t and s approach 0, the first two terms approach 1 each, and the third term approaches 0. So sum approaches 2. If t and s approach 0 independently, but maybe even faster. Wait, but if t approaches 0 and s is fixed, then x3 approaches infinity, but the first term approaches 1, second term is 1/sqrt(1 + 2s), and third term approaches 0. To maximize the sum, we want the first two terms as large as possible. So set s approaching 0 as well. So sum approaches 2. So even if we set two variables to approach 0 and the third to infinity, sum approaches 2. So for n=3, maximum is 2.Similarly, for n=4, maximum is 3, for n=5, maximum is 4, etc. So maybe for each n ≥2, the maximum is n-1. But for n=2, when variables are equal, sum is 2/sqrt(3)≈1.1547 which is greater than n-1=1. So perhaps for n=2, the maximum is 2/sqrt(3). Then, the minimal λ is:- For n=1: 1/sqrt(3)- For n=2: 2/sqrt(3)- For n ≥3: n -1But the problem states that n is a given positive integer. So we need to find λ in terms of n. Therefore, perhaps the answer is λ = max(n/sqrt(3), n -1). But we saw that for n=2, n/sqrt(3) ≈1.1547 >1, and for n=3, n -1=2 >3/sqrt(3)=sqrt(3)≈1.732. So the maximum switches at some point. Let's check when n/sqrt(3) =n -1. Solve for n:n / sqrt(3) = n -1n - n / sqrt(3) =1n (1 -1 / sqrt(3))=1n=1 / (1 -1 / sqrt(3))≈1 / (1 -0.577)≈1 /0.423≈2.366So between n=2 and n=3. Therefore, for n=2, n/sqrt(3) is larger, and for n ≥3, n -1 is larger. Hence, the minimal λ is:λ= { n/sqrt(3), if n ≤2; n -1, if n ≥3 }But the problem says "n be a given positive integer". So we need to write λ in terms of n, but the answer might be a piecewise function. However, in Olympiad problems, often the answer is uniform for all n. Maybe I need to check whether for n=3, the maximum is indeed 2 or maybe higher?Wait, let's take n=3. Suppose we set all three variables to t, t, 1/t^2. Then, as t approaches 0, sum approaches 2. But what if we set variables in another way? Let's set x1 = x2 = x3 =1. Then the sum is 3/sqrt(3)=sqrt(3)≈1.732. If we set two variables to 2 and one to 1/4. Then product is 2*2*(1/4)=1. The sum is 2/sqrt(1 + 4) +1/sqrt(1 + 2*(1/4)) =2/sqrt(5) +1/sqrt(1.5)≈2/2.236 +0.816≈0.894 +0.816≈1.710, which is still less than 2.Another test case: x1=4, x2=4, x3=1/16. Then sum is 2/sqrt(1 +8) +1/sqrt(1 + 2*(1/16))=2/3 +1/sqrt(1.125)≈0.666 +0.894≈1.56, which is still less than 2.Alternatively, set x1 =x2 =x3=1: sum≈1.732.If we set x1=1, x2=1, x3=1: sum≈1.732.If we set x1=0.1, x2=0.1, x3=100. Then sum is 2/sqrt(1 +0.2) +1/sqrt(1 +200)≈2/sqrt(1.2)≈2/1.095≈1.826 +0.070≈1.896, which is still less than 2. Wait, but as t approaches 0, sum approaches 2, so 2 is the supremum. So even though for finite t, the sum is less than 2, it can get arbitrarily close to 2 by taking t approaching 0. Therefore, the maximum is indeed 2 for n=3. Therefore, for n=3, λ=2.Similarly, for n=4, the maximum is 3, as when three variables approach 0 and one approaches infinity, sum approaches 3. Therefore, for general n, the maximum is n -1. Except when n=1 or n=2. Wait, for n=1, product is x1=1, so sum is 1/sqrt(3). So for n=1, λ=1/sqrt(3). For n=2, as we saw, the maximum is 2/sqrt(3). For n≥3, the maximum is n -1. Therefore, the minimal λ is:λ = 1/sqrt(3) when n=1,λ = 2/sqrt(3) when n=2,λ = n -1 when n ≥3.But the problem says "n be a given positive integer". Maybe they expect a single expression? But given that the behavior changes at n=3, perhaps the answer is λ = max(n/sqrt(3), n -1). But for n=1, n/sqrt(3)≈0.577, and n -1=0, but λ must be at least 1/sqrt(3). Wait, but when n=1, the problem constraints require x1=1, so the sum is fixed at 1/sqrt(3). So in that case, λ=1/sqrt(3). But for n≥2, λ is either n/sqrt(3) or n -1, whichever is larger. But for n=2, n/sqrt(3)≈1.1547 is larger than n -1=1. For n=3, n -1=2 is larger than n/sqrt(3)=sqrt(3)≈1.732.Therefore, in general, λ= max(n/sqrt(3), n -1). However, for n=1, even though n -1=0, the correct λ is 1/sqrt(3). But since n is a given positive integer, perhaps the problem assumes n ≥2? Wait, the original problem says "n be a given positive integer", so n=1 is allowed. But the minimal λ for n=1 is 1/sqrt(3). However, in the general formula, if n=1, max(1/sqrt(3), 0)=1/sqrt(3). So maybe the formula λ= max(n/sqrt(3), n -1) works for all n ≥1. Wait, for n=1, n -1=0, so max(1/sqrt(3),0)=1/sqrt(3). For n=2, max(2/sqrt(3),1)=2/sqrt(3). For n=3, max(3/sqrt(3),2)=max(sqrt(3),2)=2. For n=4, max(4/sqrt(3),3)=max≈2.309, 3)=3. So yes, this formula works.But in that case, the minimal λ is the maximum between n/sqrt(3) and n -1. But is this correct? Wait, but when n=3, we can approach 2 but never exceed it, and 2 is less than 3/sqrt(3)=sqrt(3)≈1.732? Wait, no, 2 is greater than sqrt(3)≈1.732. Wait, 2 is approximately 2.0, so max(3/sqrt(3),3 -1)=max(1.732,2)=2. So that's correct.But for n=3, the maximum possible sum is 2, which is n -1=2. For n=4, the maximum is 3, which is n -1=3. So in general, for n ≥3, the maximum is n -1. For n=1 and n=2, it's n/sqrt(3). Therefore, the answer is:λ = begin{cases}frac{n}{sqrt{3}}, & text{if } n leq 2, n - 1, & text{if } n geq 3.end{cases}But the problem asks for the smallest positive number λ such that the inequality always holds. Since the problem mentions "Three. (50 points)", maybe it's part of an exam with multiple questions, and this is question three. But regardless, the answer depends on n.However, maybe there is a more unified approach. Let me think again.Suppose we use Lagrange multipliers to maximize the function f(x1, x2, ..., xn) = sum_{i=1}^n 1/sqrt(1 + 2xi) subject to the constraint product_{i=1}^n xi =1.Set up the Lagrangian: L = sum_{i=1}^n 1/sqrt(1 + 2xi) - λ(prod_{i=1}^n xi -1).Take partial derivatives with respect to xi:dL/dxi = (-1/2)(1 + 2xi)^{-3/2} * 2 - λ(prod_{j≠i} xj) = - (1 + 2xi)^{-3/2} - λ(prod_{j≠i} xj) =0.But since all variables are equal at the maximum (if the maximum occurs at symmetric point), suppose xi =x for all i. Then product xi =x^n =1 => x=1.But when n ≥3, as we saw, the maximum occurs at a boundary point where some variables approach 0 and others approach infinity. Therefore, the critical point at xi=1 is a local minimum or saddle point? Wait, for n=2, when variables are equal, the sum is 2/sqrt(3)≈1.1547, which is a maximum. For n=3, when variables are equal, the sum is sqrt(3)≈1.732, but we can get up to 2, so the critical point is a local minimum? That seems contradictory. Wait, maybe the symmetry gives a local maximum for n=2 but a local minimum for n=3?Alternatively, when we take the derivative at the symmetric point, perhaps the second derivative test can tell us whether it's a maximum or minimum. But this might get complicated.Alternatively, perhaps using convexity. The function 1/sqrt(1 + 2x) is convex or concave? Let's check the second derivative.First derivative: d/dx [1/sqrt(1 + 2x)] = (-1/2)(1 + 2x)^{-3/2} * 2 = - (1 + 2x)^{-3/2}Second derivative: d^2/dx^2 [ ] = (3/2)(1 + 2x)^{-5/2} * 2 = 3(1 + 2x)^{-5/2} >0 for all x> -0.5. Therefore, the function is convex on x >0.Since the function is convex, the sum of convex functions is convex. Therefore, the maximum occurs at the boundary of the domain. Since the domain is constrained by the product=1, which is a convex set in the positive reals? Wait, the set {x_i >0, product x_i=1} is not convex. For example, for n=2, it's the hyperbola x1x2=1, which is not convex. So even if the function is convex, the maximum could be at the boundary.But in optimization, maxima of convex functions over convex sets are attained at the boundary. However, here the set is not convex. So the previous analysis may not apply.Alternatively, using the theory of inequalities. Since we need to maximize the sum, and given that variables can be perturbed to increase the sum, perhaps the maximum occurs when as many variables as possible are as small as possible (thereby making their reciprocal sqrt terms as large as possible, i.e., approaching 1), while the remaining variables are large enough to maintain the product constraint.So, for example, to maximize the sum, set k variables to approach 0, which makes their terms approach 1, and set the remaining n -k variables to approach infinity, making their terms approach 0. The number k is limited by the product constraint: if we set k variables to t, then the remaining n -k variables must be set to t^{-k/(n -k)} to keep the product 1. Then, as t approaches 0, the sum approaches k. Hence, the maximum possible sum is the maximum k for which such a configuration is possible.But k can be any integer from 1 to n -1. Therefore, the maximum possible k is n -1. So, setting n -1 variables to t and 1 variable to t^{-(n -1)}, which tends to infinity as t approaches 0. Then, the sum approaches n -1. However, when n=2, this gives sum approaching 1, but the actual maximum is 2/sqrt(3)≈1.1547. Therefore, this suggests that for n=2, the maximum is attained at the symmetric point, but for n≥3, the maximum is attained at the boundary. Therefore, the minimal λ is n -1 for n≥3 and 2/sqrt(3) for n=2.But how do we reconcile this? It seems that the answer depends on n. However, the problem states "n be a given positive integer". Therefore, the answer should be expressed in terms of n. So the answer is:λ = begin{cases}frac{n}{sqrt{3}}, & text{if } n =1,2, n -1, & text{if } n geq3.end{cases}But the problem doesn't specify different cases for n. Maybe there's a way to write it without piecewise function. Let's see.For n=1, λ=1/sqrt(3). For n=2, λ=2/sqrt(3). For n≥3, λ=n -1. Therefore, the minimal λ is the maximum between n/sqrt(3) and n -1. Since for n=1, max(1/sqrt(3),0)=1/sqrt(3); for n=2, max(2/sqrt(3),1)=2/sqrt(3); for n=3, max(3/sqrt(3),2)=2; for n≥4, max(n/sqrt(3),n -1)=n -1. Therefore, the answer is λ= max(n/sqrt(3),n -1). So we can write this as:λ = maxleft{ frac{n}{sqrt{3}}, n -1 right}But is this the correct answer? For n=3, 3/sqrt(3)=sqrt(3)≈1.732, and n-1=2, so max is 2. For n=4, 4/sqrt(3)≈2.309 and n -1=3, so max is 3. For n=5, 5/sqrt(3)≈2.886 and n -1=4, max is 4. So starting from n=3, n -1 is larger. For n=2, 2/sqrt(3)≈1.1547 >1. For n=1, 1/sqrt(3)≈0.577>0. So yes, the formula holds.But wait, for n=1, n -1=0, so max(n/sqrt(3),n -1)=1/sqrt(3), which is correct. So this formula works for all positive integers n.Therefore, the minimal λ is the maximum between n divided by sqrt(3) and n minus 1. Therefore, the answer is:λ = maxleft{ dfrac{n}{sqrt{3}}, n - 1 right}But in the problem statement, it's mentioned "Three. (50 points)", so maybe this is question number three, and the answer is different. Wait, but according to the analysis, this seems correct.Wait, let's verify with n=3. If λ is 2, can we confirm that the inequality holds for all x_i>0 with product 1?Suppose someone claims that for n=3, the sum is ≤2. Let's test with variables x1=x2=t, x3=1/t^2. Then sum is 2/sqrt(1 + 2t) +1/sqrt(1 + 2/t^2). As t approaches 0, the first two terms approach 1 each, and the third term approaches 0. So sum approaches 2. If we take t=0.1:sum=2/sqrt(1 +0.2) +1/sqrt(1 +2/(0.1)^2)=2/sqrt(1.2)+1/sqrt(1 +200)≈2/1.095 +1/14.177≈1.826 +0.070≈1.896 <2.If we take t=0.01:sum=2/sqrt(1 +0.02)+1/sqrt(1 +2/(0.01)^2)=2/sqrt(1.02)+1/sqrt(1 +20000)≈2/1.01 +1/141.42≈1.98 +0.007≈1.987 <2.So approaches 2 as t approaches 0, but never exceeds 2. Hence, the inequality holds.Similarly, for n=4, sum approaches 3. If we take three variables approaching 0 and one approaching infinity, the sum approaches 3, which is n -1=3. So the inequality holds with λ=3.For n=2, the maximum sum is 2/sqrt(3). If we take x1=x2=1, sum=2/sqrt(3). If we take x1 approaching 0, x2 approaching infinity, sum approaches1. Hence, 2/sqrt(3)≈1.1547 is indeed the maximum.Therefore, the conclusion is that λ= max(n/sqrt(3),n -1).But in the problem statement, the user wrote "Three. (50 points)" at the beginning. Maybe "Three" is the question number, and the rest is the problem. So the answer should be written as boxed. Since the answer is in terms of n, we need to express it using the max function or as a piecewise function. However, in many math competitions, especially those expecting answers in boxed notation, it might need to be expressed differently. But given that the problem asks for the smallest positive number λ such that the inequality always holds, and the analysis shows that λ is the maximum between n/sqrt(3) and n-1, I think the answer is boxed{maxleft{dfrac{n}{sqrt{3}}, n - 1right}}.But maybe the problem expects a single expression without the max function. Let me check the problem again. It says "Try to find the smallest positive number λ, such that the inequality [...] always holds." So λ depends on n, and the answer is a function of n. Since for different n, λ is either n/sqrt(3) or n -1, whichever is larger, so expressing it as the maximum is acceptable.Alternatively, if we rationalize the denominator, n/sqrt(3) can be written as n*sqrt(3)/3, so the answer is max(n*sqrt(3)/3, n -1). But either way, the answer is as above.Therefore, I think the correct answer is boxed{maxleft{dfrac{n}{sqrt{3}}, n - 1right}}.Final AnswerThe smallest positive number (lambda) is boxed{maxleft{dfrac{n}{sqrt{3}}, n - 1right}}.</think>Given the problem of finding the smallest positive number (lambda) such that the inequality [frac{1}{sqrt{1+2 x_{1}}} + frac{1}{sqrt{1+2 x_{2}}} + cdots + frac{1}{sqrt{1+2 x_{n}}} leqslant lambda]always holds, where (x_i in mathbf{R}^+) for (i=1,2,ldots,n) and (x_1 x_2 cdots x_n = 1), we need to analyze the maximum value of the sum under the given constraint.1. Case Analysis for Different (n): - For (n=1), the only value is (x_1 = 1), leading to (lambda = frac{1}{sqrt{3}}). - For (n=2), when (x_1 = x_2 = 1), the sum is (frac{2}{sqrt{3}}). Testing other configurations shows this is the maximum. - For (n geq 3), setting (k) variables approaching 0 and (n-k) variables approaching infinity (to maintain the product constraint) shows the sum approaches (k). The maximum (k) is (n-1), leading the sum to approach (n-1).2. General Formulation: - The function (frac{1}{sqrt{1+2x}}) is convex, and the maximum value of the sum is found at the boundary of the constraint set. - For each (n), the maximum value of the sum is the maximum between (frac{n}{sqrt{3}}) (when all variables are equal) and (n-1) (when (n-1) variables approach 0 and one approaches infinity).3. Conclusion: - The smallest (lambda) is determined by taking the maximum of (frac{n}{sqrt{3}}) and (n-1).Thus, the smallest positive number (lambda) is [boxed{maxleft{dfrac{n}{sqrt{3}}, n - 1right}}.],<think>Okay, so I need to find the smallest positive number λ such that the sum of 1 over the square root of (1 + 2x_i) from i=1 to n is always less than or equal to λ, given that all x_i are positive real numbers and their product is 1. The problem mentions n is a given positive integer. Alright, let's try to break this down.First, since the product of all x_i is 1, this is a constraint that reminds me of optimization problems under multiplicative constraints. Often, with such constraints, the AM-GM inequality comes into play, but I need to see how that applies here. Alternatively, maybe Lagrange multipliers could be useful for optimization with constraints. Hmm.But the goal is to find the maximum possible value of the sum ∑ 1/√(1 + 2x_i), given the constraint that the product of x_i is 1. Then, the smallest λ that satisfies the inequality for all such x_i would be this maximum value. So essentially, I need to maximize this sum under the given constraint.To approach this, maybe I can consider symmetry. If all the x_i are equal, then by AM-GM, that's when the variables are equal for the product constraint. So if x_1 = x_2 = ... = x_n = 1, since their product is 1. Let's check the sum in this case. Each term becomes 1/√(1 + 2*1) = 1/√3. So the sum would be n/√3. Is this the maximum? Or could there be a configuration where some x_i are larger and others smaller, leading to a higher sum?Wait, the function 1/√(1 + 2x) is a decreasing function in x. So if we decrease x_i, the value of the function increases. But since the product of all x_i is fixed at 1, if we decrease some x_i, we have to increase others to compensate. So perhaps making some x_i very small and others large could lead to a higher sum? Let me check.Suppose n=2 for simplicity. Let's take n=2, x1*x2=1. Then the sum is 1/√(1+2x1) + 1/√(1+2x2). Let's set x1 = t, so x2 = 1/t. Then the sum becomes 1/√(1 + 2t) + 1/√(1 + 2/t). Let's analyze this function of t. Maybe take the derivative with respect to t to find maximum.Let f(t) = 1/√(1 + 2t) + 1/√(1 + 2/t). Compute f'(t):First term derivative: d/dt [ (1 + 2t)^{-1/2} ] = (-1/2)(2)(1 + 2t)^{-3/2} = - (1 + 2t)^{-3/2}Second term derivative: Let u = 1 + 2/t = 1 + 2t^{-1}. Then derivative of (u)^{-1/2} is (-1/2)u^{-3/2} * (-2)t^{-2} = (1/2)*2*u^{-3/2}*t^{-2} = u^{-3/2}/t^2So f'(t) = - (1 + 2t)^{-3/2} + (1 + 2/t)^{-3/2} / t^2Set derivative to zero:- (1 + 2t)^{-3/2} + (1 + 2/t)^{-3/2} / t^2 = 0Move one term to the other side:(1 + 2/t)^{-3/2} / t^2 = (1 + 2t)^{-3/2}Multiply both sides by t^2 (assuming t ≠ 0):(1 + 2/t)^{-3/2} = t^2 (1 + 2t)^{-3/2}Take both sides to the power of -2/3 to simplify:(1 + 2/t) = [ t^2 (1 + 2t)^{-3/2} ]^{-2/3}Wait, maybe another approach. Let's let’s cross multiply:[ (1 + 2/t)^{-3/2} ] = t^2 (1 + 2t)^{-3/2}Raise both sides to the power of -2/3:(1 + 2/t) = [ t^2 (1 + 2t)^{-3/2} ]^{-2/3}Wait, this seems complicated. Let me see:Left-hand side: (1 + 2/t)Right-hand side: [ t^2 (1 + 2t)^{-3/2} ]^{-2/3} = t^{-4/3} (1 + 2t)^{1}So:1 + 2/t = t^{-4/3} (1 + 2t)Multiply both sides by t^{4/3}:t^{4/3} + 2 t^{4/3 -1} = 1 + 2tSimplify exponents:t^{4/3} + 2 t^{1/3} = 1 + 2tHmm, this looks messy. Maybe substitute z = t^{1/3}, so t = z^3.Then equation becomes:z^4 + 2 z = 1 + 2 z^3Bring all terms to left-hand side:z^4 - 2 z^3 + 2 z -1 =0Factor this polynomial. Let's try possible rational roots. By Rational Root Theorem, possible roots are ±1.Test z=1: 1 -2 +2 -1=0. Yes, z=1 is a root.So we can factor out (z -1):Use polynomial division or synthetic division.Divide z^4 -2 z^3 +0 z^2 +2 z -1 by (z -1):Coefficients: 1 | -2 | 0 | 2 | -1Bring down 1. Multiply by 1: 1Add to next coefficient: -2 +1 = -1Multiply by1: -1Add to next coefficient: 0 + (-1) = -1Multiply by1: -1Add to next coefficient: 2 + (-1) =1Multiply by1:1Add to last coefficient: -1 +1=0. So it factors as (z -1)(z^3 - z^2 - z +1)Now factor z^3 - z^2 - z +1. Group terms:(z^3 - z^2) - (z -1) = z^2(z -1) -1(z -1) = (z^2 -1)(z -1) = (z -1)(z +1)(z -1)Wait, that's not correct. Wait, if we factor z^3 - z^2 - z +1, let's try possible roots again. z=1: 1 -1 -1 +1=0. So z=1 is a root again. Divide z^3 - z^2 - z +1 by (z -1):Coefficients:1 | -1 | -1 |1Bring down 1.Multiply by1:1. Add to -1:0.Multiply by1:0. Add to -1: -1.Multiply by1: -1. Add to 1:0.So it factors as (z -1)(z^2 +0 z -1) = (z -1)(z^2 -1). Wait, z^2 -1 factors further into (z -1)(z +1). Wait, but that would make the entire polynomial (z -1)^3(z +1). But let me check:Wait, original polynomial after factoring out (z -1) twice is (z -1)^2(z +1). Wait, let me check:Wait, z^3 - z^2 - z +1 = (z -1)(z^2 -1) = (z -1)^2(z +1). Yes, because z^2 -1 is (z -1)(z +1). So then the original equation factors as (z -1)^3(z +1). Wait, but when z= -1, does that satisfy the equation?Wait, when z=-1, plug into z^4 -2 z^3 +2 z -1: (-1)^4 -2*(-1)^3 +2*(-1) -1 =1 -2*(-1) -2 -1=1 +2 -2 -1=0. Yes, z=-1 is also a root. Wait, so maybe the factorization is (z -1)^3(z +1)?Wait, no, let's retrace. Starting with z^4 -2 z^3 +2 z -1, we factored out (z -1) to get (z -1)(z^3 - z^2 - z +1). Then z^3 - z^2 - z +1 factors into (z -1)^2(z +1). So overall, the polynomial is (z -1)^3(z +1). Therefore, the roots are z=1 (triple root) and z=-1. However, z was defined as t^{1/3}, which is a real number, and since t is positive (as x_i are positive reals), z must be positive. Therefore, z=-1 is not acceptable. So the only real positive root is z=1. Therefore, t=z^3=1.Therefore, the critical point is at t=1. So when n=2, the maximum occurs at x1=x2=1, giving the sum as 2/√3. Therefore, for n=2, λ=2/√3. Is that the case?Wait, but let's test with t approaching 0. If t approaches 0, x1 approaches 0, x2 approaches infinity. Then the sum becomes 1/√(1 + 0) + 1/√(1 + ∞) = 1 + 0 =1. Which is less than 2/√3≈1.1547. Similarly, if t approaches infinity, x1 approaches infinity, x2 approaches 0, sum approaches 0 +1=1. So the maximum seems indeed at t=1. So for n=2, λ=2/√3.Hmm. So perhaps for general n, the maximum occurs when all x_i=1, giving the sum as n/√3. But is that the case? Let's test for n=1. Wait, n is a positive integer. For n=1, the product x1=1, so the sum is 1/√(1 +2*1)=1/√3≈0.577. So λ=1/√3. But that's trivial. Let's check n=3.For n=3, if all x_i=1, the sum is 3/√3=√3≈1.732. But what if we set two variables to t and one to 1/t^2 to maintain the product 1. Then sum is 2/√(1 +2t) +1/√(1 +2/t^2). Let's see if this can be larger than √3.Take t approaching 0. Then two terms go to 2/√1=2, and the third term goes to 1/√(1 + ∞)=0. So sum approaches 2, which is bigger than √3≈1.732. Wait, that's a problem. So maybe for n=3, the maximum is higher when variables are unequal. Therefore, the maximum might not occur at all x_i=1. So my initial assumption is wrong.Wait, so perhaps the maximum occurs when some variables approach zero and others approach infinity. But the sum of 1/√(1 + 2x_i) when x_i approaches zero becomes 1/√1=1, and when x_i approaches infinity, it becomes 0. So if I set k variables to approach zero and n -k variables to approach infinity, but such that the product is 1. Wait, but if k variables approach zero, then the remaining n -k variables must approach infinity in such a way that the product is 1. For example, if I take k variables as t and n -k variables as t^{-k/(n -k)} so that the product is t^k * t^{-k/(n -k)*(n -k)}=t^k * t^{-k}=1. Wait, that seems not helpful.Wait, perhaps if I take one variable approaching zero, then another variable must approach infinity to compensate, but in the case of n=3, if I take two variables approaching zero, then the third variable must approach infinity squared? Wait, let's see: suppose we have x1 = x2 = t, and x3 = 1/(t^2). Then the product is t*t*1/t^2=1. Then as t approaches 0, x3 approaches infinity. Then the sum becomes 2/√(1 + 2t) + 1/√(1 + 2/(t^2)). As t approaches 0, 2/√1 + 1/√(∞) = 2 + 0 =2. Which is larger than √3≈1.732. So indeed, the sum can approach 2. Therefore, for n=3, the maximum is 2? Wait, but when t approaches 0, the sum approaches 2, but can it get higher?Wait, let's take x1 = x2 = t, x3 = 1/t^2. Let's compute the sum as a function of t:Sum = 2/√(1 + 2t) + 1/√(1 + 2/t^2)Let’s see if this function can exceed 2. Let’s take t very small, say t=0.01:Sum ≈ 2/√1.02 + 1/√(1 + 2/0.0001) ≈ 2/1.0099 + 1/√(20001) ≈1.980 + 0.007 ≈1.987, which is less than 2. Wait, but as t approaches 0, the sum approaches 2, but never actually reaches 2. So maybe 2 is the supremum. But is there a configuration where the sum can be larger than 2?Suppose instead we take three variables: x1 = t, x2 = t, x3 = 1/t^2. Wait, that's the same as before. Alternatively, take x1 = t, x2 = t, x3 = t, but that would require t^3=1, so t=1, leading to sum 3/√3=√3. Alternatively, take x1 approaching 0, x2 approaching 0, x3 approaching infinity^2? But how?Wait, product x1x2x3=1. If x1 and x2 approach 0, then x3 must approach infinity, but how? If x1 = x2 = t, then x3 =1/t^2. As t→0, x3→infty. Sum is 2/√(1+2t) +1/√(1 + 2/t^2). As t→0, this approaches 2 +0=2. So the maximum is 2 for n=3? Wait, but 2 is greater than √3≈1.732. So for n=3, λ=2?But wait, let's try another configuration. Suppose x1 approaches 0, x2=1, x3 approaches infinity. So x1x2x3= x1*1*x3=1 => x3=1/x1. Then sum becomes 1/√(1 +0) +1/√(1 +2*1) +1/√(1 +2*(1/x1)). As x1 approaches 0, this becomes 1 +1/√3 +0≈1.577, which is less than 2. So not better.Alternatively, take two variables approaching 0 and one approaching infinity. Then sum is 2*1 +0=2. So that's the maximum. So for n=3, the maximum λ is 2.Similarly, if n=4, maybe λ=2? Wait, let's test n=4. Suppose we take two variables approaching 0, and two variables approaching infinity such that product is 1. Wait, but if we take two variables as t and two variables as 1/t, then product is t^2*(1/t)^2=1. Then sum is 2/√(1 +2t) + 2/√(1 +2/t). As t approaches 0, sum approaches 2/1 + 2/√(infty)=2+0=2. But if t=1, sum is 4/√3≈2.309. Wait, that's bigger than 2. Wait, this is conflicting.Wait, hold on. For n=4, if we set all variables to 1, sum is 4/√3≈2.309. But if we set two variables to t and two variables to 1/t, then as t approaches 0, two terms approach 1 and two terms approach 0, giving sum=2. But 2.309 is bigger. So in this case, the maximum occurs at t=1. Wait, but that contradicts the previous case for n=3.Wait, so perhaps when n is even, the maximum is higher? Wait, no, when n=3, when variables are set to t,t,1/t², then sum approaches 2. But for n=4, if variables are set to t,t,1/t,1/t, then as t approaches 0, sum approaches 2*1 + 2*0=2, but when variables are all 1, sum is 4/√3≈2.309. Therefore, for n=4, the maximum is 4/√3≈2.309.But wait, this seems inconsistent. Wait, maybe the maximum depends on n. For even n, maybe the maximum is n/√3, but for odd n, maybe the maximum is something else.Wait, let's check n=3 again. If we set two variables to t and one variable to 1/t². Then as t→0, sum approaches 2*1 +0=2. If t=1, sum is 3/√3≈1.732. But for some intermediate t, maybe the sum is higher?Wait, let's calculate the sum for t=0.1. Then x1=x2=0.1, x3=1/(0.1)^2=100. The sum is 2/√(1 +0.2) +1/√(1 +200)=2/√1.2 +1/√201≈2/1.0954 +1/14.177≈1.826 +0.070≈1.896, which is less than 2. If t=0.5, x1=x2=0.5, x3=4. Sum=2/√(1 +1) +1/√(1 +8)=2/√2 +1/3≈1.414 +0.333≈1.747, still less than 2. If t=0.01, sum≈2/√1.02 +1/√(1 +2/0.0001)=≈1.980 +0.007≈1.987. So as t approaches 0, sum approaches 2. So the maximum seems to be 2. But how can that be, when for n=4, the maximum when variables are all 1 is higher?Wait, but for n=3, the case when two variables go to 0 and one goes to infinity, sum approaches 2. But when variables are all 1, sum is 1.732. Therefore, the maximum is indeed 2 for n=3. For n=4, the maximum is when all variables are 1, giving sum≈2.309. Hmm. So maybe there's a different pattern here.Wait, perhaps the maximum occurs when as many variables as possible are minimized (i.e., approaching 0), with the remaining variables compensating to keep the product 1. Let's see. Suppose n is arbitrary. Let’s take k variables approaching 0, then the remaining n -k variables must approach infinity such that the product of all variables is 1. The sum would then approach k*1 + (n -k)*0 =k. So the maximum possible sum would be k, where k is the number of variables approaching 0. But how many variables can we set to approach 0?But we need the product to remain 1. If we take k variables as t, then the remaining n -k variables must be t^{-k/(n -k)}. So as t approaches 0, the remaining variables approach infinity only if k/(n -k) >0, which is always true if k <n. Wait, but if k approaches n, then the exponent becomes large. Wait, but if k=n, then product is t^n=1, so t=1. So we cannot set all variables to approach 0.So to maximize the sum, we can set as many variables as possible to approach 0, but with the constraint that the remaining variables must approach infinity to maintain the product=1. However, the more variables we set to approach 0, the more terms in the sum that contribute 1 each. So to maximize the sum, we need to maximize the number of variables approaching 0. However, each time we set a variable to approach 0, another variable must approach infinity, but depending on n.Wait, no, if we set k variables to t, then the remaining n -k variables must be t^{-k/(n -k)}. So if k=1, then remaining n -1 variables are t^{-1/(n -1)}. As t approaches 0, each of those variables approaches infinity. But the sum would be 1/√(1 + 2t) + (n -1)/√(1 + 2t^{-1/(n -1)}). As t approaches 0, the first term approaches 1, and each of the remaining terms approach 0. So the sum approaches 1. If k=2, the remaining n -2 variables are t^{-2/(n -2)}. Then sum is 2/√(1 + 2t) + (n -2)/√(1 + 2t^{-2/(n -2)}). As t→0, this approaches 2 +0=2. Similarly, for k=3, sum approaches 3, but only if the remaining variables can be adjusted such that their product with t^3=1. However, for k=3, the remaining variables are t^{-3/(n -3)}. So this is possible only if n >3. For example, n=4, k=2: sum approaches 2. If n=5, k=2: sum approaches 2, but if k=3, sum approaches 3. Wait, but for n=5, if k=3, then remaining 2 variables are t^{-3/2}. As t approaches 0, they approach infinity, so sum approaches 3*1 +2*0=3. Similarly, for n=6, k=3, sum approaches 3.But for this to be possible, the exponent must be positive, i.e., for k variables set to t, remaining n -k variables are t^{-k/(n -k)}. So as t→0, the remaining variables approach infinity only if -k/(n -k) <0, which is equivalent to n -k >0, which is always true as long as k <n. So in principle, for any k from 1 to n-1, we can set k variables to approach 0 and the rest to approach infinity, giving the sum approaching k. Therefore, the maximum possible sum is the maximum integer k such that k <=n. Wait, but k can be up to n-1, but the sum would be k. So the maximum possible sum is n-1? Wait, no, if we set n-1 variables to approach 0, then the last variable must be t^{-(n-1)/1}=t^{-(n-1)}. Then the sum would be (n-1)*1 +1/√(1 + 2t^{-(n-1)}). As t→0, the last term approaches 0, so sum approaches n-1. So in that case, for any n, the maximum sum could approach n-1. But wait, when n=3, this would mean the sum approaches 2, which matches our earlier result. For n=4, it would approach 3. But earlier for n=4, when all variables are 1, the sum was 4/√3≈2.309, which is less than 3. So this suggests that if we set n-1 variables to approach 0, the sum approaches n-1, which would be larger than the symmetric case. But is that really possible?Wait, let's take n=4. Suppose we set 3 variables to t and the fourth variable to t^{-3}. Then the product is t^3 * t^{-3}=1. Then the sum is 3/√(1 + 2t) +1/√(1 +2 t^{-3}). As t→0, the first three terms approach 3/√1=3, and the last term approaches 0. So the sum approaches 3. But when I tried n=4 earlier with two variables approaching 0 and two approaching infinity, sum approaches 2. But here, with three variables approaching 0 and one approaching infinity, sum approaches 3. Which is larger. So for n=4, the maximum is 3? Wait, but that contradicts the earlier example when variables are all 1 giving sum≈2.309. So maybe the maximum is n-1?Wait, let's check with n=4. If three variables approach 0 and one approaches infinity, sum approaches 3. But is that actually possible? Let me verify with specific numbers. Let x1=x2=x3=t and x4=1/t^3. Then as t→0, x4→infty. Then sum is 3/√(1 + 2t) +1/√(1 +2/t^3). Let's take t=0.01:Sum≈3/√1.02 +1/√(1 +2/(0.01)^3)=≈3/1.01 +1/√(1 +2*10^6)=≈2.97 +1/√2000001≈≈2.97 +0.0007≈≈2.97, which is approaching 3 as t→0.Similarly, for n=4, this suggests that λ=3. But earlier, when we considered two variables approaching 0 and two approaching infinity, sum approached 2, which is less than 3. So in this case, the more variables we set to approach 0, the higher the sum. Hence, for each n, the maximum sum is n-1. Therefore, the minimal λ is n-1. But that can't be right because for n=3, when we set two variables to 0 and one to infinity, sum approaches 2, which is n-1=2. For n=4, sum approaches 3, which is n-1=3. But when n=2, we saw that the maximum is 2/√3≈1.1547, but according to n-1, it would be 1. But in reality, for n=2, the maximum is achieved at t=1, giving 2/√3≈1.1547, but if we set one variable approaching 0 and the other approaching infinity, sum approaches 1. Which is less than 1.1547. Therefore, the previous reasoning is flawed.Wait, so for n=2, the maximum is achieved at the symmetric point, but for n≥3, the maximum is achieved when n-1 variables approach 0 and the last one approaches infinity. But why the discrepancy?Wait, let's check n=3 again. If we set two variables to approach 0 and one to infinity, sum approaches 2. For n=3, n-1=2. If we set all variables to 1, sum is 3/√3≈1.732, which is less than 2. So for n=3, λ=2. Similarly, for n=4, sum can approach 3. But wait, when n=2, even though n-1=1, the maximum is higher at 2/√3≈1.1547>1. So the pattern is different. Therefore, the conjecture that λ=n-1 holds for n≥3 but not for n=2. Hmm, but why?Wait, perhaps the function 1/√(1 +2x) has different concavity properties depending on the number of variables. Let's consider the general case. Suppose we have n variables, and we set k variables to t and the remaining n -k variables to t^{-k/(n -k)}. Then the sum S = k/√(1 + 2t) + (n -k)/√(1 +2 t^{-k/(n -k)}). We need to analyze the behavior as t→0.As t→0, the first term approaches k/√1 =k, and the second term approaches (n -k)/√(∞)=0. So S approaches k. Therefore, to maximize S, we need to choose the maximum k. However, the problem is that for k= n -1, then remaining variables are t^{-(n-1)/1}=t^{-(n-1)}, so S approaches n -1. For k= n -2, sum approaches n -2. So the maximum S approaches n -1 when we set k= n -1 variables to t.But for n=2, k=1, sum approaches 1, but in reality, the maximum is 2/√3≈1.1547. So for n=2, this approach doesn't hold. So what's different?Ah, because when n=2, we cannot set k=1 variable to t and the other to t^{-1/(2-1)}=t^{-1}, so as t→0, sum approaches 1 +0=1, but we can actually achieve a higher value when variables are equal. Therefore, in some cases, the maximum is attained at the symmetric point, and in others, it's attained when variables are at extremes. So how do we reconcile this?This suggests that the function's behavior is different for different n. Maybe for even n, the maximum is achieved at symmetric points, and for odd n, it's achieved at extremes. But no, for n=3, which is odd, maximum is achieved at extreme, but for n=4, even, maximum is also achieved at extreme (n-1=3). But in n=2, even, the maximum is at the symmetric point.Wait, perhaps there is a threshold where for n ≥3, the maximum is n-1, but for n=1,2, it's different. Wait, n=1 is trivial. For n=2, maximum is 2/√3. For n=3, maximum is 2. Wait, but earlier for n=3, when variables are approaching extremes, sum approaches 2. If we set all variables to 1, sum is 3/√3≈1.732. But how do we know that the supremum is actually 2? Because as t approaches 0, the sum approaches 2, but never actually reaches 2. So in the limit, it's 2, so the supremum is 2, hence λ=2 for n=3.Similarly, for n=4, the supremum is 3. So for general n, the minimal λ is n-1. But then why for n=2, this doesn't hold?Wait, perhaps the answer is different. Let me check for n=2 again. Let's see, with x1*x2=1, the maximum of 1/√(1 +2x1) +1/√(1 +2x2). If we set x1 =x, x2=1/x. Then the function is f(x)=1/√(1 +2x) +1/√(1 +2/x). Taking derivative and finding maximum at x=1, which gives 2/√3≈1.1547. But if we set x→0, then f(x)→1 +0=1. If x→infty, f(x)→0 +1=1. Therefore, the maximum is indeed at x=1. So for n=2, λ=2/√3. For n=3, λ=2. For n=4, λ=3. So the pattern seems λ= n-1 when n≥3, but for n=2, λ=2/√3. But this seems inconsistent.Wait, maybe there is a mistake in assuming that the maximum for n≥3 is n-1. Let me check for n=4.For n=4, suppose we set three variables approaching 0 and one approaching infinity. Then sum approaches 3. But if we set two variables approaching 0 and two variables approaching infinity, sum approaches 2. If we set one variable approaching 0 and three approaching infinity, sum approaches1. If we set all variables to 1, sum is4/√3≈2.309. So which is the maximum? If the sum can approach 3, then λ=3. But how?Wait, let's compute for n=4, set three variables to t and one variable to1/t^3. Then sum is3/√(1 +2t) +1/√(1 +2/t^3). Let's take t=0.1:Sum≈3/√1.2 +1/√(1 +2/0.001)=≈3/1.0954 +1/√2001≈≈2.74 +0.022≈2.76. If t=0.01:Sum≈3/√1.02 +1/√(1 +2/0.000001)=≈3/1.01 +1/√2000001≈≈2.97 +0.0007≈2.97. So as t→0, sum approaches3. Therefore, the supremum is3. But is this achievable? Since t can be made arbitrarily small, sum can be made arbitrarily close to3. Hence, λ=3 for n=4.Similarly, for n=5, λ=4. So in general, λ=n-1. But then for n=2, why is λ=2/√3 instead of1? Because when n=2, setting one variable to0 and the other to infinity gives sum=1, but setting them equal gives a higher sum. So why is it different?Ah, because for n=2, the function f(x)=1/√(1 +2x) +1/√(1 +2/x) attains its maximum atx=1, giving a higher value than the limit asx approaches0 or infinity. But for n≥3, when we set k=n-1 variables to0, the sum approachesn-1, which is greater than the symmetric case. Hence, the minimal λ is n-1 for n≥3, and for n=1 it's1/√3, for n=2 it's2/√3.But how can we confirm this?Wait, let's test for n=5. If we set four variables to0.01 and one variable to1/(0.01)^4=100000000. Then sum≈4/√1.02 +1/√(1 +2*100000000)≈4/1.01 + negligible≈3.96. So sum approaches4 as t→0. Therefore, for n=5, λ=4.Similarly, for n=3, λ=2; for n=4, λ=3; hence, the pattern is λ=n-1 for n≥2. But wait, n=2 gives λ=2/√3≈1.1547<2-1=1? Wait, no, n=2 would be λ=2-1=1 according to the pattern, but in reality it's higher. Therefore, the pattern holds only for n≥3. But that seems inconsistent.Wait, perhaps there is a different pattern. Let's think again. For each n≥1, the minimal λ is the maximum between the symmetric case and the case when n-1 variables are approaching0.For n=1: only x1=1, sum=1/√3≈0.577.For n=2: maximum of2/√3≈1.1547 (symmetric) and1 (asymmetric). So the maximum is1.1547.For n=3: maximum of3/√3≈1.732 (symmetric) and2 (asymmetric). The maximum is2.For n=4: maximum of4/√3≈2.309 (symmetric) and3 (asymmetric). The maximum is3.For n=5: maximum of5/√3≈2.886 (symmetric) and4 (asymmetric). The maximum is4.Hence, for n≥3, the maximum is n-1, but for n=2, it's higher than n-1. Therefore, the minimal λ is:- For n=1:1/√3- For n=2:2/√3- For n≥3:n-1But the problem states "n be a given positive integer", so we need a general formula. But how?Wait, maybe there's a more general pattern. Let's think for n=2, λ=2/√3≈1.1547 <2-1=1. Not really. Wait, 2/√3≈1.1547 is greater than1. So for n=2, the maximum is achieved at the symmetric point, and for n≥3, the maximum is achieved at the asymmetric case.Therefore, the answer depends on n. If n=1:1/√3; if n=2:2/√3; if n≥3:n-1. But the problem says "n be a given positive integer", so possibly the answer is different depending on n, but the problem asks to find the minimal λ such that the inequality holds for all n. Wait, no. Wait, the problem says "n be a given positive integer". So for each n, find the minimal λ. So the answer is:For each positive integer n, the minimal λ is:- If n=1:1/√3- If n=2:2/√3- If n≥3:n-1But the problem is stated as: Given n, find the minimal λ such that the inequality holds. So depending on the value of n, λ is as above. But the problem might expect a single expression. Alternatively, maybe the answer is n-1 for all n≥1, but for n=1, n-1=0, which is incorrect, since the sum is1/√3>0. So that can't be.Alternatively, maybe the answer is max(n-1, n/√3). But let's check:For n=1: max(0,1/√3)=1/√3. Correct.For n=2: max(1, 2/√3)=2/√3≈1.1547. Correct.For n=3: max(2, 3/√3≈1.732)=2. Correct.For n=4: max(3,4/√3≈2.309)=3. Correct.So the minimal λ is the maximum between n-1 and n/√3. Therefore, λ=max(n-1, n/√3).But let's verify for n=3: n-1=2 and n/√3≈1.732. So max is2. For n=4: max(3,≈2.309)=3. Yes, this seems to hold. Similarly, for n=5: max(4,≈2.886)=4.Thus, the formula is λ= max(n-1, n/√3). Therefore, for each n, compare n-1 and n/√3. The larger of the two is the minimal λ. Let's check when n-1 >n/√3.Solve n-1 >n/√3.Multiply both sides by √3:√3(n -1) >n√3 n -√3 >n(√3 -1)n >√3n> √3/(√3 -1)Calculate √3≈1.732, so √3 -1≈0.732.√3/(√3 -1)=1.732/0.732≈2.366. So for n>2.366, i.e., n≥3, n-1 >n/√3. For n=2: n-1=1, n/√3≈1.1547, so n/√3> n-1. For n=1: n-1=0, n/√3≈0.577>0.Therefore, the minimal λ is:λ= { n/√3, if n≤2; n-1, if n≥3 }But since n is a positive integer, this translates to:λ= n/√3 for n=1,2 and λ= n-1 for n≥3.But the problem statement says "three. (50 points)" and then the problem. Maybe the original problem was numbered three, but the user omitted the other parts. But the question is to find the minimal λ for given n.However, the problem might expect a single answer for all n, but that doesn't make sense because the minimal λ depends on n. But since the problem says "n be a given positive integer", then the answer should depend on n. But how to present it?Wait, maybe I made a mistake. Let me check for n=3 again. If λ=2, then the sum must always be ≤2. Suppose we take x1=x2=x3=1. Sum=3/√3=√3≈1.732≤2. If we take x1=x2=t, x3=1/t². Then sum=2/√(1+2t)+1/√(1+2/t²). For t approaching0, sum approaches2 +0=2. So equality is achieved in the limit. Therefore, λ=2 is the minimal for n=3.Similarly, for n=4, λ=3. For n=5, λ=4, etc. For n=2, λ=2/√3≈1.1547. For n=1, λ=1/√3≈0.577.Therefore, the minimal λ is:- For n=1:1/√3- For n=2:2/√3- For n≥3:n-1But the problem might expect a general formula. Since for n≥3, n-1> n/√3, and for n=1,2, n/√3> n-1. Hence, we can write λ= max(n-1, n/√3).Alternatively, since the problem mentions "three. (50 points)", maybe it's question three, and the original problem may have been in Chinese, and the user translated it, but "Three. (50 points)" is just the problem number and points.Assuming the problem is correctly presented, and the user just wants the answer boxed, likely the answer is n-1 for n≥3 and 2/√3 for n=2, but since the problem says "n be a given positive integer", and doesn't specify particular n, the answer likely depends on n.However, maybe there's a different approach. Let me consider using the Cauchy-Schwarz inequality.Let’s denote S= ∑_{i=1}^n 1/√(1 +2x_i}We need to maximize S given that ∏x_i=1.Using Lagrange multipliers. Let’s set up the Lagrangian:L= ∑_{i=1}^n 1/√(1 +2x_i} - λ(∏x_i -1)Take derivative with respect to x_k:dL/dx_k= (-1/2)(2)(1 +2x_k)^{-3/2} - λ(∏x_i)/x_k= - (1 +2x_k)^{-3/2} - λ/x_k=0Because ∏x_i=1, so the derivative of the constraint term is λ(∏x_i)/x_k=λ/x_k.Therefore, we have:- (1 +2x_k)^{-3/2} - λ/x_k=0 → (1 +2x_k)^{-3/2}= -λ/x_kBut since all terms are positive (x_k>0, λ>0), the right-hand side is negative, but left-hand side is positive. Contradiction. Therefore, Lagrange multipliers here give no solution, which suggests that the maximum is attained at the boundary of the domain, i.e., when some variables approach0 and others approach infinity.Hence, the maximum cannot be found using Lagrange multipliers in the interior points, because there are no critical points there. Thus, the maximum occurs at the boundary, which aligns with our previous analysis.Therefore, confirming that for n≥3, the maximum sum is n-1, achieved when n-1 variables approach0 and the nth variable approaches infinity. For n=2, the maximum is at the symmetric point x1=x2=1, giving λ=2/√3.Hence, the minimal λ is n-1 for n≥3 and 2/√3 for n=2.But the problem states "n be a given positive integer", so depending on n, the answer is different. However, the question is to "find the smallest positive number λ" given the problem. If it's expecting a single answer regardless of n, that's impossible. Therefore, likely the answer is n-1 for n≥3 and 2/√3 for n=1,2. But the problem doesn’t specify particular n, so we need to express λ in terms of n.But how?If we consider the maximum between n-1 and n/√3, that works for all n. For n=1,2, n/√3> n-1. For n≥3, n-1> n/√3. Therefore, λ= max(n-1, n/√3).But the problem asks for the smallest positive number λ such that the inequality always holds. Therefore, for each n, the minimal λ is the maximum between n-1 and n/√3. So the answer is λ= max(n-1, n/√3).But to confirm, let's test n=3:max(3-1,3/√3)=max(2,√3)=2≈2>1.732. Correct.For n=4: max(3,4/√3)=3>2.309. Correct.For n=2: max(1,2/√3)=2/√3≈1.1547. Correct.For n=1: max(0,1/√3)=1/√3≈0.577. Correct.Therefore, the general formula is λ= max(n-1, n/√3). Therefore, the smallest λ is the maximum of these two values.But can we write this in a cleaner form? Since for n≥3, n-1≥n/√3, and for n=1,2, n/√3> n-1. Therefore, the answer is:λ= begin{cases}frac{n}{sqrt{3}}, & text{if } n leq 2, n -1, & text{if } n geq 3.end{cases}However, the problem might expect an expression without piecewise definition. Since max(n-1, n/√3) is acceptable.Alternatively, note that n/√3 <n-1 for n≥3. Because dividing both sides by n, we get 1/√3 <1 -1/n. For n≥3, 1 -1/n ≥2/3≈0.666, while 1/√3≈0.577<0.666. Therefore, yes, for n≥3, n-1> n/√3.Hence, the minimal λ is max(n-1, n/√3).But the problem might expect this answer expressed as the maximum function. Therefore, the answer is λ= max(n -1, n/√3).But let me check for n=3: max(3 -1,3/√3)=max(2,√3)=2, since √3≈1.732<2.Yes. So the answer is λ= max(n -1, n/√3).Therefore, the smallest positive number λ is the maximum between n-1 and n divided by the square root of 3.But to write this in a boxed format, depending on the requirement. Since the problem may want the answer in terms of n, the answer is max(n -1, n/√3). However, in Chinese competitions or Olympiads, sometimes they accept piecewise functions.But given that the problem is worth 50 points and is problem three, it might require a more insightful answer.Wait, perhaps there's a different approach. Let's consider using the Hölder inequality.Let’s denote S= ∑_{i=1}^n 1/√(1 +2x_i}We want to maximize S given that ∏x_i=1.Consider applying Hölder’s inequality. Hölder's inequality states that for sequences (a_i) and (b_i), (∑a_i b_i) ≤ (∑a_i^p)^{1/p} (∑b_i^q)^{1/q} where 1/p +1/q=1.But I'm not sure how to apply this here. Alternatively, using the Cauchy-Schwarz inequality.Let’s consider writing S= ∑_{i=1}^n 1/√(1 +2x_i}We can square both sides: S²= [∑1/√(1 +2x_i)]² ≤ n ∑[1/(1 +2x_i)] by Cauchy-Schwarz.So S² ≤n ∑1/(1 +2x_i).Now, we need to bound ∑1/(1 +2x_i). Let's see if we can find the maximum of ∑1/(1 +2x_i) given ∏x_i=1.Again, maybe AM-GM. Note that 1/(1 +2x_i) ≤ something.But this approach might not be straightforward. Alternatively, consider substituting y_i=√x_i. Then since x_i>0, y_i>0. The constraint ∏x_i=1 becomes ∏y_i²=1, so (∏y_i)^2=1 ⇒ ∏y_i=1 or ∏y_i=-1, but since y_i>0, ∏y_i=1.But not sure if this helps. Let’s see:The sum S= ∑1/√(1 +2y_i²}Not obviously easier.Alternatively, use substitution: let’s set t_i=2x_i +1, then x_i=(t_i -1)/2. The constraint ∏x_i=1 becomes ∏(t_i -1)/2=1. So ∏(t_i -1)=2^n.But this might complicate things.Alternatively, maybe use the convexity or concavity of the function. The function f(x)=1/√(1 +2x) is convex or concave?Compute the second derivative. First derivative f’(x)= - (1 +2x)^{-3/2}*(2)/2= - (1 +2x)^{-3/2}Second derivative f''(x)= (3/2)(1 +2x)^{-5/2}*2= 3(1 +2x)^{-5/2} >0 for all x>0. So f(x) is convex.Therefore, by Jensen's inequality, for convex functions, the sum is maximized at the endpoints. Which aligns with our previous analysis that maximum occurs when variables are at extremes (some approaching0, others approaching infinity).Therefore, the maximum of the sum is achieved when as many variables as possible are minimized (i.e., approaching0), with the remaining variable(s) adjusted to maintain the product constraint. For n≥3, this maximum is n-1. For n=1, it's 1/√3. For n=2, it's2/√3.Therefore, consolidating, the minimal λ is:λ= begin{cases}frac{n}{sqrt{3}}, & text{if } n =1,2, n -1, & text{if } n geq 3.end{cases}Since the problem asks for the smallest positive number λ such that the inequality always holds, given any positive real numbers x_i with product1, the answer depends on n. But since n is a given positive integer, the answer is either n-1 or n/√3, whichever is larger. As we saw, for n≥3, n-1 is larger, and for n=1,2, n/√3 is larger.Thus, the answer is the maximum of n-1 and n/√3. To write this in a boxed format, we can express it as:boxed{maxleft(n-1, dfrac{n}{sqrt{3}}right)}This covers all cases since max(n-1, n/√3) gives the correct λ for each positive integer n.Final AnswerThe smallest positive number (lambda) is (boxed{maxleft(n-1, dfrac{n}{sqrt{3}}right)}).</think>Given positive real numbers ( x_i ) (for ( i = 1, 2, ldots, n )) such that their product is 1, we need to find the smallest positive number ( lambda ) such that the inequality[frac{1}{sqrt{1+2 x_{1}}} + frac{1}{sqrt{1+2 x_{2}}} + cdots + frac{1}{sqrt{1+2 x_{n}}} leqslant lambda]always holds. Key Steps and Reasoning:1. Symmetry and AM-GM Inequality: - For ( x_i = 1 ) (by AM-GM), the sum is ( frac{n}{sqrt{3}} ). - However, testing with extreme values (some ( x_i ) approaching 0 and others approaching infinity) showed higher sums for ( n geq 3 ).2. Behavior of the Function: - The function ( frac{1}{sqrt{1 + 2x}} ) is decreasing in ( x ). Setting some ( x_i ) to approach 0 increases their contribution to the sum, while others must approach infinity to maintain the product constraint.3. Critical Analysis for Different ( n ): - For ( n = 1 ), the sum is ( frac{1}{sqrt{3}} ). - For ( n = 2 ), the maximum sum is ( frac{2}{sqrt{3}} ). - For ( n geq 3 ), setting ( n-1 ) variables to approach 0 and one to approach infinity results in the sum approaching ( n-1 ).4. Verification with Hölder's and Jensen's Inequalities: - The function ( frac{1}{sqrt{1 + 2x}} ) is convex, leading to the maximum sum at the boundary values (extremes) by Jensen's inequality. Conclusion:The smallest positive number ( lambda ) is determined by the maximum of ( n-1 ) and ( frac{n}{sqrt{3}} ), depending on the value of ( n ):- For ( n = 1 ) and ( n = 2 ), the maximum is ( frac{n}{sqrt{3}} ).- For ( n geq 3 ), the maximum is ( n-1 ).Thus, the final answer is:[boxed{maxleft(n-1, dfrac{n}{sqrt{3}}right)}]]

🔑:Let 2 x_{1}=tan ^{2} theta_{1}, 2 x_{2}=tan ^{2} theta_{2}, cdots, 2 x_{n}=tan ^{2} theta_{n}, where theta_{i} in(0, pi / 2), (i=1,2, cdots, n), then multiplying these n equations gives tan ^{2} theta_{1} tan ^{2} theta_{2} cdots tan ^{2} theta_{n}=2^{n}.The original inequality Leftrightarrow cos theta_{1}+cos theta_{2}+cdots+cos theta_{n} leqslant lambda Leftrightarrow lambda geqslant max left{cos theta_{1}+cos theta_{2}+cdots+cos theta_{n}right}.When n=1, from tan ^{2} theta_{1}=2 we get cos theta_{1}=frac{sqrt{3}}{3}; when n=2, tan theta_{1} tan theta_{2}=2 Leftrightarrowbegin{array}{l}sin ^{2} theta_{1} sin ^{2} theta_{2}=4 cos ^{2} theta_{1} cos ^{2} theta_{2} Leftrightarrow cos ^{2} theta_{1}+cos ^{2} theta_{2}=1-3 cos ^{2} theta_{1} cos ^{2} theta_{2} Leftrightarrow left(cos theta_{1}+cos theta_{2}right)^{2}=-3 cos ^{2} theta_{1} cos ^{2} theta_{2}+2 cos theta_{1} cos theta_{2}+1,end{array}When cos theta_{1} cos theta_{2}=frac{1}{3}, left(cos theta_{1}+cos theta_{2}right)_{text {max }}=frac{2 sqrt{3}}{3}.When n geqslant 3, we now prove cos theta_{1}+cos theta_{2}+cdots+cos theta_{n}7, then tan ^{2} theta_{1} geqslant tan ^{2} theta_{2}> frac{7}{2}, therefore cos theta_{1} leqslant cos theta_{2} cos theta_{1}+cos theta_{2}+cdots+cos theta_{n-1}=r, contradiction! therefore n-1 is the smallest upper bound of the left side of inequality (1).When n geqslant 3, the smallest positive number lambda=n-1, in summary, the smallest positive number lambda=left{begin{array}{l}frac{sqrt{3} n}{3}(n=1,2) n-1(n geqslant 3)end{array}right..