🧠:[<think>Okay, let me try to solve this inequality system. So, there are two inequalities here. The first one is x minus 1 is greater than 0, and the second one is (2x plus 1) divided by 3 is less than or equal to 3. Hmm, let me take them one at a time.Starting with the first inequality: x - 1 > 0. To solve for x, I need to isolate x on one side. So, if I add 1 to both sides, that should do it. Let's see:x - 1 + 1 > 0 + 1 Which simplifies to x > 1. Okay, that seems straightforward. So, the first inequality tells me that x has to be greater than 1.Now moving on to the second inequality: (2x + 1)/3 ≤ 3. I need to solve this for x as well. Let's see. To get rid of the denominator 3, maybe I can multiply both sides by 3. That usually works. Let me try that:Multiply both sides by 3: (2x + 1)/3 * 3 ≤ 3 * 3 Which simplifies to 2x + 1 ≤ 9. Alright, now subtract 1 from both sides to isolate the term with x:2x + 1 - 1 ≤ 9 - 1 So, 2x ≤ 8. Then divide both sides by 2:2x/2 ≤ 8/2 Therefore, x ≤ 4. Hmm, okay. So the second inequality gives me x ≤ 4.So putting both inequalities together, the solution set should be all x that are greater than 1 and less than or equal to 4. Let me check if I did everything right.First inequality: x > 1. Yeah, adding 1 to both sides is correct. Second inequality: multiply both sides by 3, subtract 1, divide by 2. Let me verify that again. Original inequality: (2x +1)/3 ≤ 3.Multiply both sides by 3: 2x +1 ≤ 9. Subtract 1: 2x ≤ 8. Divide by 2: x ≤ 4. Yep, that seems right.So combining both, x has to be greater than 1 and less than or equal to 4. So in interval notation, that would be (1, 4]. But let me make sure there's no overlap or something I missed.Wait, the first inequality is strict: x > 1, so 1 is not included. The second inequality includes 4. So x can't be 1, but can be 4. So combining both, x is between 1 and 4, not including 1 but including 4. Yeah, that makes sense.Is there any number that satisfies one inequality but not the other? Let's pick x = 2. First inequality: 2 -1 =1 >0? Wait, 1 is not greater than 0? Wait, hold on. Wait, the first inequality is x -1 >0. So x >1. So x=2 would be 2-1=1>0? 1>0 is true, so x=2 is okay. Wait, but 1 is not greater than 0? Wait, 1 is greater than 0. Wait, 1>0 is true, right? So if x=2, then x-1=1, which is greater than 0. So x=2 is okay.Wait, maybe I confused something. Let me check x=1. If x=1, then x-1=0, which is not greater than 0, so x=1 is excluded. So x must be greater than 1, so numbers like 1.1, 1.5, up to 4.What about x=4? Let's check both inequalities. First inequality: 4 -1=3>0, which is true. Second inequality: (2*4 +1)/3=(8+1)/3=9/3=3≤3, which is true. So x=4 is included.How about x=0? First inequality: 0-1=-1>0? No, so x=0 is excluded. x=5? First inequality: 5-1=4>0. Second inequality: (10 +1)/3=11/3≈3.666>3, so 3.666 is not ≤3. So x=5 is excluded. So the upper limit is indeed 4.So the solution is all real numbers x such that x>1 and x≤4, which is (1,4] in interval notation.Wait, let me just make sure I didn't make any algebraic mistakes. For the second inequality:(2x +1)/3 ≤3 Multiply both sides by 3: 2x +1 ≤9 Subtract 1: 2x ≤8 Divide by 2: x ≤4. Yes, that's correct.First inequality: x -1 >0 x>1. Yes, that's correct.So combining both, x must be greater than 1 and less than or equal to 4. So the answer is (1,4]. I think that's right.Final AnswerThe solution to the inequality system is boxed{(1, 4]}.</think>To solve the inequality system:[left{begin{array}{l}x - 1 > 0 frac{2x + 1}{3} leq 3end{array}right.]1. Solve the first inequality ( x - 1 > 0 ): [ x - 1 > 0 implies x > 1 ]2. Solve the second inequality ( frac{2x + 1}{3} leq 3 ): [ frac{2x + 1}{3} leq 3 ] Multiply both sides by 3: [ 2x + 1 leq 9 ] Subtract 1 from both sides: [ 2x leq 8 ] Divide by 2: [ x leq 4 ]3. Combine the solutions from both inequalities: - The first inequality gives ( x > 1 ) - The second inequality gives ( x leq 4 )Therefore, the solution to the system is the intersection of these intervals, which is ( (1, 4] ).[boxed{(1, 4]}],<think>Okay, so I need to solve this system of inequalities. Let me see. The system is:1. x - 1 > 02. (2x + 1)/3 ≤ 3Alright, let's take them one at a time. Starting with the first inequality: x - 1 > 0. Hmm, to solve for x, I can just add 1 to both sides, right? That would isolate the x. So if I add 1 to both sides:x - 1 + 1 > 0 + 1Which simplifies to:x > 1Okay, so the first inequality tells me that x has to be greater than 1. Got that. Now moving on to the second inequality: (2x + 1)/3 ≤ 3. Hmm, let me think. I need to solve for x here as well. Let me start by multiplying both sides by 3 to eliminate the denominator. That's a common first step when dealing with fractions.So, multiplying both sides by 3:3 * [(2x + 1)/3] ≤ 3 * 3The left side simplifies because the 3 in the numerator and the denominator cancel out, leaving me with:2x + 1 ≤ 9Now, I can subtract 1 from both sides to get:2x + 1 - 1 ≤ 9 - 1Which simplifies to:2x ≤ 8Then, divide both sides by 2:2x/2 ≤ 8/2So:x ≤ 4Alright, so the second inequality tells me that x has to be less than or equal to 4. Now, putting both inequalities together. The first one says x > 1, and the second one says x ≤ 4. So the solution set is all x values that are greater than 1 and less than or equal to 4.Let me double-check my work to make sure I didn't make any mistakes. For the first inequality: starting with x - 1 > 0, adding 1 gives x > 1. That seems straightforward. For the second inequality: (2x + 1)/3 ≤ 3. Multiplying both sides by 3 gives 2x + 1 ≤ 9. Subtract 1: 2x ≤ 8. Divide by 2: x ≤ 4. Yep, that looks right.So combining both results, x must be greater than 1 and at the same time less than or equal to 4. Therefore, the solution is the interval (1, 4]. In inequality terms, that's 1 < x ≤ 4.Wait a second, should I check if there's any overlap or if they contradict each other? Well, since 1 is less than 4, there's definitely an overlap. All numbers greater than 1 up to 4 satisfy both inequalities. So there's no contradiction here.Let me also test a value in each interval to be sure. Let's pick x = 2. Plugging into the first inequality: 2 - 1 = 1 > 0? Well, 1 is greater than 0, so that's true. For the second inequality: (2*2 + 1)/3 = (4 + 1)/3 = 5/3 ≈ 1.666..., which is ≤ 3. So that's true. So 2 works.What about x = 0? That's outside the solution. First inequality: 0 - 1 = -1 > 0? No, that's false. So 0 doesn't work, which is correct.How about x = 4? First inequality: 4 - 1 = 3 > 0, which is true. Second inequality: (2*4 + 1)/3 = (8 + 1)/3 = 9/3 = 3 ≤ 3. That's true. So x = 4 is included, which matches the x ≤ 4.What about x = 5? First inequality: 5 - 1 = 4 > 0, which is true, but the second inequality: (10 + 1)/3 = 11/3 ≈ 3.666... which is greater than 3. So 5 doesn't work, which is correct because it's greater than 4.And x = 1? First inequality: 1 - 1 = 0 > 0? No, which is why x must be greater than 1, not equal to. So 1 is excluded, which matches the solution.Alright, so all the test points check out. I think I did this correctly. So the solution is all real numbers x such that x is greater than 1 and less than or equal to 4. In interval notation, that's (1, 4].Final AnswerThe solution to the inequality system is boxed{(1, 4]}.</think>To solve the inequality system:[left{begin{array}{l}x - 1 > 0 frac{2x + 1}{3} leq 3end{array}right.]First, solve the inequality ( x - 1 > 0 ):[x - 1 > 0 implies x > 1]Next, solve the inequality ( frac{2x + 1}{3} leq 3 ):[frac{2x + 1}{3} leq 3]Multiply both sides by 3:[2x + 1 leq 9]Subtract 1 from both sides:[2x leq 8]Divide by 2:[x leq 4]Combining the solutions from both inequalities, we get:[1 < x leq 4]In interval notation, the solution is (boxed{(1, 4]}).]

🔑:To solve the given system of inequalities, we approach each inequality step by step.1. For the first inequality x - 1 > 0: - We add 1 to both sides of the inequality to isolate x: [ x - 1 + 1 > 0 + 1 ] - Simplifying both sides gives us: [ x > 1 ]2. For the second inequality frac{2x + 1}{3} leq 3: - We multiply both sides by 3 to eliminate the denominator: [ 3 cdot frac{2x + 1}{3} leq 3 cdot 3 ] - This simplifies to: [ 2x + 1 leq 9 ] - Subtracting 1 from both sides to solve for x gives: [ 2x leq 8 ] - Dividing by 2 to isolate x: [ x leq 4 ]Combining the results from both inequalities, we find that the solution set for the system is when x is greater than 1 and less than or equal to 4. Therefore, the solution set of the inequality system is:[boxed{1 < x leq 4}]

🧠:[<think>Okay, so I need to find the measure of angle BDC in quadrilateral ABCD. The problem gives me three pieces of information:1. Angle DAB is 150 degrees.2. The sum of angles DAC and ABD is 120 degrees.3. The difference between angles DBC and ABD is 60 degrees.Hmm, let me try to visualize the quadrilateral first. Let me sketch ABCD, with points labeled in order: A, B, C, D. But I need to remember that in a quadrilateral, the order of the points matters, so the sides are AB, BC, CD, DA. The angles given are at points A and B, so maybe it's a convex quadrilateral? Not sure yet. Let me note down all the given information step by step.First, angle DAB is 150 degrees. So angle at A between sides AD and AB is 150 degrees. That's angle DAB. Then, the sum of angle DAC and angle ABD is 120 degrees. Angle DAC is the angle at A between DA and AC, which is a diagonal. Angle ABD is the angle at B between AB and BD, another diagonal. Then, the third condition is that the difference between angle DBC (angle at B between DB and BC) and angle ABD is 60 degrees. So angle DBC minus angle ABD equals 60 degrees. Hmm, this seems to involve several angles related to the diagonals AC and BD. Maybe I can use some triangle properties here. Let me try to mark all these angles on the sketch.Starting with angle DAB = 150°. So at vertex A, the angle between sides AD and AB is 150°. Now, angle DAC is part of this angle, right? Because DAC is between DA and AC, so if I draw diagonal AC from A to C, then angle DAC is part of angle DAB. Similarly, angle CAB would be the other part of angle DAB. So angle DAB = angle DAC + angle CAB = 150°. So angle DAC + angle CAB = 150°. But the problem says angle DAC + angle ABD = 120°. So angle DAC + angle ABD = 120°, and angle DAC + angle CAB = 150°. Therefore, subtracting the two equations: (angle DAC + angle CAB) - (angle DAC + angle ABD) = 150° - 120°, which simplifies to angle CAB - angle ABD = 30°. So angle CAB = angle ABD + 30°. Got that.Now, the third condition is angle DBC - angle ABD = 60°. So angle DBC = angle ABD + 60°. Hmm. Let me see.So angle DBC is at vertex B, between DB and BC, so that's part of angle ABC. Similarly, angle ABD is at vertex B, between AB and BD. So angle ABC is angle ABD + angle DBC. Since angle DBC = angle ABD + 60°, then angle ABC = angle ABD + (angle ABD + 60°) = 2 angle ABD + 60°. So angle ABC is twice angle ABD plus 60°. Interesting.Now, let me think about the triangles involved. Let's look at triangle ABD. Wait, but BD is a diagonal. Hmm. Maybe triangle ABC or triangle BDC? Let me see.Wait, angle CAB is angle at A between CA and AB. And angle ABD is angle at B between AB and BD. If I can relate these angles through some triangle. Maybe triangle ABD? Let me see. In triangle ABD, the sum of angles should be 180°. But angle at A is angle BAD = 150°, angle at B is angle ABD, and angle at D is angle ADB. So 150° + angle ABD + angle ADB = 180°, which would mean angle ABD + angle ADB = 30°. But I don't know angle ADB yet.Alternatively, maybe triangle ABC. If I consider triangle ABC, angles at A, B, and C. But I don't know much about angles at C. Wait, but angle CAB is angle at A in triangle ABC, angle ABC is angle at B, and angle BCA is angle at C. Sum of angles is 180°. So angle CAB + angle ABC + angle BCA = 180°. We have angle CAB = angle ABD + 30°, angle ABC = 2 angle ABD + 60°, so substituting, we get (angle ABD + 30°) + (2 angle ABD + 60°) + angle BCA = 180°, which simplifies to 3 angle ABD + 90° + angle BCA = 180°, so 3 angle ABD + angle BCA = 90°, hence angle BCA = 90° - 3 angle ABD. Hmm, not sure if that helps yet.Alternatively, maybe looking at triangle BDC. The angle we need is angle BDC. In triangle BDC, the angles are angle BDC, angle DBC, and angle BCD. So angle DBC + angle BCD + angle BDC = 180°. If we can find angle DBC and angle BCD, then we can find angle BDC. We know angle DBC = angle ABD + 60°, so if we can express angle BCD in terms of angle ABD, then maybe.But angle BCD is part of angle BCA (if point C is connected via diagonal AC). Wait, angle BCA is angle at C between BC and CA. So angle BCD is part of angle BCA if D is on the other side. Wait, maybe not. Hmm, this is getting a bit confusing. Let me try to label the angles step by step.Let me denote angle ABD as x degrees. Then, angle DBC = x + 60°, as given. Then angle ABC = angle ABD + angle DBC = x + (x + 60°) = 2x + 60°, as before. Also, angle CAB = angle ABD + 30° = x + 30°, from earlier. So in triangle ABC, angles are:- At A: angle CAB = x + 30°- At B: angle ABC = 2x + 60°- At C: angle BCA = 90° - 3x (from above)So angle BCA = 90° - 3x. Hmm. Now, angle BCA is at vertex C between BC and CA. If we can relate this to angle BCD, which is part of angle BCA if point D is somewhere such that CD is connected. Wait, maybe angle BCD is the same as angle BCA? That would be if points D and A are on the same side with respect to BC. But in a quadrilateral, points are connected in order, so ABCD, so BC is connected to CD. So angle BCD is the angle at C between BC and CD, which is different from angle BCA (which is between BC and CA). So angle BCD is different from angle BCA. Therefore, angle BCA and angle BCD are different angles. So that complicates things.Alternatively, maybe consider triangle ADC. If we can find some angles there. But angle at A is angle DAC, which is part of angle DAB = 150°. So angle DAC is given to be part of the sum angle DAC + angle ABD = 120°, so angle DAC = 120° - angle ABD = 120° - x. Since angle DAB = angle DAC + angle CAB = 150°, then angle CAB = 150° - angle DAC = 150° - (120° - x) = 30° + x. Which matches what we had before, since angle CAB = x + 30°. So that's consistent.So angle DAC = 120° - x. Then, in triangle ADC, angles at A is angle DAC = 120° - x, angle at D is angle ADC, and angle at C is angle ACD. But I don't know angles at D or C. Hmm. Maybe not helpful.Alternatively, looking at triangle ABD. As I mentioned before, in triangle ABD, angles at A is 150°, at B is x (angle ABD), and at D is angle ADB. So 150 + x + angle ADB = 180 => angle ADB = 30 - x. Hmm, angle ADB = 30° - x. That's interesting. So angle at D in triangle ABD is 30° - x. Maybe this can help us in another triangle.What about triangle BDC? The angles in triangle BDC are angle DBC = x + 60°, angle BDC (which we need to find), and angle BCD. If we can find angle BCD, then we can solve for angle BDC.Alternatively, maybe triangle BCD? Let me see. Wait, if we can find angle BCD. How?Alternatively, think about the quadrilateral ABCD. The sum of all interior angles is 360°. We know angle DAB = 150°, angle ABC = 2x + 60°, and we need angles at C and D. But angle at D is angle ADC, and angle at C is angle BCD. Wait, but in quadrilaterals, the angles are at each vertex. So angle at A is 150°, angle at B is 2x + 60°, angle at C is angle BCD + angle BCA? Wait, no. In quadrilateral ABCD, angle at C is angle BCD + angle BCA? Wait, no. Because in a quadrilateral, each angle is formed by two adjacent sides. So angle at C is between BC and CD. Similarly, angle at D is between CD and DA. So angle at C is angle BCD, and angle at D is angle CDA. So sum of angles: 150° (A) + (2x + 60°) (B) + angle BCD (C) + angle CDA (D) = 360°. So angle BCD + angle CDA = 360° - 150° - (2x + 60°) = 360° - 150° - 2x - 60° = 150° - 2x. So angle BCD + angle CDA = 150° - 2x.But how does that help? If I can find expressions for angle BCD and angle CDA.Wait, angle CDA is the angle at D between CD and DA. In triangle ADC, angle at D is angle ADC (which is angle CDA in the quadrilateral), angle at A is angle DAC = 120° - x, angle at C is angle ACD. So sum of angles in triangle ADC: (120° - x) + angle ADC + angle ACD = 180°. So angle ADC + angle ACD = 60° + x. But angle ADC is the same as angle CDA in the quadrilateral, which we already mentioned. So angle CDA + angle ACD = 60° + x.But angle ACD is part of angle BCA in triangle ABC. Wait, angle BCA in triangle ABC is 90° - 3x. But angle ACD is part of angle BCA. Wait, angle BCA is the angle at C between BC and CA, and angle ACD is the angle at C between CA and CD. So angle BCA = angle BCD + angle ACD? Wait, no. If you have point C connected to B, A, and D, then angle at C between BC and CD is angle BCD, and angle at C between BC and CA is angle BCA, and between CA and CD is angle ACD. So angle BCA and angle ACD are adjacent angles at point C, so their sum is equal to angle BCD? Wait, no. Wait, if you have point C with three lines: CB, CA, CD. Then the angles at C would be angle BCA (between BC and CA), angle ACD (between CA and CD), and angle BCD (between BC and CD). But actually, angle BCD is the angle between BC and CD, which is adjacent to angle DCE if there's an extension, but here in quadrilateral, angle BCD is just that angle. Wait, maybe the three angles at point C: angle BCA, angle ACD, and angle BCD. Wait, no, angle BCA and angle ACD are adjacent and form angle BCD? No, that might not be the case. Wait, maybe angle BCD is equal to angle BCA + angle ACD. Let me visualize this. If from point C, we have lines to B, A, and D. Then, angle between BC and CD is angle BCD. The angle between BC and CA is angle BCA. The angle between CA and CD is angle ACD. So angle BCD is actually angle BCA + angle ACD. Is that correct? Yes, because if you have two adjacent angles, their sum is the total angle between the outer sides. So angle BCD = angle BCA + angle ACD. Therefore, angle BCD = (90° - 3x) + angle ACD. But from triangle ADC, angle ACD = 60° + x - angle CDA. So angle BCD = (90° - 3x) + (60° + x - angle CDA) = 150° - 2x - angle CDA. But from the quadrilateral, angle BCD + angle CDA = 150° - 2x. So substituting angle BCD from above: (150° - 2x - angle CDA) + angle CDA = 150° - 2x, which checks out. So that's consistent but doesn't give new information.Hmm, maybe need another approach. Let's see, since we have angle ADB = 30° - x from triangle ABD. Maybe consider triangle ADC. Wait, angle ADC is angle CDA in the quadrilateral, which we can denote as y. Then, from triangle ADC: angle DAC = 120° - x, angle ADC = y, angle ACD = 60° + x - y. Then, angle BCD = angle BCA + angle ACD = (90° - 3x) + (60° + x - y) = 150° - 2x - y. But from quadrilateral angle sum: angle BCD + angle CDA = 150° - 2x. So substituting angle BCD = 150° - 2x - y and angle CDA = y, we get 150° - 2x - y + y = 150° - 2x, which is true. So again, no new info.Perhaps we need to involve triangle BDC. In triangle BDC, angles are angle DBC = x + 60°, angle BDC (which we need to find, let's call it z), and angle BCD = 150° - 2x - y. But angle BCD is also equal to angle BCA + angle ACD, which is (90° - 3x) + (60° + x - y) = 150° - 2x - y. So in triangle BDC: (x + 60°) + z + (150° - 2x - y) = 180°. Simplifying: x + 60 + z + 150 - 2x - y = 180 => (-x) + z + 210 - y = 180 => -x + z - y = -30 => z = x + y - 30. So angle BDC = x + y - 30°. But we need to express z in terms of known variables or find a relation between x and y.From quadrilateral angle sum, angle BCD + angle CDA = 150° - 2x, which is 150° - 2x = (150° - 2x - y) + y = 150° - 2x. Again, not helpful. Wait, maybe from triangle ABD, angle ADB = 30° - x. Then, in triangle ADC, angle ADC = y, angle DAC = 120° - x, angle ACD = 60° + x - y. Also, angle ADB = 30° - x is part of triangle ABD. Is there a way to relate angle ADB to angle ADC or something else?Alternatively, consider triangle ABD and triangle CBD. Not sure. Wait, maybe the diagonals intersect at some point? Wait, in the problem statement, are the diagonals AC and BD intersecting? If so, maybe using the intersecting angles. But the problem doesn't mention the intersection point. Maybe that's overcomplicating.Alternatively, consider using the Law of Sines or Law of Cosines in some triangles. Since we have various angles and maybe can relate sides. But since no side lengths are given, maybe express ratios.Let me try that. In triangle ABD, using the Law of Sines: AB/sin(angle ADB) = AD/sin(angle ABD) = BD/sin(angle BAD). So AB/sin(30° - x) = AD/sin(x) = BD/sin(150°). Similarly, in triangle ABC, using Law of Sines: AB/sin(angle BCA) = BC/sin(angle CAB) = AC/sin(angle ABC). So AB/sin(90° - 3x) = BC/sin(x + 30°) = AC/sin(2x + 60°). Maybe relate these ratios?Also, in triangle BDC, using Law of Sines: BD/sin(angle BCD) = BC/sin(angle BDC) = CD/sin(angle DBC). So BD/sin(150° - 2x - y) = BC/sin(z) = CD/sin(x + 60°). But I don't know CD or BC or BD. Hmm.Alternatively, let's see if we can relate BD from triangle ABD and BD from triangle BDC. From triangle ABD: BD = AB * sin(x) / sin(150°). Since in Law of Sines, BD / sin(angle BAD) = AB / sin(angle ADB). So BD = AB * sin(angle BAD) / sin(angle ADB). Wait, angle BAD is 150°, angle ADB is 30° - x. So BD = AB * sin(150°) / sin(30° - x). Similarly, in triangle ABC, AB / sin(angle BCA) = BC / sin(angle CAB). So BC = AB * sin(angle CAB) / sin(angle BCA) = AB * sin(x + 30°) / sin(90° - 3x).Now, in triangle BDC, BD / sin(angle BCD) = BC / sin(z). So BD = BC * sin(angle BCD) / sin(z). Substitute BD from above and BC from here:AB * sin(150°) / sin(30° - x) = [AB * sin(x + 30°) / sin(90° - 3x)] * sin(angle BCD) / sin(z)Simplify AB cancels out:sin(150°) / sin(30° - x) = [sin(x + 30°) / sin(90° - 3x)] * sin(angle BCD) / sin(z)But angle BCD is 150° - 2x - y, and we have angle BCD + angle CDA = 150° - 2x, so angle CDA = y = 150° - 2x - angle BCD. Wait, but angle BCD = 150° - 2x - y. So substituting this into angle BCD, we get angle BCD = 150° - 2x - (150° - 2x - angle BCD). Wait, that leads to angle BCD = angle BCD, which is circular. Not helpful.Alternatively, since angle CDA = y and angle CDA + angle BCD = 150° - 2x, then angle BCD = 150° - 2x - y, so in the equation above:sin(150°) / sin(30° - x) = [sin(x + 30°) / sin(90° - 3x)] * sin(150° - 2x - y) / sin(z)But z = angle BDC = x + y - 30°, from earlier.So sin(150°) / sin(30° - x) = [sin(x + 30°) / sin(90° - 3x)] * sin(150° - 2x - y) / sin(x + y - 30°)This seems complicated. Maybe need another equation involving y.From triangle ADC, angles sum to 180°: angle DAC + angle ADC + angle ACD = 180°, which is (120° - x) + y + (60° + x - y) = 180°, simplifying: 120 - x + y + 60 + x - y = 180 => 180° = 180°. So that's just an identity, doesn't help.Hmm, maybe this approach is too algebraically intensive. Let me see if I can assign a variable to x and try to find its value.Let me denote angle ABD as x. Then, angle DBC = x + 60°, angle CAB = x + 30°, angle ABC = 2x + 60°, angle BCA = 90° - 3x, angle DAC = 120° - x, angle ADB = 30° - x.Now, in triangle ABC, sides:AB / sin(angle BCA) = BC / sin(angle CAB) = AC / sin(angle ABC)So AB / sin(90° - 3x) = BC / sin(x + 30°) = AC / sin(2x + 60°)Similarly, in triangle ABD:AB / sin(angle ADB) = BD / sin(angle BAD) = AD / sin(angle ABD)So AB / sin(30° - x) = BD / sin(150°) = AD / sin(x)Therefore, BD = AB * sin(150°) / sin(30° - x)Similarly, in triangle BDC:BD / sin(angle BCD) = BC / sin(angle BDC) = CD / sin(angle DBC)But BD and BC can be related through the expressions above.From triangle ABC, BC = AB * sin(x + 30°) / sin(90° - 3x)From triangle ABD, BD = AB * sin(150°) / sin(30° - x)So in triangle BDC, BD / sin(angle BCD) = BC / sin(angle BDC)Substituting BD and BC:[AB * sin(150°) / sin(30° - x)] / sin(angle BCD) = [AB * sin(x + 30°) / sin(90° - 3x)] / sin(z)Simplify AB cancels:sin(150°) / [sin(30° - x) * sin(angle BCD)] = sin(x + 30°) / [sin(90° - 3x) * sin(z)]But angle BCD = 150° - 2x - y, and z = x + y - 30°, but we need to express angle BCD in terms of x. Wait, angle BCD is also part of quadrilateral angle sum. If only we can express y in terms of x.Wait, angle CDA = y. From triangle ADC, angle CDA + angle ACD + angle DAC = 180°, so y + angle ACD + (120° - x) = 180°, so angle ACD = 60° + x - y. But angle ACD is part of angle BCA. Wait, angle BCA = angle BCD + angle ACD? No, earlier we thought angle BCD = angle BCA + angle ACD. Wait, earlier conclusion was angle BCD = angle BCA + angle ACD. Let me verify.At point C, the angles around point C are:- angle BCA (between BC and CA)- angle ACD (between CA and CD)- angle BCD (between BC and CD)But angle BCD is the angle between BC and CD, which is formed by the two adjacent angles if we split it with CA. So angle BCD = angle BCA + angle ACD. Yes, that's correct. Therefore:angle BCD = angle BCA + angle ACD = (90° - 3x) + (60° + x - y) = 150° - 2x - y.But from quadrilateral angle sum, angle BCD + angle CDA = 150° - 2x. So (150° - 2x - y) + y = 150° - 2x, which holds. Therefore, angle BCD is 150° - 2x - y, but since angle BCD + angle CDA = 150° - 2x, angle CDA = y = 150° - 2x - angle BCD. But since angle BCD = 150° - 2x - y, this is again circular. So, perhaps we need another equation.Wait, from triangle BDC: angle BDC = z = x + y - 30°, as established earlier.But angle CDA = y. From triangle ADC, angle ACD = 60° + x - y.But angle ACD is part of angle BCA. Wait, angle BCA = 90° - 3x. So angle ACD = angle BCA - angle something? Wait, no. Wait, angle BCA is 90° - 3x, and angle ACD is 60° + x - y. But angle BCA and angle ACD are adjacent at point C, separated by CA. So angle BCA + angle ACD = angle BCD. Which we already have.Hmm. Maybe this is getting too convoluted. Let's try plugging in possible values for x. Since angles can't be negative, let's see:From triangle ABD, angle ADB = 30° - x. Since angles must be positive, 30° - x > 0 => x < 30°. Also, angle ABD = x > 0. So x is between 0° and 30°.From angle BCA = 90° - 3x. Since angle BCA must be positive, 90° - 3x > 0 => x < 30°, which is consistent with above.Also, angle DBC = x + 60°, but angle DBC is part of angle ABC = 2x + 60°. Since angle ABC must be less than 180°, 2x + 60° < 180° => 2x < 120° => x < 60°, which is already satisfied since x < 30°.Additionally, angle CAB = x + 30°, which must be positive, which it is since x > 0.Also, in triangle ABC, angle BCA = 90° - 3x must be positive, so x < 30°, as above.So x is between 0° and 30°. Let's try to find x.Looking back at the problem, perhaps there's a way to use the fact that angle DAB is 150°, and maybe construct some auxiliary lines. For example, if I extend some sides or draw diagonals.Alternatively, maybe use the Law of Sines in triangles ABD and CBD.In triangle ABD:AB / sin(30° - x) = BD / sin(150°)In triangle BDC:BD / sin(angle BCD) = BC / sin(z)And in triangle ABC:BC = AB * sin(x + 30°) / sin(90° - 3x)So BD = AB * sin(150°) / sin(30° - x)Then, substituting BD and BC into the second equation:[ AB * sin(150°) / sin(30° - x) ] / sin(angle BCD) = [ AB * sin(x + 30°) / sin(90° - 3x) ] / sin(z)Cancel AB:[ sin(150°) / sin(30° - x) ] / sin(angle BCD) = [ sin(x + 30°) / sin(90° - 3x) ] / sin(z)Multiply both sides by sin(angle BCD) and sin(z):sin(150°) * sin(z) / sin(30° - x) = sin(x + 30°) * sin(angle BCD) / sin(90° - 3x)But angle BCD = 150° - 2x - y, and z = x + y - 30°. Since angle CDA = y, and angle CDA is part of triangle ADC, which we know angle ACD = 60° + x - y.Alternatively, recall from triangle ADC, angle ACD = 60° + x - y, and angle BCA = 90° - 3x. Since angle BCA and angle ACD are adjacent at point C, angle BCA + angle ACD = angle BCD. Therefore:angle BCD = (90° - 3x) + (60° + x - y) = 150° - 2x - yBut from quadrilateral angle sum, angle BCD + angle CDA = 150° - 2x, and angle CDA = y. So 150° - 2x - y + y = 150° - 2x, which is just an identity.This seems like we're going in circles. Maybe we can express y in terms of x from another equation.Wait, from triangle ADC: angle DAC = 120° - x, angle ADC = y, angle ACD = 60° + x - y. So, using Law of Sines in triangle ADC:AD / sin(angle ACD) = AC / sin(angle ADC) = CD / sin(angle DAC)Similarly, in triangle ABC: AC / sin(angle ABC) = AB / sin(angle BCA) = BC / sin(angle CAB)So AC = AB * sin(2x + 60°) / sin(90° - 3x)From triangle ADC: AC = AD * sin(y) / sin(60° + x - y)So equating the two expressions for AC:AB * sin(2x + 60°) / sin(90° - 3x) = AD * sin(y) / sin(60° + x - y)From triangle ABD: AD / sin(x) = AB / sin(30° - x). So AD = AB * sin(x) / sin(30° - x)Substitute AD into equation:AB * sin(2x + 60°) / sin(90° - 3x) = [ AB * sin(x) / sin(30° - x) ] * sin(y) / sin(60° + x - y)Cancel AB:sin(2x + 60°) / sin(90° - 3x) = [ sin(x) / sin(30° - x) ] * [ sin(y) / sin(60° + x - y) ]This is getting really complicated. Maybe there's a simpler way.Wait, let me consider the possibility that x is a specific value. Since x is between 0° and 30°, maybe x = 15°, which would make angle DBC = 15° + 60° = 75°, angle ABD = 15°, angle CAB = 15° + 30° = 45°, angle ABC = 2*15° + 60° = 90°, angle BCA = 90° - 3*15° = 45°, angle DAC = 120° - 15° = 105°, angle ADB = 30° - 15° = 15°, angle BCD = 150° - 2*15° - y = 120° - y. Then, quadrilateral angles: angle BCD + angle CDA = 150° - 2*15° = 120°, so angle CDA = y = 120° - angle BCD. If angle BCD = 120° - y, then angle CDA = y = 120° - (120° - y) => y = y. Still circular.But if we check triangle ABC with x=15°, angles are 45°, 90°, 45°, which is a 45-90-45 triangle. So sides would be AB, BC, AC. If AB is 1, then BC = AB * sin(45°) / sin(45°) = 1, AC = AB * sin(90°) / sin(45°) = √2. Then in triangle ABD, AD = AB * sin(15°) / sin(15°) = 1 (since AD / sin(15°) = AB / sin(15°)), so AD=1. Then, BD = AB * sin(150°) / sin(15°) = 1 * 0.5 / sin(15°). Sin(15°) is approximately 0.2588, so BD ≈ 0.5 / 0.2588 ≈ 1.931. Then, in triangle BDC, BD ≈1.931, BC=1, angle DBC=75°, angle BCD=120° - y, angle BDC=z=15° + y - 30°= y -15°. Using Law of Sines: BD / sin(angle BCD) = BC / sin(z). So 1.931 / sin(120° - y) = 1 / sin(y -15°). Then, 1.931 sin(y -15°) = sin(120° - y). Let me check if there's a y that satisfies this. Let me try y = 45°, then left side: 1.931 sin(30°) ≈ 1.931 * 0.5 ≈ 0.9655. Right side: sin(75°) ≈ 0.9659. Close enough. So y ≈45°, z=45° -15°=30°. Then angle BDC=30°. So maybe the answer is 30°? Let me check if this works.If x=15°, y=45°, z=30°, then:- angle ABD=15°, angle DBC=75°, angle ABC=90°, angle CAB=45°, angle BCA=45°, angle DAC=105°, angle ADB=15°, angle CDA=45°, angle BCD=75°, angle BDC=30°.Check quadrilateral angles: 150° (A) + 90° (B) + 75° (C) + 45° (D) = 360°, which sums to 360°, correct.Check triangle ADC: angle DAC=105°, angle ADC=45°, angle ACD=60° + 15° -45°=30°, sum 180°, correct.Check triangle BDC: angle DBC=75°, angle BCD=75°, angle BDC=30°, sum 180°, correct.All angles check out. So angle BDC=30°. So maybe the answer is 30 degrees.But wait, I assumed x=15°, is there a reason for that? Or is there a different value?Alternatively, let's suppose angle BDC=30°, then work backwards. If z=30°, then from z = x + y -30°, 30°=x + y -30°, so x + y=60°. From quadrilateral angle sum, angle BCD + angle CDA=150° -2x, which is 75° +45°=120°=150° -2x => 2x=30° =>x=15°, which matches. So if x=15°, then angle BDC=30°, which seems consistent.Alternatively, is there another possible solution? Let me check x=10°. Then angle DBC=70°, angle ABD=10°, angle CAB=40°, angle ABC=80°, angle BCA=90° -30°=60°, angle DAC=110°, angle ADB=20°, angle BCD=150° -20 - y, but angle BCD + y=150° -20=130°, so angle BCD=130° - y. Then in triangle BDC: angle DBC=70°, angle BCD=130° - y, angle BDC=10 + y -30= y -20. Sum:70 +130 - y + y -20=180, which is 70 +130 -20=180, 180=180. So again, consistent but not helpful.But in this case, using Law of Sines:From triangle ABD: BD = AB * sin(150°)/sin(20°). From triangle ABC: BC = AB * sin(40°)/sin(60°). Then in triangle BDC: BD/sin(130° - y)= BC/sin(y -20). But angle BCD=130° - y and angle BDC= y -20. Also, from quadrilateral angle sum, angle BCD + y=130°, so angle BCD=130° - y, which is consistent. So sin(BD)/sin(angle BCD)= BC/sin(angle BDC). Plugging in BD and BC:[ AB * sin(150°)/sin(20°) ] / sin(130° - y) = [ AB * sin(40°)/sin(60°) ] / sin(y -20°)Cancel AB:sin(150°)/(sin(20°) sin(130° - y)) = sin(40°)/(sin(60°) sin(y -20°))But sin(150°)=0.5, sin(60°)=√3/2≈0.866, sin(40°)≈0.6428, sin(20°)≈0.3420.So:0.5 / (0.3420 * sin(130° - y)) ≈ 0.6428 / (0.866 * sin(y -20°))Cross-multiplying:0.5 * 0.866 * sin(y -20°) ≈ 0.6428 * 0.3420 * sin(130° - y)0.433 * sin(y -20°) ≈ 0.2197 * sin(130° - y)Let’s approximate:Left side: ~0.433 sin(y -20°)Right side: ~0.2197 sin(130° - y)Let me try y=45°:Left side:0.433 sin(25°)≈0.433*0.4226≈0.183Right side:0.2197 sin(85°)≈0.2197*0.9962≈0.218Not equal. Try y=50°:Left side:0.433 sin(30°)=0.433*0.5≈0.2165Right side:0.2197 sin(80°)=0.2197*0.9848≈0.216Close enough. So y≈50°, then angle BDC=50 -20=30°, same as before. Wait, but if x=10°, y=50°, then angle BDC=30°. But this would imply angle BDC is always 30°, regardless of x? But how?Wait, but in the case when x=15°, angle BDC=30°, when x=10°, also angle BDC=30°. Is angle BDC always 30° regardless of x? That can't be unless there's a unique solution.Wait, this suggests that angle BDC is always 30°, which contradicts unless it's indeed the case. Wait, let's try with x=20°. Then angle ABD=20°, angle DBC=80°, angle CAB=50°, angle ABC=100°, angle BCA=90°-60°=30°, angle DAC=100°, angle ADB=10°, angle BCD=150°-40 - y=110° - y. From quadrilateral sum: angle BCD + angle CDA=150°-40=110°, so angle CDA= y=110°-angle BCD. Which is again, angle BCD=110°-y. Then in triangle BDC: angle DBC=80°, angle BCD=110°-y, angle BDC=20+y-30= y -10°. Sum:80 +110 - y + y -10=180, which holds.Using Law of Sines:BD / sin(angle BCD)= BC / sin(angle BDC)BD from triangle ABD: AB * sin(150°)/sin(10°). BC from triangle ABC: AB * sin(50°)/sin(30°). So:[ AB * 0.5 / sin(10°) ] / sin(110° - y) = [ AB * sin(50°) / 0.5 ] / sin(y -10°)Cancel AB:0.5 / (sin(10°) sin(110° - y)) = sin(50°) / (0.5 sin(y -10°))Multiply both sides by 0.5 sin(y -10°):0.25 / (sin(10°) sin(110° - y)) = sin(50°) sin(y -10°)Hmm, this is complicated. Let's plug in y=60°, then angle BDC=50°, but this might not work. Wait, if x=20°, and angle BDC= y -10°, and angle BCD=110° - y. Let's try y=60°, angle BDC=50°, angle BCD=50°. Then triangle BDC angles:80°,50°,50°, sum 180°, works. Let's check Law of Sines:BD / sin(50°)= BC / sin(50°), which implies BD=BC. From triangle ABD, BD=AB * sin(150°)/sin(10°)=AB*0.5 /0.1736≈AB*2.88. From triangle ABC, BC=AB*sin(50°)/0.5≈AB*0.7660 /0.5≈AB*1.532. So BD≈2.88 AB, BC≈1.532 AB, which are not equal. So this contradicts. Therefore, y=60° is not valid.Alternatively, trying y=40°, angle BDC=30°, angle BCD=70°. Then, BD / sin(70°)= BC / sin(30°). From triangle ABD, BD=AB*0.5/sin(10°)≈2.88 AB. From triangle ABC, BC=AB*sin(50°)/0.5≈1.532 AB. Then, BD / sin(70°)=2.88 AB /0.9397≈3.064 AB. BC / sin(30°)=1.532 AB /0.5=3.064 AB. So this works! So when y=40°, angle BDC=30°, angle BCD=70°, BD / sin(70°)=BC / sin(30°), and this holds. So in this case, angle BDC=30°.So regardless of x, angle BDC=30°. Therefore, the answer is 30 degrees. This must be the case. The problem must have angle BDC=30°, regardless of the value of x. Therefore, the answer is boxed{30} degrees.Final Answerboxed{30}</think>Given quadrilateral (ABCD) with the following conditions:1. (angle DAB = 150^circ)2. (angle DAC + angle ABD = 120^circ)3. (angle DBC - angle ABD = 60^circ)We need to find (angle BDC).1. Angle Relationships: - From (angle DAB = 150^circ), we know (angle DAC + angle CAB = 150^circ). - Given (angle DAC + angle ABD = 120^circ), subtracting these equations gives (angle CAB - angle ABD = 30^circ), so (angle CAB = angle ABD + 30^circ). - Given (angle DBC - angle ABD = 60^circ), we have (angle DBC = angle ABD + 60^circ).2. Triangle Analysis: - Let (angle ABD = x). Then, (angle DBC = x + 60^circ), (angle CAB = x + 30^circ), and (angle ABC = 2x + 60^circ). - In (triangle ABC), the angles are (angle CAB = x + 30^circ), (angle ABC = 2x + 60^circ), and (angle BCA = 90^circ - 3x).3. Using Law of Sines and Quadrilateral Angle Sum: - Considering the angles around point (C) and the quadrilateral angle sum, we find that (angle BCD + angle CDA = 150^circ - 2x). - Through detailed analysis using the Law of Sines in triangles (ABD), (ABC), and (BDC), we find that regardless of the value of (x), (angle BDC) consistently resolves to (30^circ).Thus, the measure of (angle BDC) is (boxed{30}).,<think>Okay, so I need to find the measure of angle BDC in quadrilateral ABCD. Let me try to visualize the quadrilateral first. Maybe I should draw a rough sketch to get a sense of where the points are. Let me label the quadrilateral ABCD with the given angles.First, it's given that angle DAB is 150 degrees. That's the angle at vertex A between sides DA and AB. Then, there are two other angle conditions: angle DAC plus angle ABD equals 120 degrees, and angle DBC minus angle ABD equals 60 degrees. Hmm, okay. So we have some relationships between these angles. My goal is to find angle BDC, which is the angle at vertex D between sides BD and DC.Let me start by trying to note down all the given information:1. angle DAB = 150°2. angle DAC + angle ABD = 120°3. angle DBC - angle ABD = 60°Need to find angle BDC.Let me assign variables to some of these angles to make it easier. Let's let angle DAC = x and angle ABD = y. Then, according to the second condition, x + y = 120°. That's equation (1).From the third condition, angle DBC - angle ABD = 60°, so angle DBC = y + 60°. Let's call angle DBC = z, so z = y + 60°. That's equation (3).Also, angle DAB is 150°, which is angle at A between DA and AB. Since angle DAC is x, then angle CAB = angle DAB - angle DAC = 150° - x. So angle CAB = 150° - x.Now, let's consider triangle ABC. Wait, but ABCD is a quadrilateral, so maybe triangle ABC is part of it. But angle ABD is y, which is in triangle ABD. Wait, maybe I need to consider triangles ABD and BDC?Wait, angle ABD is in triangle ABD, and angle DBC is in triangle DBC. Maybe triangles ABD and BDC share the side BD. Maybe I can use some triangle angle properties or laws of sines or cosines?But first, let's see if we can find the values of x and y. From equation (1): x + y = 120°. If we can find another equation involving x and y, we can solve for them.Looking at angle CAB = 150° - x. Then, in triangle ABC, we have angles at A, B, and C. Wait, but we might not know angle ABC or angle BCA. Hmm. Maybe triangle ABD?Wait, angle ABD is y, angle BAD is angle DAB = 150°, so in triangle ABD, angles at A, B, and D. Wait, angle at A is 150°, angle at B is y, so angle at D in triangle ABD would be 180° - 150° - y = 30° - y. But is that correct?Wait, angle BAD is 150°, angle ABD is y, so the third angle in triangle ABD is angle ADB = 180° - 150° - y = 30° - y. So angle ADB = 30° - y. Hmm.Now, perhaps looking at triangle BDC. The angles we might need are angle DBC = z = y + 60°, angle BDC (which we need to find), and angle BCD. But we don't know angle BCD. Alternatively, maybe quadrilateral angles.Wait, in quadrilateral ABCD, the sum of all interior angles is 360°. So angle at A (150°) + angle at B + angle at C + angle at D = 360°. But we don't know angles at B, C, D. Maybe we can express them in terms of other variables.Alternatively, maybe considering triangles connected by the diagonal BD. Since BD splits the quadrilateral into triangles ABD and BCD.In triangle ABD, we have angles at A: 150°, at B: y°, at D: 30° - y°, as previously established.In triangle BCD, we have angle at B: z = y + 60°, angle at D: angle BDC (which we need to find), and angle at C. So if we can find angle at C or some other angles, maybe we can relate them.Alternatively, perhaps looking at triangle ADC. Wait, angle DAC is x°, angle at A in triangle ADC is x°, but we don't know other angles. Hmm.Wait, perhaps considering triangle ABC. Let me see. In triangle ABC, angle at A is angle CAB = 150° - x°, angle at B is angle ABC. But angle ABC can be split into angle ABD and angle DBC. Since angle ABD = y and angle DBC = z = y + 60°, so angle ABC = y + z = y + y + 60° = 2y + 60°. Therefore, in triangle ABC, angles are:At A: 150° - xAt B: 2y + 60°At C: ?Sum of angles in triangle is 180°, so angle at C in triangle ABC is 180° - (150° - x + 2y + 60°) = 180° - 210° + x - 2y = -30° + x - 2y. Hmm, angle at C is x - 2y - 30°. That seems problematic because angles can't be negative. So either my reasoning is flawed, or I need to check this.Wait, let's re-express the angles in triangle ABC. Wait, angle at A in triangle ABC is angle CAB = 150° - x. Angle at B is angle ABC, which is angle ABD + angle DBC = y + z. Since angle DBC is z = y + 60°, so angle ABC = y + (y + 60°) = 2y + 60°, as I had before.So angle at C would be 180° - (angle at A + angle at B) = 180° - (150° - x + 2y + 60°) = 180° - 210° + x - 2y = -30° + x - 2y. Wait, this gives angle at C as x - 2y - 30°, which must be a positive angle. Therefore, x - 2y - 30° > 0. But since we have x + y = 120°, we can write x = 120° - y. Substituting into this inequality:(120° - y) - 2y - 30° > 0 => 90° - 3y > 0 => 3y < 90° => y < 30°. So angle ABD = y must be less than 30°. That's a constraint.But this angle at C is negative unless 3y < 90°, i.e., y < 30°. So angle at C is x - 2y - 30°, which must be positive. Therefore, x - 2y - 30° > 0. Given x = 120° - y, substitute:120° - y - 2y - 30° > 0 => 90° - 3y > 0 => y < 30°. So that's consistent.So angle at C in triangle ABC is x - 2y - 30°, which is (120° - y) - 2y - 30° = 90° - 3y. So angle C is 90° - 3y. But angle C in triangle ABC is also angle BCD in quadrilateral ABCD. Wait, is that correct? Because in the quadrilateral ABCD, angle at C is angle BCD plus angle ACB or something else? Wait, maybe not.Wait, in quadrilateral ABCD, angle at C is angle BCD. But in triangle ABC, angle at C is angle ACB. Wait, unless points A, B, C, D are arranged such that angle BCD is the same as angle ACB. That might not necessarily be the case unless D lies on some specific position.Hmm, perhaps I need to reconsider. Maybe triangle BCD: angle at C is angle BCD, which is different from angle ACB in triangle ABC. So perhaps angle at C in triangle ABC is angle ACB, and angle at C in quadrilateral ABCD is angle BCD. So unless those are the same, which they might not be. Therefore, my previous assumption might be incorrect.This seems complicated. Maybe another approach. Let's consider triangle ADC. In triangle ADC, we have angle at A: angle DAC = x°, angle at D: angle ADC, and angle at C: angle ACD. But unless we have more information, this might not help.Alternatively, let's consider triangle BDC, which is part of the quadrilateral. In triangle BDC, angles are:At B: angle DBC = z = y + 60°At D: angle BDC = ? (what we need to find)At C: angle BCD = ?Sum of angles in triangle BDC is 180°, so angle BDC + z + angle BCD = 180°. Therefore, angle BDC = 180° - z - angle BCD. But we need to find angle BCD.Alternatively, let's consider triangle ABD and triangle BDC. If I can relate sides using the Law of Sines or something.In triangle ABD:angle at A: 150°angle at B: y°angle at D: 30° - y°So by Law of Sines,AB / sin(30° - y) = BD / sin(150°) = AD / sin(y)Similarly, in triangle BDC:angle at B: z = y + 60°angle at D: angle BDC = ?angle at C: angle BCD = ?Law of Sines:BD / sin(angle BCD) = BC / sin(angle BDC) = CD / sin(z)But since we don't know the sides, this might not directly help. Unless we can express some sides in terms of others.Alternatively, maybe there is a point where lines intersect or some cyclic quadrilateral properties? But I don't have information about sides or other angles.Wait, perhaps using the fact that angle DAC + angle ABD = 120°, which is x + y = 120°, and angle DBC = y + 60°, so z = y + 60°.Given that angle at A (DAB) is 150°, angle DAC is x, so angle CAB is 150° - x. Which is also angle at A in triangle ABC.In triangle ABC, angles:A: 150° - xB: 2y + 60°C: 90° - 3ySum is 150° - x + 2y + 60° + 90° - 3y = 150° + 60° + 90° - x - y = 300° - (x + y). But x + y = 120°, so 300° - 120° = 180°, which checks out. So that's good.But angle C in triangle ABC is 90° - 3y°, which must be positive. So 90° - 3y > 0 => y < 30°, which is consistent with previous result.Now, angle at C in triangle ABC is angle ACB = 90° - 3y. Is this angle related to angle BCD in quadrilateral ABCD? Maybe not directly, unless point D is located such that DC is extending from C in some way. This is getting a bit confusing.Wait, maybe if I consider triangle ADC. In triangle ADC, angle at A is x°, angle at D is angle ADC, and angle at C is angle ACD. Then, in quadrilateral ABCD, angle at C is angle BCD + angle ACD? Not necessarily, because angle at C in the quadrilateral is determined by the two adjacent sides CB and CD. So angle BCD is part of the quadrilateral angle at C. But unless we know how the quadrilateral is structured, it's hard to relate.Alternatively, maybe using the fact that the sum of all angles in quadrilateral ABCD is 360°, so:angle at A (150°) + angle at B + angle at C + angle at D = 360°But angle at B in the quadrilateral is angle ABC + angle ABD. Wait, no. Angle at B is the internal angle of the quadrilateral, which is angle ABC. Wait, no. Wait, in a quadrilateral, each vertex has one internal angle. So angle at B is angle ABC. Wait, but earlier, we had angle ABC = 2y + 60°, from triangle ABC.Similarly, angle at C in quadrilateral is angle BCD, which is angle in triangle BCD. Wait, maybe not. Wait, angle at C is formed by sides BC and CD, so it's angle BCD. So angle at C is angle BCD. Similarly, angle at D is angle ADC plus angle BDC? No, angle at D is formed by sides CD and DA, so in the quadrilateral, angle at D is angle ADC. But we have angle BDC as part of triangle BDC. Hmm, this is confusing.Wait, maybe I need to clarify the structure of the quadrilateral. Let's assume that quadrilateral ABCD is convex. Then, the internal angles at each vertex are the angles between the adjacent sides. So:- Angle at A: angle DAB = 150°- Angle at B: angle ABC- Angle at C: angle BCD- Angle at D: angle CDABut in triangle ABD, we found angle ADB = 30° - y. Angle ADB is part of angle ADC at vertex D. So angle ADC is angle ADB + angle BDC? If the quadrilateral is convex, then yes. Because from point D, side DA connects to A, and side DC connects to C. So if BD is a diagonal, then angle ADC is the angle between DA and DC, which would be composed of angle ADB and angle BDC if BD is a diagonal. Wait, not exactly. Let me think.Actually, angle ADC is the internal angle at D between sides AD and DC. If BD is a diagonal, then angle ADC is not directly composed of angle ADB and BDC unless points A, B, C, D are arranged in a specific way. Hmm, maybe not. So angle ADC is separate from angle BDC.This is getting too vague. Maybe I need to use the fact that in triangle ABD and triangle BDC, we can relate angles and sides.From triangle ABD, using Law of Sines:AB / sin(angle ADB) = BD / sin(angle BAD) = AD / sin(angle ABD)So,AB / sin(30° - y) = BD / sin(150°) = AD / sin(y)Similarly, in triangle BDC:BD / sin(angle BCD) = BC / sin(angle BDC) = CD / sin(angle DBC)But unless we have some relations between the sides AB, BC, CD, DA, this might not help. Maybe if we assume some lengths? Not sure.Alternatively, maybe using the Law of Sines in triangle ABC. In triangle ABC, angles:A: 150° - xB: 2y + 60°C: 90° - 3yLaw of Sines:AB / sin(angle ACB) = BC / sin(angle BAC) = AC / sin(angle ABC)So,AB / sin(90° - 3y) = BC / sin(150° - x) = AC / sin(2y + 60°)But again, without knowing side lengths, this might not help.Wait, but maybe ratios can help. For example, from triangle ABD:AB / sin(30° - y) = BD / sin(150°)So BD = AB * sin(150°) / sin(30° - y)Similarly, in triangle ABC:BC / sin(150° - x) = AB / sin(90° - 3y)So BC = AB * sin(150° - x) / sin(90° - 3y)Similarly, in triangle BDC:BD / sin(angle BCD) = BC / sin(angle BDC)So BD / sin(angle BCD) = BC / sin(angle BDC)Thus, BD / BC = sin(angle BCD) / sin(angle BDC)But BD and BC can be expressed in terms of AB and angles from the previous equations.From triangle ABD: BD = AB * sin(150°) / sin(30° - y) = AB * (1/2) / sin(30° - y) = AB / [2 sin(30° - y)]From triangle ABC: BC = AB * sin(150° - x) / sin(90° - 3y)So BD / BC = [AB / (2 sin(30° - y))] / [AB * sin(150° - x) / sin(90° - 3y)] = [1 / (2 sin(30° - y))] * [sin(90° - 3y) / sin(150° - x)] = [sin(90° - 3y)] / [2 sin(30° - y) sin(150° - x)]But BD / BC = sin(angle BCD) / sin(angle BDC)Therefore,sin(angle BCD) / sin(angle BDC) = [sin(90° - 3y)] / [2 sin(30° - y) sin(150° - x)]This is getting really complicated. Maybe there's a better way.Wait, let me recall that x + y = 120°, so x = 120° - y. So we can substitute x in terms of y.So angle at A in triangle ABC is 150° - x = 150° - (120° - y) = 30° + ySo angle BAC = 30° + yAngle at B in triangle ABC is 2y + 60°Angle at C in triangle ABC is 90° - 3yAlso, angle BDC is what we need to find. Let's denote angle BDC as θ.In triangle BDC, angles are:angle DBC = z = y + 60°angle BDC = θangle BCD = 180° - z - θ = 180° - (y + 60°) - θ = 120° - y - θSo angle BCD = 120° - y - θBut angle BCD in quadrilateral ABCD is the same as angle BCD in triangle BDC, right? Because angle at C is between BC and CD, which is angle BCD. So angle BCD = 120° - y - θBut earlier, in triangle ABC, angle at C (angle ACB) is 90° - 3y. Is angle ACB related to angle BCD? If point D is placed such that DC is a continuation or something, maybe angle ACB and angle BCD are supplementary or something. But I need to visualize the quadrilateral.Alternatively, if we consider quadrilateral ABCD, the angles at C are angle BCD (from triangle BDC) and angle ACB (from triangle ABC). But unless DC is a straight line from C to D, which it is, but how are these angles related?Wait, maybe the sum of angle ACB and angle BCD is 180° if points A, B, C, D are arranged in a certain way. But unless ABCD is a convex quadrilateral, and points are ordered A-B-C-D-A, then angles at C would be angle BCD and angle ACB. But in a convex quadrilateral, those angles are on different sides. So angle ACB is part of triangle ABC, and angle BCD is part of the quadrilateral. They might not be directly related unless there's some specific configuration.This is really tricky. Maybe there's a different approach. Let's see.We know that in triangle ABC:angle at A: 30° + yangle at B: 2y + 60°angle at C: 90° - 3yFrom triangle ABC, using Law of Sines:AB / sin(90° - 3y) = BC / sin(30° + y) = AC / sin(2y + 60°)Similarly, from triangle ABD:AB / sin(30° - y) = BD / sin(150°) = AD / sin(y)And from triangle BDC:BD / sin(angle BCD) = BC / sin(theta) = CD / sin(y + 60°)But angle BCD is 120° - y - theta, as established.This seems too involved. Maybe instead of using Law of Sines, use some trigonometric identities or substitutions.Alternatively, maybe assume specific values for y. Since we have constraints that y < 30°, and x = 120° - y, so x > 90°, since y < 30°. So angle DAC is more than 90°, which might mean that point D is somewhere such that angle DAC is obtuse.Alternatively, maybe try to assign a value to y and see if it works. For example, let me assume y = 20°, then x = 100°. Then check if the angles make sense.If y = 20°, then x = 100°.Then in triangle ABD:angle at D is 30° - y = 10°, angle at A = 150°, angle at B = 20°, sum is 180°, okay.In triangle ABC:angle at A = 30° + y = 50°, angle at B = 2y + 60° = 100°, angle at C = 90° - 3y = 30°, sum 50 + 100 + 30 = 180°, good.In triangle BDC:angle DBC = y + 60° = 80°, angle BDC = theta, angle BCD = 120° - y - theta = 100° - theta.But angle BCD in the quadrilateral is 100° - theta. However, angle C in triangle ABC is 30°, which is angle ACB. If angle BCD is 100° - theta, then how does that relate to angle ACB?Is there a relation between angle ACB (30°) and angle BCD (100° - theta)? If line DC is such that angle BCD is 100° - theta, and angle ACB is 30°, unless those angles are supplementary or something. But 30° + (100° - theta) = 130° - theta. Not sure.Alternatively, maybe quadrilateral angle at C is angle BCD. So in the quadrilateral, angle at C is angle BCD = 100° - theta. Then, sum of angles in quadrilateral ABCD is 150° (at A) + angle at B (which is angle ABC = 2y + 60° = 100°) + angle at C (100° - theta) + angle at D (angle ADC). Wait, angle at D is angle ADC. How is angle ADC related?In triangle ADC, angles are angle DAC = x = 100°, angle ADC = ?, angle ACD = ?Wait, angle ACD is part of angle BCD. If angle BCD = 100° - theta, then angle ACD = angle BCD - angle ACB? Wait, maybe not. If in quadrilateral ABCD, at point C, the angle is between BC and CD, which is angle BCD. In triangle ABC, angle ACB is between BC and AC. Unless AC and CD are the same line, which they are not. So angle ACD is different from angle BCD.This is getting too convoluted. Maybe another approach.Let me consider the sum of the angles in quadrilateral ABCD:angle A + angle B + angle C + angle D = 360°We have angle A = 150°, angle B = angle ABC = 2y + 60°, angle C = angle BCD = 120° - y - theta, angle D = angle ADC. But angle ADC is part of triangle ADC. In triangle ADC, angles are angle DAC = x = 120° - y, angle ADC, angle ACD. So angle ADC = 180° - x - angle ACD. But angle ACD is part of angle BCD. Wait, angle ACD = angle BCD - angle BCA? If that's the case, then angle ACD = (120° - y - theta) - (90° - 3y) = 120° - y - theta - 90° + 3y = 30° + 2y - theta.Therefore, angle ADC = 180° - x - (30° + 2y - theta) = 180° - (120° - y) - 30° - 2y + theta = 180° - 120° + y - 30° - 2y + theta = 30° - y + theta.So angle ADC = 30° - y + theta.Therefore, the sum of angles in quadrilateral ABCD is:angle A (150°) + angle B (2y + 60°) + angle C (120° - y - theta) + angle D (30° - y + theta) = 150 + 2y + 60 + 120 - y - theta + 30 - y + theta = 150 + 60 + 120 + 30 + 2y - y - y + (-theta + theta) = 360 + 0 + 0 = 360°, which checks out. So the sum works, which is good.But this doesn't help us find theta, since all terms cancel out. So this approach isn't helpful.Hmm. Maybe we need to find a relation between theta and y from another equation. Let's see.Earlier, we had BD expressed from triangle ABD and BC expressed from triangle ABC. Then in triangle BDC, using Law of Sines, BD / BC = sin(angle BCD) / sin(theta)We have BD = AB / [2 sin(30° - y)]BC = AB * sin(150° - x) / sin(90° - 3y) = AB * sin(150° - (120° - y)) / sin(90° - 3y) = AB * sin(30° + y) / sin(90° - 3y)Therefore, BD / BC = [AB / (2 sin(30° - y))] / [AB * sin(30° + y) / sin(90° - 3y)] = [1 / (2 sin(30° - y))] * [sin(90° - 3y) / sin(30° + y)] = [sin(90° - 3y)] / [2 sin(30° - y) sin(30° + y)]Also, from Law of Sines in triangle BDC:BD / BC = sin(angle BCD) / sin(theta)But angle BCD = 120° - y - theta, so:sin(angle BCD) = sin(120° - y - theta)Therefore:[sin(90° - 3y)] / [2 sin(30° - y) sin(30° + y)] = sin(120° - y - theta) / sin(theta)This is a complicated equation, but maybe we can simplify it.First, note that sin(90° - 3y) = cos(3y)Also, sin(30° - y) sin(30° + y) can be expanded using the identity sin(A - B) sin(A + B) = sin²A - sin²B. Here, A = 30°, B = y. So:sin(30° - y) sin(30° + y) = sin²30° - sin²y = (1/2)² - sin²y = 1/4 - sin²yAlternatively, using product-to-sum:sin(30° - y) sin(30° + y) = [cos(2y) - cos(60°)] / 2 = [cos(2y) - 0.5] / 2 = (cos(2y) - 0.5)/2But maybe not helpful.Also, note that angle BCD = 120° - y - theta, so sin(angle BCD) = sin(120° - y - theta). Let's denote phi = theta + y, so sin(120° - phi) = sin(120° - phi). Hmm, not sure.Alternatively, maybe try to assign numerical values. Let's suppose y = 20°, then x = 100°, as before.Then compute LHS:sin(90° - 3y) / [2 sin(30° - y) sin(30° + y)] = sin(90° - 60°) / [2 sin(10°) sin(50°)] = sin(30°) / [2 sin(10°) sin(50°)] = 0.5 / [2 * sin(10°) sin(50°)]Calculate sin(10°) sin(50°). Using identity: sin A sin B = [cos(A - B) - cos(A + B)] / 2So sin(10°) sin(50°) = [cos(40°) - cos(60°)] / 2 = [cos(40°) - 0.5]/2cos(40°) ≈ 0.7660So [0.7660 - 0.5]/2 ≈ 0.2660 / 2 ≈ 0.1330Therefore, LHS ≈ 0.5 / (2 * 0.1330) ≈ 0.5 / 0.266 ≈ 1.879RHS: sin(120° - y - theta)/sin(theta) = sin(120° - 20° - theta)/sin(theta) = sin(100° - theta)/sin(theta)We need to solve for theta such that sin(100° - theta)/sin(theta) ≈ 1.879Using approximate value:Let’s test theta = 30°:sin(70°)/sin(30°) ≈ 0.9397 / 0.5 = 1.8794 ≈ 1.879. So theta ≈ 30°Wow, that works! So if y = 20°, then theta = 30°. So angle BDC = 30°. Is this a coincidence or the actual answer?But we need to verify if y = 20° is correct. Wait, but y was assumed to be 20°, but we don't know if that's the actual value. However, the problem doesn't give numerical values for y; it's a general problem. So maybe the answer is always 30°, regardless of y?Wait, let's check another value. Suppose y = 10°, so x = 110°.Compute LHS:sin(90° - 3y) = sin(60°) ≈ 0.8660sin(30° - y) = sin(20°) ≈ 0.3420sin(30° + y) = sin(40°) ≈ 0.6428Thus, LHS = 0.8660 / [2 * 0.3420 * 0.6428] ≈ 0.8660 / (2 * 0.2197) ≈ 0.8660 / 0.4394 ≈ 1.970RHS: sin(120° - 10° - theta)/sin(theta) = sin(110° - theta)/sin(theta)We need sin(110° - theta)/sin(theta) ≈ 1.970Let’s test theta = 30°:sin(80°)/sin(30°) ≈ 0.9848 / 0.5 = 1.9696 ≈ 1.970. Close enough. So theta ≈ 30°So regardless of the value of y (as long as constraints are satisfied), theta ≈ 30°. Therefore, angle BDC is 30°. So maybe the answer is 30°.Therefore, regardless of the value of y, angle BDC is 30°, which is boxed{30} degrees.But wait, need to confirm if this holds generally. Let's suppose theta = 30°, then:From the equation:sin(90° - 3y)/[2 sin(30° - y) sin(30° + y)] = sin(120° - y - 30°)/sin(30°)Simplify RHS:sin(90° - y)/0.5 = 2 sin(90° - y) = 2 cos(y)LHS:sin(90° - 3y)/[2 sin(30° - y) sin(30° + y)] = cos(3y)/[2*(sin(30° - y) sin(30° + y))]Using identity: sin(A - B) sin(A + B) = sin²A - sin²BSo denominator: 2*(sin²30° - sin²y) = 2*(1/4 - sin²y) = 1/2 - 2 sin²yNumerator: cos(3y)Thus, LHS = cos(3y) / (1/2 - 2 sin²y)We need to show that cos(3y)/(1/2 - 2 sin²y) = 2 cos(y)Multiply both sides by (1/2 - 2 sin²y):cos(3y) = 2 cos(y) (1/2 - 2 sin²y) = cos(y) - 4 sin²y cos(y)But cos(3y) = 4 cos³y - 3 cosy. Let's verify RHS:cos(y) - 4 sin²y cos(y) = cos(y) [1 - 4 sin²y] = cos(y) [1 - 4 (1 - cos²y)] = cos(y) [1 - 4 + 4 cos²y] = cos(y) [-3 + 4 cos²y] = 4 cos³y - 3 cosy, which is equal to cos(3y). Therefore, identity holds.Therefore, equation holds true for any y. Hence, angle BDC is indeed 30°, regardless of the value of y (as long as it's within constraints). Therefore, the answer is 30°.Final Answerboxed{30}</think>Given quadrilateral (ABCD) with the following conditions:1. (angle DAB = 150^circ)2. (angle DAC + angle ABD = 120^circ)3. (angle DBC - angle ABD = 60^circ)We need to find (angle BDC).1. Let (angle DAC = x) and (angle ABD = y). From the given conditions, we have: [ x + y = 120^circ quad text{(1)} ] [ angle DBC = y + 60^circ quad text{(2)} ]2. In triangle (ABD), the angles are: - (angle DAB = 150^circ) - (angle ABD = y) - (angle ADB = 180^circ - 150^circ - y = 30^circ - y)3. In triangle (ABC), the angles are: - (angle CAB = 150^circ - x) - (angle ABC = 2y + 60^circ) - (angle ACB = 90^circ - 3y)4. In triangle (BDC), the angles are: - (angle DBC = y + 60^circ) - (angle BDC = theta) (to be found) - (angle BCD = 180^circ - (y + 60^circ) - theta = 120^circ - y - theta)5. Using the Law of Sines in triangles (ABD) and (BDC), and verifying the trigonometric identities, we find that regardless of the value of (y), (theta = 30^circ).Thus, the measure of (angle BDC) is (boxed{30}).]

🔑:75. Let widehat{A B D}=alpha, widehat{B D C}=varphi. By the condition, widehat{D A C}= =120^{circ}-alpha, widehat{B A C}=30^{circ}+alpha, widehat{A D B}=30^{circ}-alpha, widehat{D B C}=60^{circ}+alpha. Using the Law of Sines for triangles A B C, B C D, A C D, we getbegin{aligned}& frac{|B C|}{|A C|}=frac{sin left(30^{circ}+alpharight)}{sin left(60^{circ}+2 alpharight)}=frac{1}{2 cos left(30^{circ}+alpharight)}, & frac{|D C|}{|B C|}=frac{sin left(60^{circ}+alpharight)}{sin varphi}, & frac{|A C|}{|D C|}=frac{sin left(30^{circ}-alpha+varphiright)}{sin left(120^{circ}-alpharight)} .end{aligned}Multiplying these equations, we havesin left(30^{circ}-alpha+varphiright)=sin left(30^{circ}+alpha+varphiright)-sin left(30^{circ}+alpha-varphiright) RightarrowRightarrow sin left(30^{circ}-alpha+varphiright)+sin left(30^{circ}+alpha-varphiright)=sin left(30^{circ}+alpha+varphiright) RightarrowRightarrow cos (varphi-alpha)=sin left(30^{circ}+alpha+varphiright) RightarrowRightarrow sin left(90^{circ}-varphi+alpharight)-sin left(30^{circ}+alpha+varphiright)==2 cos left(60^{circ}+alpharight) sin left(30^{circ}-varphiright)=0 ;thus, varphi=30^{circ}.

🧠:[<think>Okay, so I need to find the value of the function f(x) at x₀ = 1000. They gave me f(0) = 1, and for any x, the equality f(x + 2) = f(x) + 4x + 2 holds. Hmm, let's see. Alright, let's start by understanding the problem.First, the function f is defined recursively. The value at x + 2 depends on the value at x, plus some linear term 4x + 2. Since we need to find f(1000), and we know f(0) = 1, maybe we can find a pattern or formula for f(x) in general. That way, plugging in x = 1000 will be straightforward.Let me think. Let's start by computing some values of f(x) for small x to see if I can spot a pattern. Starting from x = 0:f(0) = 1 (given)Now, to find f(2), we use the recursive formula with x = 0:f(0 + 2) = f(0) + 4*0 + 2 => f(2) = 1 + 0 + 2 = 3Then, f(4) can be found using x = 2:f(2 + 2) = f(2) + 4*2 + 2 => f(4) = 3 + 8 + 2 = 13Similarly, f(6):f(4 + 2) = f(4) + 4*4 + 2 = 13 + 16 + 2 = 31Wait, hold on, maybe let's tabulate these:x | f(x)0 | 12 | 34 | 136 | 318 | ?Wait, let me compute f(8):f(6 + 2) = f(6) + 4*6 + 2 = 31 + 24 + 2 = 57Hmm, so the values are 1, 3, 13, 31, 57...Looking at these numbers, I wonder if there's a quadratic relationship here. Let's check the differences between consecutive terms.From f(0) to f(2): 3 - 1 = 2From f(2) to f(4): 13 - 3 = 10From f(4) to f(6): 31 - 13 = 18From f(6) to f(8): 57 - 31 = 26So the differences are 2, 10, 18, 26,...Wait, the differences themselves are increasing by 8 each time. Hmm, 2, 10 (which is 2 + 8), 18 (10 + 8), 26 (18 + 8)... So the second differences are constant (8). That suggests that the function f(x) is a quadratic function in terms of x. Because if the second differences are constant, then the function is quadratic.So if f(x) is quadratic, let's assume f(x) = ax² + bx + c.But wait, the recursive formula is given for x and x + 2. So maybe we need to express the recursive equation in terms of the quadratic function.Let me try to express f(x + 2) - f(x) = 4x + 2. So, substituting the quadratic form:[a(x + 2)² + b(x + 2) + c] - [ax² + bx + c] = 4x + 2Let's expand that:a(x² + 4x + 4) + b(x + 2) + c - ax² - bx - c = 4x + 2Simplify:a x² + 4a x + 4a + b x + 2b + c - a x² - b x - c = 4x + 2The x² terms cancel, the x terms: 4a x + b x - b x = 4a xConstants: 4a + 2bSo overall:4a x + (4a + 2b) = 4x + 2Therefore, equating coefficients:4a = 4 => a = 14a + 2b = 2 => 4*1 + 2b = 2 => 4 + 2b = 2 => 2b = -2 => b = -1So, a = 1, b = -1. Then, we can use the initial condition f(0) = c = 1. Therefore, c = 1.Thus, the quadratic function is f(x) = x² - x + 1.Wait, let's check this with the earlier values.At x = 0: 0 - 0 + 1 = 1 ✔️x = 2: 4 - 2 + 1 = 3 ✔️x = 4: 16 - 4 + 1 = 13 ✔️x = 6: 36 - 6 + 1 = 31 ✔️x = 8: 64 - 8 + 1 = 57 ✔️Perfect, that matches all the computed values. So the general formula is f(x) = x² - x + 1.Therefore, f(1000) would be 1000² - 1000 + 1 = 1,000,000 - 1000 + 1 = 999,001.But wait, let me verify this formula with the recursive equation again to make sure I didn't make a mistake.Let me take f(x) = x² - x + 1. Then f(x + 2) = (x + 2)² - (x + 2) + 1 = x² + 4x + 4 - x - 2 + 1 = x² + 3x + 3.On the other hand, f(x) + 4x + 2 = x² - x + 1 + 4x + 2 = x² + 3x + 3. Which matches f(x + 2). So the formula is correct.Therefore, the value at x = 1000 is indeed 1000² - 1000 + 1 = 999,001.But let me check another approach to confirm. Since the recursion is every 2 steps, maybe we can express x in terms of steps of 2. Let's set x = 2n, where n is an integer. So maybe for even x, we can write n = x/2. Let's see.Wait, but 1000 is even, so n = 500. Then, maybe f(2n) can be expressed as a function of n. Let's see.Starting from f(0) = 1. Then f(2) = 1 + 4*0 + 2 = 3. Then f(4) = 3 + 4*2 + 2 = 3 + 8 + 2 = 13. So for each step, adding 4*(2(n-1)) + 2? Wait, maybe.Alternatively, since we already have the closed-form formula, perhaps we can compute f(1000) directly. But to confirm, let's try another way.Suppose we consider the recursion f(x + 2) - f(x) = 4x + 2. This is a linear recurrence relation. The homogeneous solution would be found by solving f(x + 2) - f(x) = 0. The characteristic equation would be r² - 1 = 0, so roots at r = 1 and r = -1. Therefore, the homogeneous solution is A(1)^x + B(-1)^x.But since the nonhomogeneous term is 4x + 2, which is a linear polynomial, we can assume a particular solution of the form Cx + D. Let's plug that into the recurrence:For the particular solution, f_p(x) = Cx + D. Then f_p(x + 2) - f_p(x) = C(x + 2) + D - (Cx + D) = 2C. This must equal 4x + 2. Wait, but 2C is a constant, and the RHS is 4x + 2. So this approach doesn't work because the particular solution can't be a linear polynomial if the nonhomogeneous term is linear. Wait, maybe we need a quadratic particular solution. Let's try f_p(x) = ax² + bx + c.Then, f_p(x + 2) - f_p(x) = a(x + 2)^2 + b(x + 2) + c - [ax² + bx + c] = a(x² + 4x + 4) + b(x + 2) + c - ax² - bx - c = 4a x + 4a + 2b. Set this equal to 4x + 2.Therefore, equating coefficients:4a = 4 => a = 14a + 2b = 2 => 4*1 + 2b = 2 => 2b = -2 => b = -1So the particular solution is x² - x + c. Wait, but when we subtract f_p(x + 2) - f_p(x), the constant term 4a + 2b was set to 2. But c cancels out. So we can choose c arbitrarily? Wait, but in the homogeneous solution, we can have constants. Wait, maybe the particular solution is x² - x, and the homogeneous solution is A + B(-1)^x. Then the general solution is f(x) = x² - x + A + B(-1)^x.But we have the initial condition f(0) = 1. Let's check x=0: 0 - 0 + A + B(-1)^0 = A + B = 1.But we need another initial condition. Let's compute f(2) = 3. From the formula: 2² - 2 + A + B(-1)^2 = 4 - 2 + A + B = 2 + A + B. We already know f(2) = 3, so 2 + (A + B) = 3. But since A + B = 1, this gives 2 + 1 = 3, which is correct. Hmm, but we need another condition to solve for A and B. Wait, maybe we can compute f(4) = 13. Let's plug in x=4: 16 - 4 + A + B(-1)^4 = 12 + A + B. Which equals 13, so 12 + (A + B) = 13 => A + B = 1, which is consistent. Wait, so all even x terms give A + B, and odd x terms? Wait, but our original recursion is only defined for x and x + 2. So maybe the function is only defined for even x? Wait, but the problem says "for any x" the equality holds. Wait, but x₀ = 1000 is even. Maybe the function is defined for all integers x, even and odd. But if so, then maybe we need to define f(1) as well? But the problem doesn't give us f(1), only f(0) = 1.Wait, maybe the function is defined for all real numbers x? But the problem mentions x₀ = 1000, but the recursion is for any x. Hmm. Wait, the problem statement says "for any x the equality f(x+2)=f(x)+4x+2 holds". So x can be any real number? Or any integer? The problem doesn't specify, but since we're asked to find f(1000), which is an integer, perhaps x is an integer. However, the problem mentions "the point x₀=1000", which might imply x is a real number. Hmm, but the recursive formula is f(x + 2) = f(x) + 4x + 2. If x is a real number, then this is a functional equation. But if x is an integer, then it's a recurrence relation.Wait, the original problem is in Chinese, but the translation here says "for any x the equality holds". So likely, x is a real number, so it's a functional equation, not just a recurrence. Therefore, f is a function defined on real numbers satisfying f(x + 2) = f(x) + 4x + 2 for all real x, and f(0) = 1. Then, we need to find f(1000).In that case, our previous approach of assuming a quadratic function is valid, because if f is a quadratic function, then it will satisfy the functional equation for all real x. So, as we did earlier, solving f(x) = x² - x + 1 works for all real x, including x = 1000. Therefore, the answer is 1000² - 1000 + 1 = 999,001.Alternatively, if we consider the functional equation f(x + 2) = f(x) + 4x + 2, we can think of this as a first-order linear difference equation with step 2. The general solution can be found by finding the homogeneous solution and a particular solution.The homogeneous equation is f(x + 2) - f(x) = 0. The solutions to this are functions periodic with period 2. However, since we're dealing with functions over the reals, the solutions aren't as straightforward as in the integer case. However, if we assume f is a polynomial, then the only periodic polynomials are constant functions. So the homogeneous solution would be a constant. But in our earlier approach, when we found a particular solution, which was quadratic, and then the homogeneous solution would be a constant. However, in the general solution for a difference equation, the homogeneous solution can include functions that satisfy the homogeneous equation. But over real numbers, for a difference equation with step 2, the homogeneous solutions can be more complex. However, if we restrict ourselves to polynomial solutions, then only constants are homogeneous solutions.But given that the nonhomogeneous term is linear (4x + 2), we look for a particular solution as a quadratic polynomial, which we found earlier. Therefore, the general solution would be f(x) = x² - x + C, where C is a constant. Then, using the initial condition f(0) = 1, we find C = 1. Therefore, f(x) = x² - x + 1. Hence, f(1000) = 1000² - 1000 + 1 = 999,001.Alternatively, another method: telescope the recursion. Since f(x + 2) = f(x) + 4x + 2, we can express f(1000) in terms of f(998), then f(998) in terms of f(996), and so on, until we reach f(0). Since 1000 is even, we can write 1000 = 2*500. Therefore, we can write:f(1000) = f(998) + 4*(998) + 2f(998) = f(996) + 4*(996) + 2...f(2) = f(0) + 4*0 + 2So, if we add all these equations up, starting from f(1000) down to f(2), each term f(2k) is expressed as f(2(k-1)) + 4*2(k-1) + 2. So, when we telescope, we get:f(1000) = f(0) + Σ [from k=0 to 499] (4*(2k) + 2)Wait, let's check. Each step from f(0) to f(2) adds 4*0 + 2.From f(2) to f(4) adds 4*2 + 2.From f(4) to f(6) adds 4*4 + 2....From f(998) to f(1000) adds 4*998 + 2.Therefore, the total sum is the sum of 4*(2k) + 2 for k from 0 to 499 (since 2*499 = 998). Therefore, let's compute this sum:Sum = Σ [k=0 to 499] (8k + 2) = 8 * Σk + 2*500 = 8*(499*500)/2 + 1000 = 8*(124,750) + 1000 = 998,000 + 1000 = 999,000Therefore, f(1000) = f(0) + 999,000 = 1 + 999,000 = 999,001. Which matches the previous result.So both methods agree. Therefore, the answer is 999,001.Just to recap, the key steps were recognizing the recurrence relation as a functional equation leading to a quadratic function, solving for its coefficients using the given initial condition, and verifying through telescoping the sum. Both approaches led to the same answer, so I'm confident it's correct.Final AnswerThe value of the function at ( x_0 = 1000 ) is boxed{999001}.</think>Given the function ( f(x) ) with the initial condition ( f(0) = 1 ) and the recurrence relation ( f(x + 2) = f(x) + 4x + 2 ) for any ( x ), we need to find the value of ( f(1000) ).First, we recognize that the recurrence relation suggests a quadratic function. We assume ( f(x) ) is a quadratic function of the form ( f(x) = ax^2 + bx + c ). Substituting this into the recurrence relation:[f(x + 2) - f(x) = 4x + 2]Expanding and simplifying the left-hand side:[[a(x + 2)^2 + b(x + 2) + c] - [ax^2 + bx + c] = 4ax + 4a + 2b]Equating this to the right-hand side ( 4x + 2 ):[4a = 4 implies a = 1][4a + 2b = 2 implies 4(1) + 2b = 2 implies 2b = -2 implies b = -1]Using the initial condition ( f(0) = 1 ):[c = 1]Thus, the quadratic function is ( f(x) = x^2 - x + 1 ).To verify, we check the values for small ( x ):- ( f(0) = 0^2 - 0 + 1 = 1 )- ( f(2) = 2^2 - 2 + 1 = 3 )- ( f(4) = 4^2 - 4 + 1 = 13 )- These values match the recurrence relation calculations.Finally, substituting ( x = 1000 ):[f(1000) = 1000^2 - 1000 + 1 = 1000000 - 1000 + 1 = 999001]Thus, the value of the function at ( x_0 = 1000 ) is (boxed{999001}).,<think>Okay, so I need to find the value of the function f(x) at x₀ = 1000, given that f(0) = 1 and for any x, the equality f(x + 2) = f(x) + 4x + 2 holds. Hmm, let's see. First, I remember that this kind of problem involves recurrence relations. The function is defined recursively here, with each step increasing x by 2. Since we know f(0) = 1, maybe I can find a pattern or formula by calculating some values step by step and then generalize it to reach x = 1000. Let me start by computing f(2). Using the given equality, if x = 0, then f(0 + 2) = f(0) + 4*0 + 2. That would be f(2) = 1 + 0 + 2 = 3. Okay, so f(2) is 3. Next, let's compute f(4). If x = 2, then f(2 + 2) = f(2) + 4*2 + 2. So f(4) = 3 + 8 + 2 = 13. Hmm, f(4) is 13. Continuing this way, f(6) would be f(4 + 2) = f(4) + 4*4 + 2. So f(6) = 13 + 16 + 2 = 31. Similarly, f(8) = f(6) + 4*6 + 2 = 31 + 24 + 2 = 57. Wait, this seems like a pattern where each time we add 4x + 2, but x here is increasing by 2 each step. Let me check the differences between consecutive terms. From f(0) = 1 to f(2) = 3, the difference is 2. Then from f(2) to f(4) is 10 (3 to 13). Then from f(4) to f(6) is 18 (13 to 31). Then from f(6) to f(8) is 26 (31 to 57). So the differences themselves are increasing by 8 each time? Wait, 2, 10, 18, 26... each time adding 8. Hmm, that's interesting. But maybe instead of looking at differences, I should try to find a closed-form formula for f(x). Since the recurrence relation is linear, perhaps f(x) is a quadratic function. Let me assume that f(x) is a quadratic function of the form f(x) = ax² + bx + c. Then I can use the given recurrence relation to solve for a, b, c. Given f(x + 2) = f(x) + 4x + 2, substituting the quadratic form into this equation:a(x + 2)² + b(x + 2) + c = ax² + bx + c + 4x + 2.Let's expand the left side:a(x² + 4x + 4) + b(x + 2) + c = ax² + 4ax + 4a + bx + 2b + c.So left side becomes: ax² + (4a + b)x + (4a + 2b + c).The right side is: ax² + bx + c + 4x + 2 = ax² + (b + 4)x + (c + 2).Now, set the coefficients equal for corresponding terms:For x² terms: a = a. That's okay, no new info.For x terms: 4a + b = b + 4. Subtract b from both sides: 4a = 4. So a = 1.For constant terms: 4a + 2b + c = c + 2. Subtract c from both sides: 4a + 2b = 2. Since we already found a = 1, substitute that in: 4*1 + 2b = 2 → 4 + 2b = 2 → 2b = -2 → b = -1.Now, we have a = 1, b = -1. To find c, use the initial condition f(0) = 1. When x = 0, f(0) = a*0 + b*0 + c = c = 1. So c = 1.Therefore, the quadratic function is f(x) = x² - x + 1.Wait, let's check if this works with the values I calculated earlier.For x = 0: 0 - 0 + 1 = 1. Correct.x = 2: 4 - 2 + 1 = 3. Correct.x = 4: 16 - 4 + 1 = 13. Correct.x = 6: 36 - 6 + 1 = 31. Correct.x = 8: 64 - 8 + 1 = 57. Correct. Okay, seems like this quadratic function satisfies the given conditions.Therefore, the general formula is f(x) = x² - x + 1. Therefore, f(1000) would be 1000² - 1000 + 1 = 1,000,000 - 1,000 + 1 = 999,001.But wait, just to be thorough, let me verify with the recurrence relation once more. Let's take x = 6. Then f(8) should be f(6) + 4*6 + 2. According to our formula, f(6) = 31, so f(8) should be 31 + 24 + 2 = 57. Which matches f(8) = 57. So the formula holds.Another check: Let's pick x = 10. f(10) = 100 - 10 + 1 = 91. But according to the recurrence, f(10) = f(8) + 4*8 + 2 = 57 + 32 + 2 = 91. Perfect. So the formula works.Therefore, f(x) = x² - x + 1 is the correct function, and substituting x = 1000 gives 1000² - 1000 + 1. 1000 squared is 1,000,000, subtract 1000 gives 999,000, add 1 gives 999,001. Therefore, the answer should be 999001.Alternatively, since the function is quadratic, maybe there's another way to approach it. Let me think. The recurrence is f(x + 2) - f(x) = 4x + 2. That's similar to a sequence with step 2. So, the difference between terms two apart is 4x + 2.If we think of x as being even numbers, since we're stepping by 2 each time. Let me let x = 2n, where n is an integer. Then f(2n + 2) - f(2n) = 4*(2n) + 2 = 8n + 2.So, if we consider n as a variable, then f(2(n + 1)) - f(2n) = 8n + 2. This is a linear recurrence in terms of n. Then, f(2n) is a quadratic function in n. Let me check that.Suppose f(2n) = An² + Bn + C. Then, f(2(n + 1)) - f(2n) = A(n + 1)² + B(n + 1) + C - [An² + Bn + C] = A(2n + 1) + B = 2An + (A + B). According to the recurrence, this should equal 8n + 2. Therefore, we have:2A = 8 ⇒ A = 4A + B = 2 ⇒ 4 + B = 2 ⇒ B = -2Therefore, f(2n) = 4n² - 2n + C. Now, use the initial condition f(0) = 1. When n = 0, f(0) = 4*0 -2*0 + C = C = 1. So f(2n) = 4n² - 2n + 1.Now, express this in terms of x. Since x = 2n, n = x/2. Therefore,f(x) = 4*(x/2)² - 2*(x/2) + 1 = 4*(x²/4) - x + 1 = x² - x + 1.Same result as before. So that confirms the formula again. Therefore, for x = 1000, which is even, n = 500, so f(1000) = 1000² - 1000 + 1 = 999,001.Alternatively, if I didn't notice the quadratic nature, maybe I could have summed the recurrence relation. Since each step from x to x + 2 adds 4x + 2. So starting from x = 0, to get to x = 1000, we need 500 steps (since 1000 / 2 = 500). Therefore, f(1000) = f(0) + sum_{k=0}^{499} [4*(2k) + 2]. Wait, let me check that.Wait, actually, each time we increment x by 2, starting from x = 0, so after one step, x = 2, after two steps x = 4, ..., after 500 steps x = 1000. So the total number of terms in the sum would be 500. The terms would be for x = 0, 2, 4, ..., 998. Each term is 4x + 2. Therefore, the sum is sum_{x=0,2,4,...,998} (4x + 2). Let me compute this sum.First, factor out the 2 step. Let x = 2k, where k ranges from 0 to 499. Then the sum becomes sum_{k=0}^{499} [4*(2k) + 2] = sum_{k=0}^{499} (8k + 2) = 8*sum_{k=0}^{499} k + 2*500.Sum_{k=0}^{499} k = (499*500)/2 = (499*250) = 124,750.Therefore, the total sum is 8*124,750 + 1000. Let's compute that:8*124,750 = 998,000998,000 + 1000 = 999,000.Therefore, f(1000) = f(0) + 999,000 = 1 + 999,000 = 999,001. Which matches the earlier result.So regardless of the method, the answer is 999,001. So that seems solid.Another way: Since the recurrence is f(x + 2) - f(x) = 4x + 2, this is a linear nonhomogeneous recurrence relation. The homogeneous solution would satisfy f(x + 2) - f(x) = 0, whose solutions are periodic functions with period 2. But since we are dealing with a function over the integers, and the nonhomogeneous term is linear, the particular solution should be a quadratic function, as we found earlier.Thus, combining the homogeneous and particular solutions, the general solution is f(x) = A + B*(-1)^x + quadratic term. But since the nonhomogeneous term is linear, and the homogeneous equation has solutions with period 2, but in our case, the function is defined for all x, not just integers. Wait, actually, in the problem statement, it's for "any x", which might mean real numbers? Wait, the problem says "for any x the equality f(x + 2) = f(x) + 4x + 2 holds". So x is a real number. Then, we need a function defined on real numbers, satisfying this functional equation.But the initial condition is given at x = 0. Hmm. But earlier, we assumed it's a quadratic function. Let's verify if the functional equation holds for all real numbers x.Suppose f(x) = x² - x + 1. Then f(x + 2) = (x + 2)² - (x + 2) + 1 = x² + 4x + 4 - x - 2 + 1 = x² + 3x + 3. On the other hand, f(x) + 4x + 2 = x² - x + 1 + 4x + 2 = x² + 3x + 3. Therefore, equality holds for all real numbers x. Therefore, the quadratic function is indeed the solution for all real x, so f(1000) = 1000² - 1000 + 1 = 999,001.Therefore, regardless of the approach, the answer is 999,001.Final Answerboxed{999001}</think>Given the function ( f(x) ) with the initial condition ( f(0) = 1 ) and the recurrence relation ( f(x + 2) = f(x) + 4x + 2 ) for any ( x ), we need to find the value of ( f(1000) ).First, we assume ( f(x) ) is a quadratic function of the form ( f(x) = ax^2 + bx + c ). Substituting this into the recurrence relation, we get:[a(x + 2)^2 + b(x + 2) + c = ax^2 + bx + c + 4x + 2]Expanding and simplifying both sides, we find:[ax^2 + (4a + b)x + (4a + 2b + c) = ax^2 + (b + 4)x + (c + 2)]By equating coefficients, we determine ( a = 1 ), ( b = -1 ), and using the initial condition ( f(0) = 1 ), we find ( c = 1 ). Thus, the function is:[f(x) = x^2 - x + 1]To verify, we check the recurrence relation and initial conditions, confirming it holds for all ( x ). Therefore, substituting ( x = 1000 ):[f(1000) = 1000^2 - 1000 + 1 = 1000000 - 1000 + 1 = 999001]Thus, the value of the function ( f(x) ) at ( x_0 = 1000 ) is:[boxed{999001}]]

🔑:Answer: 999001Solution: In the equation f(x+2)-f(x)=4 x+2, we will substitute for x the numbers 0,2,4, ldots, 998. We get:begin{aligned}& f(2)-f(0)=4 cdot 0+2 & f(4)-f(2)=4 cdot 2+2end{aligned}f(1000)-f(998)=4 cdot 998+2Adding the equations, we get: f(1000)-f(0)=4 cdot(0+2+4+cdots+998)+2 cdot 500=4. frac{998 cdot 500}{2}+1000=999000. Then f(1000)=999001.

🧠:[<think>Okay, so I need to find the limit as n approaches infinity of the sum from k=0 to n-1 of sqrt(n² - k²) divided by n². Hmm, let me think. First, let me write down the expression again to make sure I have it right:limₙ→∞ Σₖ₌₀^{n-1} [√(n² - k²)/n²]Alright, so we're summing terms of the form sqrt(n² - k²) over n², starting from k=0 up to k=n-1. Then taking the limit as n becomes very large. I remember that when dealing with sums like this, especially when taking a limit as n goes to infinity, it might help to interpret the sum as a Riemann sum. Riemann sums approximate integrals, so maybe this sum can be turned into an integral which would be easier to evaluate. Let me recall how Riemann sums work.A Riemann sum for a function f(x) over an interval [a, b] is given by Σₖ₌₁^N f(xₖ*) Δx, where Δx is the width of each subinterval and xₖ* is a sample point in each subinterval. If we let the number of subintervals N go to infinity, the Riemann sum approaches the integral from a to b of f(x) dx.So, let's see if we can manipulate the given sum into a form that resembles a Riemann sum. Let's start by analyzing the term inside the sum:sqrt(n² - k²)/n²Let's factor out n² inside the square root:sqrt(n²(1 - (k²/n²))) / n² = sqrt(n²) * sqrt(1 - (k²/n²)) / n² = n * sqrt(1 - (k²/n²)) / n² = sqrt(1 - (k²/n²)) / nSo, the term simplifies to sqrt(1 - (k/n)²) * (1/n). That's interesting. So the entire sum becomes:Σₖ₌₀^{n-1} sqrt(1 - (k/n)²) * (1/n)This looks more like a Riemann sum. Let's check the components. The (1/n) term is like the Δx in the Riemann sum, and the sqrt(1 - (k/n)²) is the function evaluated at the sample points xₖ = k/n. However, in Riemann sums, the sample points can be xₖ = a + kΔx, but here, if we set xₖ = k/n, then the interval would start at x=0 when k=0 and go up to x=(n-1)/n when k=n-1. As n approaches infinity, the upper limit approaches 1. So, this seems like a Riemann sum for the integral from 0 to 1 of sqrt(1 - x²) dx, using right endpoints. Wait, but in our case, the sum goes from k=0 to k=n-1, which are the points x = 0, 1/n, 2/n, ..., (n-1)/n. So, those are left endpoints if we consider the interval from 0 to 1 divided into n subintervals. Wait, but each subinterval has width Δx = 1/n. So, the sum is a left Riemann sum for the integral from 0 to 1 of sqrt(1 - x²) dx. But as n approaches infinity, both left and right Riemann sums converge to the same value, the integral. So, the limit should be equal to the integral from 0 to 1 of sqrt(1 - x²) dx.But let me verify this carefully. Let me consider the sum S_n = Σₖ₌₀^{n-1} sqrt(1 - (k/n)^2) * (1/n). If I set xₖ = k/n, then S_n is the sum of f(xₖ) * Δx, where f(x) = sqrt(1 - x²) and Δx = 1/n. Therefore, this is indeed a Riemann sum for f(x) over [0, 1] with left endpoints. Since f(x) is continuous on [0, 1], the Riemann sum converges to the integral as n approaches infinity. Therefore,limₙ→∞ S_n = ∫₀¹ sqrt(1 - x²) dxSo, now I need to compute that integral. The integral of sqrt(1 - x²) dx from 0 to 1 is a standard integral, right? It's the area of a quarter of the unit circle. Because the equation y = sqrt(1 - x²) describes the upper half of the unit circle x² + y² = 1. So, integrating from 0 to 1 would give the area of a quarter of the circle. Since the area of the unit circle is π, a quarter of that is π/4. Therefore, the integral from 0 to 1 of sqrt(1 - x²) dx is π/4.Therefore, the limit should be π/4. Let me double-check this to make sure I didn't make any mistakes. Let me verify the steps again.1. Original expression: sum from k=0 to n-1 of sqrt(n² - k²)/n².2. Simplify each term: sqrt(n² - k²)/n² = sqrt(1 - (k/n)^2)/n.3. Therefore, the sum becomes (1/n) * sum from k=0 to n-1 of sqrt(1 - (k/n)^2).4. Recognize that this is a Riemann sum for the integral of sqrt(1 - x²) from 0 to 1 with Δx = 1/n and xₖ = k/n.5. As n approaches infinity, the Riemann sum approaches the integral, which is π/4.Yes, this seems correct. Wait, but let me check the integral computation again. The integral from 0 to 1 of sqrt(1 - x²) dx. Let me compute it using substitution. Let x = sinθ, so dx = cosθ dθ. When x=0, θ=0; when x=1, θ=π/2. Then the integral becomes:∫₀^{π/2} sqrt(1 - sin²θ) * cosθ dθ = ∫₀^{π/2} cosθ * cosθ dθ = ∫₀^{π/2} cos²θ dθUsing the identity cos²θ = (1 + cos2θ)/2:∫₀^{π/2} (1 + cos2θ)/2 dθ = (1/2) ∫₀^{π/2} 1 dθ + (1/2) ∫₀^{π/2} cos2θ dθCompute the first integral: (1/2)(π/2 - 0) = π/4Second integral: (1/2)*(1/2) sin2θ evaluated from 0 to π/2: (1/4)(sinπ - sin0) = 0Thus, total integral is π/4 + 0 = π/4. Yep, that checks out.So, the limit is π/4. Therefore, the answer should be π/4. Wait, but hold on. Let me think again about the original sum. Is there any possibility that I misapplied the Riemann sum?Wait, the original sum is from k=0 to n-1. So, xₖ = k/n, which gives points 0, 1/n, 2/n, ..., (n-1)/n. So, the last term is at x = (n-1)/n = 1 - 1/n. So, the interval is [0, 1 - 1/n], but as n approaches infinity, 1 - 1/n approaches 1. So, the Riemann sum over [0,1) which is the same as [0,1] since the function is continuous and the single point at 1 doesn't affect the integral. Therefore, the Riemann sum converges to the integral over [0,1], which is π/4. So, everything seems to check out.Therefore, the limit is π/4. So, the answer is π/4, which can be written as boxed{dfrac{pi}{4}}.But let me just verify with a small n value to see if the approximation makes sense. For example, take n=2. Then the sum is from k=0 to 1:Term k=0: sqrt(4 - 0)/4 = 2/4 = 0.5Term k=1: sqrt(4 - 1)/4 = sqrt(3)/4 ≈ 0.4330Sum ≈ 0.5 + 0.4330 ≈ 0.9330. Compare to π/4 ≈ 0.7854. Hmm, but with n=2, the approximation isn't that close. Maybe n=1000, but obviously, I can't compute that here. But as n increases, the sum should approach π/4. Let me check n=100:The sum is (1/100) * sum_{k=0}^{99} sqrt(1 - (k/100)^2). This would be an approximation of the integral. The integral is π/4 ≈ 0.7854. If we compute the Riemann sum with n=100, we should get something close to that. Since it's a left Riemann sum, and the function is decreasing on [0,1], the left Riemann sum will overestimate the integral. Indeed, because each rectangle's height is the left endpoint, which is higher than the average as the function is decreasing. So, the Riemann sum should be larger than π/4. For example, with n=2, as above, we had 0.9330, which is higher than π/4. For n=100, it's going to be closer but still a bit higher. But as n approaches infinity, the overestimate diminishes and approaches the integral. Therefore, the limit is indeed π/4. So, I think that's correct.Another way to think about it: maybe using variable substitution. Let me set k = xn, where x is between 0 and 1. Then, as n approaches infinity, the sum can be approximated by the integral from x=0 to x=1 of sqrt(1 - x²) dx, which again is π/4. This substitution is essentially the same as recognizing the Riemann sum. So, that's another way to see it.Alternatively, maybe expanding the square root? Let's see. sqrt(n² - k²) = n sqrt(1 - (k/n)^2). So, perhaps expand sqrt(1 - (k/n)^2) as a Taylor series? But since we have a sum over k, that might complicate things. But if we think of the sum as an approximation to the integral, then the leading term is the integral, and higher-order terms vanish as n becomes large. Therefore, the limit is the integral.Therefore, after considering different approaches and verifying the integral, I feel confident that the answer is π/4.Final Answerboxed{dfrac{pi}{4}}</think>To find the limit (lim _{n rightarrow infty} sum_{k=0}^{n-1} frac{sqrt{n^{2}-k^{2}}}{n^{2}}), we start by simplifying the term inside the summation:[frac{sqrt{n^2 - k^2}}{n^2} = frac{sqrt{n^2(1 - (k/n)^2)}}{n^2} = frac{n sqrt{1 - (k/n)^2}}{n^2} = frac{sqrt{1 - (k/n)^2}}{n}]Thus, the given sum can be rewritten as:[sum_{k=0}^{n-1} frac{sqrt{1 - (k/n)^2}}{n}]This is a Riemann sum for the integral of the function (f(x) = sqrt{1 - x^2}) over the interval ([0, 1]) with subintervals of width (frac{1}{n}). As (n) approaches infinity, the Riemann sum converges to the integral:[lim_{n to infty} sum_{k=0}^{n-1} frac{sqrt{1 - (k/n)^2}}{n} = int_{0}^{1} sqrt{1 - x^2} , dx]The integral (int_{0}^{1} sqrt{1 - x^2} , dx) represents the area of a quarter of the unit circle. The area of the unit circle is (pi), so the area of a quarter-circle is (frac{pi}{4}). Therefore, the value of the integral is (frac{pi}{4}).Hence, the limit is:[boxed{dfrac{pi}{4}}],<think>Okay, so I have this limit to compute: the limit as n approaches infinity of the sum from k = 0 to n - 1 of sqrt(n² - k²) divided by n². Hmm, let me write that out to make sure I understand it correctly.The expression is:limₙ→∞ [Σₖ₌₀^{n-1} (√(n² - k²)/n²)]Alright, so first, maybe I can simplify the expression inside the sum. Let's see, the term inside the sum is sqrt(n² - k²)/n². If I factor out n² from inside the square root, that might help. Let's try:sqrt(n² - k²) = sqrt(n²(1 - (k²/n²))) = n * sqrt(1 - (k²/n²))So substituting back into the term:n * sqrt(1 - (k²/n²)) / n² = sqrt(1 - (k²/n²) / nTherefore, the entire sum becomes:Σₖ₌₀^{n-1} [sqrt(1 - (k²/n²) / n]Wait, let me check that again. When factoring out n² inside the sqrt, it becomes n*sqrt(1 - (k/n)²), right? Because k²/n² is (k/n)². So sqrt(n² - k²) is n*sqrt(1 - (k/n)²). Then dividing by n² gives sqrt(1 - (k/n)²)/n. Therefore, each term in the sum is sqrt(1 - (k/n)²)/n, and the sum is from k = 0 to n - 1.So, the sum is (1/n) * Σₖ₌₀^{n-1} sqrt(1 - (k/n)²). Hmm, this looks familiar. It resembles a Riemann sum. Let me recall that a Riemann sum for a function f(x) over an interval [a, b] is given by Σₖ f(x_k)Δx, where Δx is (b - a)/n and x_k are the sample points. In this case, if we consider the interval [0, 1], for example, then Δx would be 1/n, and x_k would be k/n. Wait, our term here is sqrt(1 - (k/n)^2) * (1/n), which is exactly the Riemann sum for the function f(x) = sqrt(1 - x²) over the interval [0, 1], with right endpoints.But wait, the sum is from k = 0 to n - 1, so the sample points are x_k = k/n, starting from 0 up to (n - 1)/n. So as n approaches infinity, the upper limit approaches 1. Therefore, the Riemann sum should approximate the integral of sqrt(1 - x²) from 0 to 1. So, the limit would be ∫₀¹ sqrt(1 - x²) dx.But wait, let me confirm that. The Riemann sum using right endpoints for the integral from a to b is Σ f(a + kΔx)Δx, where Δx = (b - a)/n. In our case, if the interval is [0, 1], then Δx = 1/n, and the sample points are x_k = 0 + kΔx = k/n. So the sum Σₖ₌₀^{n - 1} f(k/n) * Δx is indeed the right Riemann sum, which approaches the integral as n → ∞. Therefore, the limit should be ∫₀¹ sqrt(1 - x²) dx.Okay, so if that's the case, then the answer is the integral of sqrt(1 - x²) from 0 to 1. But I remember that the integral of sqrt(1 - x²) is related to the area of a semicircle. Specifically, the integral from 0 to 1 of sqrt(1 - x²) dx is equal to π/4, since the area of the quarter-circle in the first quadrant is (1/4)πr², and here r = 1, so it's π/4.Wait, let me check that again. The integral from 0 to 1 of sqrt(1 - x²) dx is indeed a quarter of the area of a unit circle. The full unit circle has area π, so the quarter-circle (first quadrant) has area π/4. Therefore, the integral is π/4. Therefore, the limit should be π/4.But hold on, let me make sure there are no mistakes in translating the sum to the integral. Let me check again step by step.Original sum: Σₖ₌₀^{n - 1} [sqrt(n² - k²)/n²]We factored out n² inside the sqrt to get n*sqrt(1 - (k/n)^2), then divided by n², so each term is sqrt(1 - (k/n)^2)/n.Therefore, the sum becomes Σₖ₌₀^{n - 1} [sqrt(1 - (k/n)^2) * (1/n)]Yes, and as n → ∞, this is a Riemann sum for the integral of sqrt(1 - x²) from x = 0 to x = 1, using right endpoints with Δx = 1/n. Therefore, the limit should indeed be ∫₀¹ sqrt(1 - x²) dx = π/4.But wait, another thought: the Riemann sum using right endpoints for an integral from a to b would have the form Σ f(x_k)Δx where x_k = a + kΔx. But in our case, when k goes from 0 to n - 1, x_k = k/n goes from 0 to (n - 1)/n, which is 1 - 1/n. So as n approaches infinity, the upper limit approaches 1, so the Riemann sum approaches the integral from 0 to 1, even though the last term is at x = 1 - 1/n. Therefore, the approximation still holds.Therefore, the conclusion is that the limit is π/4.But just to make sure, let's compute the integral explicitly. Let me compute ∫₀¹ sqrt(1 - x²) dx.To compute this integral, we can use substitution. Let x = sinθ, so dx = cosθ dθ. When x = 0, θ = 0; when x = 1, θ = π/2. The integral becomes:∫₀^{π/2} sqrt(1 - sin²θ) * cosθ dθ = ∫₀^{π/2} cosθ * cosθ dθ = ∫₀^{π/2} cos²θ dθUsing the identity cos²θ = (1 + cos2θ)/2, the integral becomes:∫₀^{π/2} (1 + cos2θ)/2 dθ = (1/2) ∫₀^{π/2} 1 dθ + (1/2) ∫₀^{π/2} cos2θ dθCalculating the first integral: (1/2)[θ]₀^{π/2} = (1/2)(π/2 - 0) = π/4Calculating the second integral: (1/2)[(1/2)sin2θ]₀^{π/2} = (1/4)[sinπ - sin0] = (1/4)(0 - 0) = 0Therefore, the total integral is π/4 + 0 = π/4. So that's correct.Therefore, the original limit is π/4. So the answer should be π/4, which can be written as boxed{dfrac{pi}{4}}.Wait, but just to be thorough, let me consider if there might be any step where I made a miscalculation or assumption that's not valid.First, the conversion from the sum to the integral: the key step is recognizing that the sum is a Riemann sum. Each term is of the form f(k/n) * (1/n), where f(x) = sqrt(1 - x²). As n approaches infinity, the sum approaches the integral of f(x) from 0 to 1. This is standard for Riemann sums, so that step seems valid.Another possible concern: the original sum is from k = 0 to n - 1, which translates to x_k = k/n from 0 to (n - 1)/n. So as n grows, the upper limit approaches 1 from the left. However, since the function sqrt(1 - x²) is continuous on [0, 1], the difference between the Riemann sum up to (n - 1)/n and up to 1 becomes negligible as n approaches infinity. Therefore, the limit is indeed the integral from 0 to 1.Alternatively, if we considered the sum from k = 1 to n, then x_k would be k/n, which goes up to 1. But in our case, it's k = 0 to n - 1. However, since we're missing only the term at x = 1, which is sqrt(1 - 1²) = 0, that term would be zero anyway. Therefore, the sum from k = 0 to n - 1 is the same as the sum from k = 0 to n, since the nth term is zero. Therefore, there's no difference in the limit.Therefore, all steps seem valid, and the conclusion that the limit is π/4 is correct.Final AnswerThe limit is boxed{dfrac{pi}{4}}.</think>To compute the limit (lim _{n rightarrow infty} sum_{k=0}^{n-1} frac{sqrt{n^{2}-k^{2}}}{n^{2}}), we start by simplifying the term inside the sum:[frac{sqrt{n^2 - k^2}}{n^2} = frac{n sqrt{1 - left(frac{k}{n}right)^2}}{n^2} = frac{sqrt{1 - left(frac{k}{n}right)^2}}{n}]Thus, the sum can be rewritten as:[sum_{k=0}^{n-1} frac{sqrt{1 - left(frac{k}{n}right)^2}}{n}]This is a Riemann sum for the function (f(x) = sqrt{1 - x^2}) over the interval ([0, 1]) with right endpoints (x_k = frac{k}{n}) and (Delta x = frac{1}{n}). As (n to infty), the Riemann sum converges to the integral:[int_0^1 sqrt{1 - x^2} , dx]To evaluate this integral, we use the substitution (x = sin theta), (dx = cos theta , dtheta). The limits of integration change from (x = 0) to (theta = 0) and from (x = 1) to (theta = frac{pi}{2}). The integral becomes:[int_0^{pi/2} cos^2 theta , dtheta]Using the identity (cos^2 theta = frac{1 + cos 2theta}{2}), we get:[int_0^{pi/2} frac{1 + cos 2theta}{2} , dtheta = frac{1}{2} int_0^{pi/2} 1 , dtheta + frac{1}{2} int_0^{pi/2} cos 2theta , dtheta]Evaluating these integrals:[frac{1}{2} left[ theta right]_0^{pi/2} = frac{pi}{4}][frac{1}{2} left[ frac{sin 2theta}{2} right]_0^{pi/2} = 0]Thus, the integral evaluates to (frac{pi}{4}). Therefore, the limit is:[boxed{dfrac{pi}{4}}]]

🔑:29.50. Answer: pi / 4. The considered sum can be written as sum_{k=0}^{n-1} frac{1}{n} sqrt{1-left(frac{k}{n}right)^{2}}. As n rightarrow infty, this sum tends to kappa int_{0}^{1} sqrt{1-x^{2}} d x=frac{pi}{4}.

🧠:[<think>Okay, let's try to figure out this problem. So, there's a confectioner who has to make a certain number of sweets over three days. The problem tells us how much she made each day in fractions of the remaining order, and on the third day, she made 40 sweets to finish the order. We need to find out the total number of sweets ordered. Hmm, okay, let's break this down step by step.First, let me note down the key information:1. On the first day, she made 2/5 of the ordered sweets.2. On the second day, she made 5/9 of the remaining sweets.3. On the third day, she made the remaining 40 sweets.We need to find the total number of sweets ordered. Let's denote the total number of sweets ordered as ( x ). That seems like a good starting point. So, ( x ) is the total order.Alright, on the first day, she makes ( frac{2}{5} ) of the total order. So, that would be ( frac{2}{5}x ). Then, the remaining sweets after the first day would be the total order minus what she made on the first day. So, remaining after day 1 is ( x - frac{2}{5}x ). Let me compute that:( x - frac{2}{5}x = frac{5}{5}x - frac{2}{5}x = frac{3}{5}x ). So, after the first day, she has ( frac{3}{5}x ) left to make.Then, on the second day, she makes ( frac{5}{9} ) of the remaining sweets. The remaining sweets after day 1 are ( frac{3}{5}x ), so she makes ( frac{5}{9} times frac{3}{5}x ). Let's compute that:First, multiply the fractions: ( frac{5}{9} times frac{3}{5} ). The 5 in the numerator and denominator cancels out, and 3 divided by 9 is ( frac{1}{3} ), so that simplifies to ( frac{1}{3}x ). Therefore, on the second day, she makes ( frac{1}{3}x ) sweets.Then, the remaining sweets after the second day would be the remaining after day 1 minus what she made on day 2. So:Remaining after day 2 = ( frac{3}{5}x - frac{1}{3}x ).To subtract these fractions, I need a common denominator. The denominators are 5 and 3, so the least common denominator is 15. Converting both fractions:( frac{3}{5}x = frac{9}{15}x )( frac{1}{3}x = frac{5}{15}x )Subtracting: ( frac{9}{15}x - frac{5}{15}x = frac{4}{15}x )So, after the second day, she has ( frac{4}{15}x ) sweets left to make.On the third day, she makes the remaining 40 sweets. Therefore, the remaining after day 2 is equal to 40. So:( frac{4}{15}x = 40 )To find ( x ), we can solve this equation:Multiply both sides by 15:( 4x = 40 times 15 )Calculate the right side:40 multiplied by 15 is 600. So:( 4x = 600 )Divide both sides by 4:( x = 600 / 4 )Which is:( x = 150 )So, the total number of sweets ordered is 150.Wait a second, let me verify this again step by step to make sure I didn't make any mistakes.Starting with total order ( x = 150 ).First day: ( frac{2}{5} times 150 = frac{300}{5} = 60 ). So, made 60, remaining 150 - 60 = 90.Second day: ( frac{5}{9} times 90 = frac{450}{9} = 50 ). So, made 50, remaining 90 - 50 = 40.Third day: makes the remaining 40, which matches the problem statement.Yup, that checks out. So, 150 is the correct answer. Hmm, but let me try another way to verify, perhaps using algebraic fractions without plugging numbers.So, let's recap algebraically:Total ordered = x.After day 1: remaining = x - (2/5)x = (3/5)x.After day 2: remaining = (3/5)x - (5/9)*(3/5)x.Simplify:(3/5)x * [1 - 5/9] = (3/5)x * (4/9) = (12/45)x = (4/15)x.Then, this remaining is 40, so (4/15)x = 40 => x = (40 * 15)/4 = 150.Yep, same result. So that's solid. So, 150 is the answer. I think that's correct.But just to make sure, let's think of another perspective. Suppose the total order is 150. So:First day: 2/5 of 150 is 60, remaining 90.Second day: 5/9 of 90 is 50, remaining 40.Third day: 40. Perfect. So, all steps add up. So, 150 is the answer.Therefore, the confectioner was ordered to make 150 sweets.Final AnswerThe confectioner was ordered to make boxed{150} sweets.</think>Let ( x ) be the total number of sweets ordered.1. On the first day, the confectioner made ( frac{2}{5}x ) sweets. The remaining sweets after the first day are: [ x - frac{2}{5}x = frac{3}{5}x ]2. On the second day, she made ( frac{5}{9} ) of the remaining sweets. The amount made on the second day is: [ frac{5}{9} times frac{3}{5}x = frac{1}{3}x ] The remaining sweets after the second day are: [ frac{3}{5}x - frac{1}{3}x = frac{9}{15}x - frac{5}{15}x = frac{4}{15}x ]3. On the third day, she made the remaining 40 sweets. Therefore: [ frac{4}{15}x = 40 ] Solving for ( x ): [ x = 40 times frac{15}{4} = 40 times 3.75 = 150 ]Verification:- First day: ( frac{2}{5} times 150 = 60 ), remaining ( 150 - 60 = 90 )- Second day: ( frac{5}{9} times 90 = 50 ), remaining ( 90 - 50 = 40 )- Third day: 40 sweetsThus, the total number of sweets ordered is (boxed{150}).,<think>Okay, let's see. There's this problem where a confectioner has to make a certain number of sweets over three days. On the first day, she makes 2/5 of the ordered sweets. Then, on the second day, she makes 5/9 of the remaining sweets. Finally, on the third day, she makes the remaining 40 sweets. The question is asking how many sweets were ordered in total.Hmm, okay. So, first, let me try to break this down step by step. Let's denote the total number of sweets ordered as, say, x. That's the number we need to find.On the first day, she made 2/5 of the total order. So, the amount made on day one is (2/5)x. That leaves a remainder of sweets still to be made. The remaining sweets after day one would be the total minus what was made on day one. So, that's x - (2/5)x. Let me compute that: x is the same as 5/5 x, right? So, 5/5x - 2/5x = 3/5x. So, after the first day, she has 3/5 of the original order left to make.Moving on to the second day. She makes 5/9 of the remaining sweets. The remaining sweets after day one are 3/5x. So, on day two, she makes (5/9) of that. Let me calculate that: (5/9)*(3/5x). Let's see, the 5 in the numerator and the 5 in the denominator cancel out, leaving (3/9)x, which simplifies to (1/3)x. Wait, is that right? Wait, no, hold on. Wait, 5/9 of the remaining 3/5x. So, 5/9 * 3/5x. Multiply the numerators: 5*3=15. Multiply denominators: 9*5=45. So, that's 15/45x, which reduces to 1/3x. Okay, so she makes 1/3x on the second day. Therefore, the remaining sweets after day two would be the previous remainder minus day two's production. So, after day one, it was 3/5x, subtract day two's 1/3x. So, 3/5x - 1/3x. To subtract those fractions, I need a common denominator. 15 is the LCD. So, 3/5x is 9/15x, and 1/3x is 5/15x. So, 9/15x - 5/15x = 4/15x. Therefore, after day two, she has 4/15x left to make. Then, on the third day, she makes the remaining 40 sweets. So, that 4/15x equals 40. Therefore, 4/15x = 40. To solve for x, we can multiply both sides by 15/4. So, x = 40*(15/4). Let's compute that: 40 divided by 4 is 10, then 10 times 15 is 150. So, x = 150. Therefore, the total number of sweets ordered should be 150.Wait, let me verify this. Let's check each day's production to see if it adds up.Total order: 150.First day: 2/5 of 150. That's (2/5)*150 = 60. So, she makes 60 sweets. Remaining: 150 - 60 = 90.Second day: 5/9 of the remaining 90. So, (5/9)*90 = 50. She makes 50 sweets. Remaining: 90 - 50 = 40.Third day: the remaining 40. That matches the problem statement. So, yes, 60 + 50 + 40 = 150. That checks out.Therefore, the answer should be 150. So, the confectioner was ordered to make 150 sweets.Final AnswerThe confectioner was ordered to make boxed{150} sweets.</think>Let ( x ) be the total number of sweets ordered.1. On the first day, the confectioner made (frac{2}{5}) of the ordered sweets: [ text{Sweets made on the first day} = frac{2}{5}x ] The remaining sweets after the first day: [ x - frac{2}{5}x = frac{3}{5}x ]2. On the second day, she made (frac{5}{9}) of the remaining sweets: [ text{Sweets made on the second day} = frac{5}{9} left( frac{3}{5}x right) = frac{1}{3}x ] The remaining sweets after the second day: [ frac{3}{5}x - frac{1}{3}x = frac{9}{15}x - frac{5}{15}x = frac{4}{15}x ]3. On the third day, she made the remaining 40 sweets: [ frac{4}{15}x = 40 ] Solving for ( x ): [ x = 40 times frac{15}{4} = 40 times 3.75 = 150 ]Verification:- First day: (frac{2}{5} times 150 = 60)- Remaining after first day: (150 - 60 = 90)- Second day: (frac{5}{9} times 90 = 50)- Remaining after second day: (90 - 50 = 40)- Third day: 40 sweetsAll steps check out, confirming the total order was 150 sweets.[boxed{150}]]

🔑:Solution. Let x denote the number of sweets ordered. Then on the first day, frac{2}{5} x of the sweets were made, on the second day, frac{5}{9}left(x-frac{2}{5} xright) were made, and on the third day, 40 sweets were made. Therefore,frac{2}{5} x+frac{5}{9}left(x-frac{2}{5} xright)+40=xBy solving this equation, it is found that the number of sweets ordered is x=150.

🧠:[<think>Okay, let's see. I need to find the value of 'a' such that 5^a + 5^(a+1) equals the square root of 4500. Hmm. Alright, let's start by writing down the equation:5^a + 5^(a+1) = √4500First, I remember that when you have exponents with the same base, there might be a way to combine them. Let's look at the left side. The terms are 5^a and 5^(a+1). Since 5^(a+1) is the same as 5^a multiplied by 5^1, right? Because when you add exponents, you multiply the bases, but here it's 5^(a+1) which is 5^a * 5. So maybe I can factor out 5^a from both terms.Let me try that:5^a + 5^(a+1) = 5^a + 5^a * 5 = 5^a (1 + 5) = 5^a * 6So that simplifies the left side to 6 * 5^a. Cool, that seems manageable. Then the equation becomes:6 * 5^a = √4500Now, I need to solve for 'a'. Let's first simplify the right side, √4500. Let me break down 4500 into factors to see if I can simplify the square root.4500. Let's see, 4500 divided by 100 is 45, so 4500 = 45 * 100. And 100 is 10^2, which is a perfect square. So √4500 = √(45 * 100) = √45 * √100 = √45 * 10. Then √45 can be simplified further. 45 is 9 * 5, and 9 is a perfect square. So √45 = √(9*5) = 3√5. Therefore, √4500 = 3√5 * 10 = 30√5.So now, the equation is:6 * 5^a = 30√5Hmm, let's write that again:6 * 5^a = 30√5I can divide both sides by 6 to isolate 5^a:5^a = (30√5)/6Simplify the right side: 30 divided by 6 is 5, so:5^a = 5√5Alright, so 5^a equals 5 times the square root of 5. Let me express √5 as 5^(1/2). Then:5^a = 5 * 5^(1/2)When multiplying with the same base, we add exponents. So 5 is 5^1, and 5^(1/2) is as is. Therefore:5^a = 5^(1 + 1/2) = 5^(3/2)So now we have 5^a = 5^(3/2). Since the bases are the same and positive (and not equal to 1), the exponents must be equal. Therefore:a = 3/2Wait, let me check that again. Starting from:5^a = 5√5Expressed as exponents:5^a = 5^1 * 5^(1/2) = 5^(1 + 1/2) = 5^(3/2)Yes, so that's correct. Therefore, a = 3/2. Hmm, 3/2 is 1.5. Let me verify the original equation with a = 3/2 to ensure there are no mistakes.First, calculate 5^(3/2). That's the same as √(5^3) = √(125) ≈ 11.1803.Then 5^(3/2 + 1) = 5^(5/2) = √(5^5) = √(3125) ≈ 55.9017.Adding those two: 11.1803 + 55.9017 ≈ 67.082.Now, check the right side: √4500. Let's compute √4500. 4500 is 100 * 45, so √4500 = √100 * √45 = 10 * √45 ≈ 10 * 6.7082 ≈ 67.082. So that matches. So the answer a = 3/2 is correct.Alternatively, if I didn't want to approximate, I could verify algebraically:Left side: 5^(3/2) + 5^(3/2 + 1) = 5^(3/2) + 5^(5/2) = 5^(3/2) (1 + 5^(1)) = 5^(3/2) * 6Right side: √4500 = 30√5 as earlier. So 5^(3/2) * 6 = 6 * 5^(3/2). Let's express 30√5 as 6 * 5^(3/2). Let's check:30√5 = 6 * 5 * √5 = 6 * (5^1) * (5^(1/2)) = 6 * 5^(3/2). Yes, that's the same as the left side. So everything checks out.Therefore, the value of a is 3/2. Which is 1.5 in decimal, but since the question probably expects a fractional answer, 3/2 is the way to go.Final AnswerThe value of ( a ) is boxed{dfrac{3}{2}}.</think>To determine the value of ( a ) for which ( 5^a + 5^{a+1} = sqrt{4500} ), we start by simplifying the left side of the equation:[5^a + 5^{a+1}]We can factor out ( 5^a ):[5^a + 5^a cdot 5 = 5^a (1 + 5) = 5^a cdot 6]Thus, the equation becomes:[6 cdot 5^a = sqrt{4500}]Next, we simplify the right side. We know that:[sqrt{4500} = sqrt{45 cdot 100} = sqrt{45} cdot sqrt{100} = 3sqrt{5} cdot 10 = 30sqrt{5}]Substituting this back into the equation, we get:[6 cdot 5^a = 30sqrt{5}]Dividing both sides by 6:[5^a = 5sqrt{5}]Expressing ( 5sqrt{5} ) in terms of exponents:[5sqrt{5} = 5^1 cdot 5^{1/2} = 5^{1 + 1/2} = 5^{3/2}]Therefore, we have:[5^a = 5^{3/2}]Since the bases are the same, the exponents must be equal:[a = frac{3}{2}]Thus, the value of ( a ) is (boxed{dfrac{3}{2}}).,<think>Okay, so I have this equation to solve: 5^a + 5^{a+1} = sqrt(4500). Hmm, I need to find the value of 'a' that satisfies this. Let me start by breaking down the problem step by step. First, I remember that when dealing with exponents, especially with the same base, it's often helpful to factor out common terms. Here, both terms on the left side have 5 raised to some power. Specifically, 5^a and 5^{a+1}. Let me see if I can factor out 5^a from both terms. So, 5^a + 5^{a+1} can be written as 5^a * (1 + 5). Because 5^{a+1} is the same as 5^a * 5^1, right? So factoring out 5^a gives me 5^a times (1 + 5). That simplifies the left side to 5^a * 6. So now the equation is 6 * 5^a = sqrt(4500). Cool, that seems simpler. Next, I need to solve for 'a'. Let me write that equation again: 6 * 5^a = sqrt(4500). First, maybe I can isolate 5^a. To do that, I can divide both sides by 6. So, 5^a = sqrt(4500) / 6. Now, I need to compute sqrt(4500). Let me simplify sqrt(4500). Let's break down 4500 into prime factors to see if I can simplify the square root. 4500. Well, 4500 divided by 100 is 45, so 4500 = 45 * 100. 100 is 10^2, so sqrt(100) is 10. Then sqrt(45) is sqrt(9*5) which is 3*sqrt(5). So putting it together, sqrt(4500) = sqrt(45*100) = sqrt(45)*sqrt(100) = 3*sqrt(5)*10 = 30*sqrt(5). So sqrt(4500) is 30*sqrt(5). Therefore, substituting back into the equation: 5^a = (30*sqrt(5))/6. Simplifying the right side: 30 divided by 6 is 5, so 5*sqrt(5). Therefore, 5^a = 5*sqrt(5). Hmm, okay. Now, 5*sqrt(5) can be written as 5^1 * 5^(1/2) because sqrt(5) is 5^(1/2). When multiplying exponents with the same base, we add the exponents. So 5^1 * 5^(1/2) = 5^(1 + 1/2) = 5^(3/2). Therefore, 5^a = 5^(3/2). Since the bases are the same and the equation holds, the exponents must be equal. So, a = 3/2. Wait, that seems straightforward. Let me verify my steps to make sure I didn't make a mistake. Starting from the original equation: 5^a + 5^{a+1} = sqrt(4500). Factored left side as 5^a(1 + 5) = 6*5^a. Then divided both sides by 6 to get 5^a = sqrt(4500)/6. Simplifying sqrt(4500) step by step: sqrt(4500) = sqrt(45 * 100) = sqrt(45)*sqrt(100) = 3*sqrt(5)*10 = 30*sqrt(5). Divided by 6 gives 5*sqrt(5). Then recognized that 5*sqrt(5) is 5^(3/2). Therefore, exponents must be equal, so a = 3/2. Let me check with a calculator. Let's compute both sides. First, compute the left side when a = 3/2: 5^(3/2) + 5^(3/2 + 1) = 5^(3/2) + 5^(5/2). 5^(3/2) is sqrt(5^3) = sqrt(125) ≈ 11.1803. 5^(5/2) is sqrt(5^5) = sqrt(3125) ≈ 55.9017. Adding them together: 11.1803 + 55.9017 ≈ 67.082. Now, compute sqrt(4500). 4500 is 45*100, so sqrt(4500) = sqrt(45)*10 ≈ 6.7082*10 ≈ 67.082. Yes, both sides equal approximately 67.082, so that checks out. Therefore, a = 3/2 is indeed the correct solution. Alternatively, maybe I can approach this using logarithms. Let's see. Starting from 5^a = 5^(3/2). But since the bases are equal, exponents must be equal. So a = 3/2. Alternatively, if I didn't recognize that 5*sqrt(5) is 5^(3/2), I could take the logarithm base 5 of both sides. Let's try that. Starting from 5^a = 5*sqrt(5). Taking log base 5: log_5(5^a) = log_5(5*sqrt(5)). Left side simplifies to a. Right side: log_5(5) + log_5(sqrt(5)) = 1 + (1/2) because sqrt(5) is 5^(1/2). Therefore, log_5(5*sqrt(5)) = 1 + 1/2 = 3/2. Therefore, a = 3/2. Same result. So, all methods converge to a = 3/2. Therefore, the answer is 3/2. Alternatively, 3/2 can be written as 1.5, but since the question doesn't specify the format, but given the original equation uses fractions in exponents, 3/2 is probably the expected answer. Wait, just to recap all steps once more: Original equation: 5^a + 5^{a+1} = sqrt(4500)Factor left side: 5^a(1 + 5) = 6*5^aThus, 6*5^a = sqrt(4500) => 5^a = sqrt(4500)/6Simplify sqrt(4500): sqrt(4500) = sqrt(100*45) = 10*sqrt(9*5) = 10*3*sqrt(5) = 30*sqrt(5)Therefore, 5^a = 30*sqrt(5)/6 = 5*sqrt(5) = 5^(3/2)Thus, a = 3/2. All steps are solid, cross-checked with logarithms and decimal approximations. Therefore, confident that the answer is 3/2.Final AnswerThe value of ( a ) is boxed{dfrac{3}{2}}.</think>To determine the value of ( a ) for which ( 5^{a} + 5^{a+1} = sqrt{4500} ), we start by factoring the left side of the equation:[5^{a} + 5^{a+1} = 5^{a} (1 + 5) = 6 cdot 5^{a}]This simplifies the equation to:[6 cdot 5^{a} = sqrt{4500}]Next, we isolate ( 5^{a} ) by dividing both sides by 6:[5^{a} = frac{sqrt{4500}}{6}]We simplify ( sqrt{4500} ):[sqrt{4500} = sqrt{45 cdot 100} = sqrt{45} cdot sqrt{100} = 3sqrt{5} cdot 10 = 30sqrt{5}]Substituting this back into the equation, we get:[5^{a} = frac{30sqrt{5}}{6} = 5sqrt{5}]Recognizing that ( 5sqrt{5} ) can be written as ( 5^{1} cdot 5^{1/2} = 5^{3/2} ), we equate the exponents:[5^{a} = 5^{3/2}]Thus, the value of ( a ) is:[a = frac{3}{2}]The final answer is (boxed{dfrac{3}{2}}).]

🔑:Using exponent laws,begin{aligned}5^{a}+5^{a+1} & =sqrt{4500} 5^{a}left(1+5^{1}right) & =30 sqrt{5} 5^{a} cdot 6 & =30 sqrt{5} 5^{a} & =5 sqrt{5} 5^{a} & =5^{1} cdot 5^{1 / 2} 5^{a} & =5^{3 / 2}end{aligned}and so a=frac{3}{2}.ANSWER: frac{3}{2}