🧠:[<think>Okay, so I need to find the largest natural number that is less than the sum of the squares of its digits. Hmm, let me start by understanding the problem properly. First, let's break down the problem statement. A natural number is a positive integer (1, 2, 3, ...). We need to find the largest such number, let's call it N, where N is less than the sum of the squares of its digits. So, if N has digits d1, d2, ..., dk, then we want N < (d1² + d2² + ... + dk²). Our goal is to find the biggest N where this inequality holds.Let me think about how to approach this. Maybe start by checking some numbers and see when this condition is true. But since we need the largest one, maybe there's a pattern or a point beyond which all numbers don't satisfy the condition. So perhaps numbers above a certain size can't satisfy N < sum of squares of digits. Then, our job is to find that threshold and check the numbers below it in descending order until we find the largest one that meets the condition.To get a sense, let's consider the maximum possible sum of the squares of the digits for a number with a given number of digits. For example, a 1-digit number: the maximum sum is 9² = 81. So for a 1-digit number N, N < 81. But since the maximum 1-digit number is 9, and 9 < 81 is true. But wait, all 1-digit numbers satisfy this condition because even 9 is less than 81. But we need the largest natural number, so maybe multi-digit numbers.Wait, hold on. Let me correct that. If N is a 1-digit number, then the sum of the squares of its digits is just N². So the condition becomes N < N². Which simplifies to N² - N > 0 => N(N - 1) > 0. Since N is a natural number, this holds true for all N > 1. So for 1-digit numbers, all numbers from 2 to 9 satisfy N < sum of squares of digits. But since 9 is the largest 1-digit number, and it's less than 81, which is 9², that's true.But we need the largest natural number overall, not just among 1-digit numbers. So we need to check 2-digit, 3-digit, etc., numbers. Let's consider 2-digit numbers. The maximum sum of squares here is 9² + 9² = 81 + 81 = 162. So a 2-digit number can go up to 99. 99 is less than 162? Let's check: 99 < 162. Yes, that's true. So 99 would satisfy the condition. But wait, does it? The sum of its digits squared is 9² + 9² = 81 + 81 = 162. So 99 < 162. So yes, 99 satisfies the condition. But is 99 the largest number satisfying this?Wait, but 162 is a 3-digit number. Let me check if 162 satisfies the condition. The sum of its digits squared is 1² + 6² + 2² = 1 + 36 + 4 = 41. So 162 < 41? No, 162 is not less than 41. So 162 doesn't satisfy the condition. Therefore, numbers with more digits might not automatically satisfy the condition.So perhaps as numbers get larger, the sum of the squares of their digits can't keep up with the number itself. So we need to find the point where the number becomes larger than the sum of its digits squared. Let's analyze that.For a number with k digits, the maximum sum of squares of digits is k * 9² = 81k. The minimum number with k digits is 10^(k-1). So we need to find the k where 81k < 10^(k-1). Because once 81k is less than 10^(k-1), then even the maximum possible sum of squares for a k-digit number is less than the smallest k-digit number. Therefore, beyond that k, no k-digit number will satisfy the condition N < sum of squares of digits. Let's find that k.Let's compute for k=1: 81*1=81, 10^(1-1)=1. 81 >1. So 1-digit numbers can satisfy.k=2: 81*2=162, 10^(2-1)=10. 162>10. So 2-digit numbers can satisfy.k=3: 81*3=243, 10^(3-1)=100. 243>100. Still possible.k=4: 81*4=324, 10^(4-1)=1000. 324 < 1000. So here, for k=4, the maximum sum of squares (324) is less than the minimum 4-digit number (1000). Therefore, no 4-digit number can satisfy the condition N < sum of squares of digits. Therefore, the number we're looking for must have at most 3 digits.Therefore, the largest possible candidate is a 3-digit number. So we need to check 3-digit numbers starting from 999 downwards until we find the largest one where N < sum of squares of digits.But checking all 999 numbers would take too long. Maybe there's a smarter way. Let's see.Alternatively, note that for numbers greater than or equal to 1000, as we saw, they can't satisfy the condition. So our upper bound is 999. Let's start checking from 999 and go down until we find the first one that satisfies N < sum of squares of digits.But let's compute the sum of squares of digits for 999: 9² + 9² + 9² = 81 + 81 + 81 = 243. Then 999 < 243? No, 999 is much larger. So 999 doesn't satisfy.Then 998: sum is 9² + 9² + 8² = 81 + 81 + 64 = 226. 998 < 226? No. Similarly, 997: sum is 81 + 81 + 49 = 211. 997 < 211? No. This pattern continues. Each time we subtract 1 from the units digit, the sum decreases by 2*previous digit -1 (since (n-1)^2 = n² - 2n +1). But the number itself is decreasing by 1. However, the sum is decreasing much more slowly than the number. Wait, actually, the number is decreasing by 1 each time, but the sum could decrease by more or less depending on the digits.Wait, maybe starting from 999 is not the right approach. Because even 100 is 100, and its sum is 1 + 0 + 0 =1. So 100 is not less than 1. So maybe the numbers we need are somewhere in between. Wait, but we need the largest N, so even if numbers like 99 (which is 2-digit) satisfy, but if there's a larger 3-digit number that also satisfies, then that would be our answer.But from the previous check, 999 doesn't satisfy. Let's check 199: sum is 1 + 81 + 81 = 163. 199 < 163? No. 179: 1 + 49 + 81 = 131. 179 <131? No. Wait, maybe numbers with lower digits?Wait, perhaps the maximum sum for a 3-digit number is 243 (all 9s), but even 243 is less than 1000. Wait, but the maximum sum is 243, so if a 3-digit number is less than 243, then maybe it can satisfy the condition. So numbers from 100 up to 243. Wait, but 243 is the maximum sum. So for a 3-digit number N, the sum of squares is at most 243. Therefore, if N is less than 243, then N could potentially be less than the sum. But N is a 3-digit number, so the minimum N is 100, and the maximum N we need to check is 243. Wait, but 243 is a 3-digit number. Let's check 243: sum is 4 + 16 + 9 = 29. So 243 <29? No. So even though the sum can be up to 243, the actual sum for 243 is 29, which is way lower. Therefore, even if the maximum possible sum is 243, individual numbers might have sums much lower.Therefore, perhaps we need to check all numbers from 243 downwards, but even that is a lot. Alternatively, maybe there's a certain point where numbers below a certain value can satisfy the condition. Let's think about this.Suppose N is a 3-digit number: abc (a, b, c are digits). Then N = 100a +10b + c. The sum S = a² + b² + c². We need N < S. So 100a +10b + c < a² + b² + c².Rearranging terms: a² -100a + b² -10b + c² -c >0.Hmm, this might not be the easiest way. Maybe instead, think of the inequality per digit. Let's see.Alternatively, think of how much each digit contributes to the number versus to the sum. For the hundreds digit 'a', it contributes 100a to N and a² to S. So the difference is 100a - a². Similarly, the tens digit contributes 10b vs b², and the units digit contributes c vs c². So for each digit, we can compute how much it's "costing" in terms of the inequality. To have N < S, the sum of the squares needs to make up for the "cost" of each digit.So for the hundreds digit, 100a - a² must be less than the combined contributions from the other digits. But maybe this is too vague.Alternatively, let's note that for a 3-digit number, the hundreds digit is the main contributor to N. For example, if a=1, then 100a =100. The sum S must be greater than 100. So S >=101. Let's see if that's possible. For a=1, S=1 + b² + c². So 1 + b² + c² > 100 +10b +c. So 1 + b² + c² -10b -c >100. Hmm, that seems difficult. Let's plug in possible values.Take a=1. Then to get S >100, we need b and c such that 1 + b² +c² >100. So b² + c² >=100. Let's see, the maximum for b and c is 9 each. So 9² +9²=81+81=162. So 1 +162=163. So the maximum S when a=1 is 163. So N=199 gives S=1+81+81=163. Then N=199 vs S=163. 199 <163? No. So even the maximum possible S when a=1 is 163, which is still less than N=199. So all numbers with a=1 (100-199) have S <=163, which is less than N (which is at least 100). So when a=1, even the maximum S is 163, but the minimum N is 100, but actually, for N=100, S=1, which is way less. So all numbers from 100 to 199 can't satisfy N < S. Because even the maximum S in that range (163) is less than the corresponding N (199). Therefore, none of the 100-199 numbers satisfy the condition.Therefore, we can skip all numbers from 100-199. Now moving to a=2. Then N ranges from 200-299. Let's check if any of these numbers satisfy N < S.For a=2, S =4 +b² +c². The maximum S here is 4 +81 +81=166. So the maximum S for a=2 is 166. The minimum N is 200. So 166 <200. Therefore, even the maximum S in the 200-299 range is 166, which is less than the minimum N=200. Therefore, none of the numbers from 200-299 can satisfy N < S.Similarly, for a=3, N ranges from 300-399. S=9 +b² +c². Maximum S=9+81+81=171 <300. So no numbers here either.Similarly, for a=4: S=16 +b² +c². Maximum 16+81+81=178 <400.a=5: 25 +81 +81=187 <500.Same logic applies for a=6,7,8,9. So all 3-digit numbers (from 100 to 999) have S <=243, but N starts at 100. But even 243 < 100? No, 243 is greater than 100, but N=243 is 243, and S for N=243 is 2² +4² +3² =4 +16 +9=29. So N=243, S=29. 243 <29? No. So S can be as high as 243 (if the number is 999), but N=999, S=243. So 999 <243? No. So all 3-digit numbers have S <=243, but N is at least 100. But the problem is, even for the maximum S=243, which occurs at N=999, the S=243 is still less than N=999. Therefore, no 3-digit number satisfies N < S.Wait, but earlier we saw that 99 (a 2-digit number) has S=162, and 99 <162. So 99 satisfies. But the question is asking for the largest natural number, so maybe 99 is the answer. But wait, are there any 3-digit numbers that satisfy?Wait, earlier analysis said that all 3-digit numbers have S <=243, and the minimum N is 100, but in reality, for 3-digit numbers, even the maximum S is 243, which is less than N=999. But maybe there's a 3-digit number where N < S. Wait, but 100 <= N <=999, and S <=243. So N must be less than S, so N has to be less than 243. So possible N is from 100 to 242. But even then, let's check N=199, S=163. 199 <163? No. N=163, S=1+36+9=46. 163 <46? No. Wait, how about smaller numbers. Let's check N=122: S=1 +4 +4=9. 122 <9? No.Wait a second, maybe even in the 3-digit numbers, none satisfy N < S. Then the answer must be a 2-digit number. But we need to confirm that.But 99 is 99 < 162 (sum of squares 9²+9²=162). So 99 works. Now, is there a larger 2-digit number that works? No, because 99 is the largest 2-digit number. But wait, maybe a 3-digit number that's lower than 99? No, because all 3-digit numbers are 100 and above, which is higher than 99. So 99 is a candidate.But wait, wait. Let's check 99: sum of squares is 162. 99 <162. So 99 is valid. Is there a higher number than 99 where N < sum of squares? But 99 is the highest 2-digit number, and all 3-digit numbers don't satisfy. So 99 might be the answer.Wait, but let's check if there are any 2-digit numbers higher than 99. There aren't. 99 is the highest 2-digit number. So if all 3-digit numbers don't satisfy, then 99 is the answer. But wait, let me check some 2-digit numbers:Take 89: sum is 64 +81=145. 89 <145: yes. 98: 81 +64=145. 98 <145: yes. But 99 gives a higher N, so 99 is better.But wait, let's check 199: sum is 1 +81 +81=163. N=199 vs S=163. 199 <163? No. So 199 doesn't satisfy. So 99 is still the candidate.But let me check another angle. Wait, what about numbers like 123: 1 +4 +9=14. 123 <14? No. How about 100:1 <100. No. So 3-digit numbers are out. So the answer is 99.But wait, wait. Wait, let me check a 3-digit number like 100: sum is 1, which is less than 100. 101: sum is 1+0+1=2 <101. So all 3-digit numbers are out. Then 2-digit numbers: let's check.The largest 2-digit number is 99, sum is 162, which is greater than 99. So 99 is valid. Now, is 99 the largest natural number where N < sum of squares of digits? Let me check numbers around 99. For example, 99 is the largest 2-digit number, so the next number is 100, which is invalid. So yes, 99 is the answer.But let me verify with a few other 2-digit numbers. For example, 89: sum is 8² +9²=64+81=145. 89 <145: yes. 79: 49 +81=130. 79 <130: yes. So lower numbers also satisfy. But 99 is the largest.Wait, but let's confirm that there isn't a higher 2-digit number than 99. But 99 is the highest. So 99 is the answer. But hold on, in the problem statement, it says "the largest natural number that is less than the sum of the squares of its digits". So the answer should be 99. But wait, let me check once again.Wait, but what about numbers with more digits? For example, 1234 is a 4-digit number, but as established earlier, for numbers with 4 or more digits, the sum of squares can't exceed 81*4=324, while the number itself is at least 1000. So 1000 >324, so no. So 3-digit numbers max sum 243, which is less than 100. Wait, 243 is a sum, but the number itself is 100 to 999. So 100-999 numbers have sums from 1 to 243. So all 3-digit numbers have N >=100, sum <=243. But 100 <=N<=999. So the sum can be up to 243. Therefore, if N is less than 243, maybe N=199, sum=163. 199 <163? No. So even the sum of 243 is for N=999, which is 999 <243? No. So no 3-digit numbers work.Thus, 99 is the answer.Wait, but wait, let me check 199 again. Wait, 199's sum is 1 + 81 + 81 =163. 199 is not less than 163. So no. How about 163? N=163, sum is 1 +36 +9=46. 163 <46? No. So 163 is not. How about 100: sum=1. No.Wait, maybe the answer is 99, but let's check 1-digit numbers. The problem says "natural number", so 1 is included. For 1-digit numbers:N=1: sum=1. 1 <1? No.N=2: sum=4. 2 <4? Yes.N=3:9>3. Yes....N=9:81>9. Yes.But the largest 1-digit number is 9. But 9 is less than 81, so 9 satisfies. But 9 is a 1-digit number, but 99 is a 2-digit number which is larger. So 99 is larger than 9, and it satisfies. Therefore, 99 is the answer.Wait, but the problem is to find the largest natural number where N is less than the sum of the squares of its digits. So between 9 and 99, 99 is larger and satisfies, so 99 is the answer.Therefore, I think the answer is 99. But let me check a few more 2-digit numbers to be thorough. For example, 99 is the largest. Let's check 95: 81 +25=106. 95 <106: yes. 96:81 +36=117. 96 <117: yes. 97:81 +49=130. 97 <130: yes. 98:81 +64=145. 98 <145: yes. 99:81 +81=162. 99 <162: yes. So all these numbers satisfy. Since 99 is the largest 2-digit number, and all 3-digit numbers don't satisfy, then 99 is indeed the largest natural number that satisfies the condition.Therefore, the answer is 99.Final AnswerThe largest natural number that is less than the sum of the squares of its digits is boxed{99}.</think>To find the largest natural number ( N ) such that ( N ) is less than the sum of the squares of its digits, we start by analyzing the problem for numbers with different digit lengths.1. 1-digit numbers: For any 1-digit number ( N ), the sum of the squares of its digits is ( N^2 ). The condition ( N < N^2 ) holds for all ( N geq 2 ). The largest 1-digit number is 9, which satisfies ( 9 < 81 ).2. 2-digit numbers: The maximum sum of the squares of the digits for a 2-digit number is ( 9^2 + 9^2 = 162 ). The largest 2-digit number is 99, which gives a sum of ( 9^2 + 9^2 = 162 ). Since ( 99 < 162 ), 99 is a candidate.3. 3-digit numbers: For a 3-digit number, the maximum sum of the squares of its digits is ( 9^2 + 9^2 + 9^2 = 243 ). However, the smallest 3-digit number is 100, and even the maximum sum of 243 is less than the smallest 3-digit number (100). Therefore, no 3-digit number can satisfy the condition ( N < text{sum of squares of its digits} ).4. Numbers with more than 3 digits: For numbers with ( k geq 4 ) digits, the maximum sum of the squares of the digits is ( 81k ), which is significantly less than the smallest ( k )-digit number ( 10^{k-1} ). Thus, no numbers with 4 or more digits can satisfy the condition.After verifying all possibilities, the largest number that satisfies the condition is the 2-digit number 99, as it is less than the sum of the squares of its digits (162).[boxed{99}],<think>Alright, let me try to figure out this problem: Find the largest natural number that is less than the sum of the squares of its digits. Hmm, okay. So, I need to find a number where the sum of the squares of its digits is bigger than the number itself, and we need the largest such number. Let me break this down.First, let me make sure I understand the problem correctly. A natural number is a positive integer (1, 2, 3, ...). We need to find the largest number n such that n < sum of the squares of its digits. For example, if the number is 12, the sum of squares of digits would be 1^2 + 2^2 = 1 + 4 = 5. But 12 is not less than 5, so 12 wouldn't qualify. On the other hand, if we take 7, the sum is 7^2 = 49, and 7 < 49, so 7 would qualify. But obviously, 7 is a small number. We need the largest such number.Wait, but if we take a two-digit number, say 99: sum of squares is 81 + 81 = 162. 99 is less than 162? No, 162 is larger than 99, so 99 would satisfy the condition. But 99 is a two-digit number. But that can't be right because maybe even larger numbers satisfy this. Wait, but maybe for larger numbers, the sum of squares of their digits doesn't keep up with the number itself. Let's check a three-digit number. For example, 100: sum of squares is 1 + 0 + 0 = 1. 100 is not less than 1. 999: sum of squares is 81 + 81 + 81 = 243. 999 is not less than 243. So, 999 is way bigger. So maybe the number is somewhere in the two-digit or three-digit range. Hmm.Wait, maybe I need to find a balance between the number of digits and the sum of their squares. Let's think about the maximum possible sum of squares for a number with d digits. For a d-digit number, the maximum sum of squares would be d*9^2 = 81d. The minimum d-digit number is 10^{d-1}. So, we need to find the largest d where 81d >= 10^{d-1}. Let's check for different values of d.For d=1: 81*1 = 81, 10^{0}=1. So, 81 >=1, which is true. But since we are looking for numbers where the sum is greater than the number, for d=1, numbers 1 to 9: sum of squares is the square of the number itself. So, for numbers 1 to 9, the sum of squares is equal to the number squared, so the sum is larger than the number only when the number is 1 (1 < 1? No, 1 is not less than 1). Wait, actually, for n=1: sum is 1^2=1, so 1 is not less than 1. For n=2: sum is 4, so 2 <4, which is true. Similarly, up to n=9: sum is 81, and 9 <81. So all single-digit numbers from 2 to 9 satisfy the condition. But we need the largest number overall, not just single-digit. So maybe there are larger numbers with more digits that also satisfy the condition.Wait, for two-digit numbers: Let's take 99. Sum is 81+81=162. 99 <162: yes, so 99 satisfies. Then 100: sum is 1, which is way less. But 99 is a two-digit number, and 162 is larger than 99. How about three-digit numbers? Let's take 199: sum is 1 +81 +81=163. 199 vs 163: 163 <199, so 199 does not satisfy. How about 111: sum is 1+1+1=3 <111. So, maybe three-digit numbers don't satisfy. Let's check some three-digit numbers. 999: sum is 243. 999 is not less than 243. How about 179: sum is 1 +49 +81=131. 179 vs 131: 131 <179, so no. How about 129: sum is 1 +4 +81=86 <129. Not enough. Hmm. How about 99: sum is 162. 99 <162, so 99 is good. Let's check 100: sum is 1. 100 is larger. So maybe two-digit numbers are the candidates here. But wait, 99 is the largest two-digit number. But is there a three-digit number that could satisfy the condition? Maybe not. But let's check.Wait, what's the maximum sum for a three-digit number? 3*81=243. So the maximum possible sum is 243. Therefore, if a three-digit number is less than 243, then it could satisfy. So numbers from 100 up to 243. Let's check the upper bound: 243. Sum of squares: 4 + 16 + 9= 29. 29 <243, so no. Wait, 243's sum is 2^2 +4^2 +3^2=4+16+9=29. 29 is much smaller than 243. So 243 itself doesn't work. How about 199: sum is 1+81+81=163. 163 is less than 199. How about 163: sum is 1+36+9=46 <163. Not good. Wait, so maybe there's a three-digit number where the sum of squares is more than the number. Let me try 130: sum is 1+9+0=10 <130. 150: 1+25+0=26 <150. 200: 4+0+0=4 <200. 123:1+4+9=14 <123. 111: 3 <111. Hmm. How about 99, sum is 162. 99 is two-digit. Wait, maybe 99 is the largest two-digit number. How about numbers between 100 and 162? Let's check 150: sum is 1+25+0=26 <150. 130:10. 140:1+16+0=17. 120:1+4+0=5. 160:1+36+0=37. 159:1+25+81=107. 107 vs 159: 107 <159. 159 doesn't work. 150:26. 145:1+16+25=42 <145. 135:1+9+25=35 <135. 129:1+4+81=86 <129. 125:1+4+25=30 <125. Hmm. How about 99: sum is 162, which is bigger than 99. Let's check 163: but 163 is a three-digit number with sum 1+36+9=46. 46 <163. So no. Maybe there are no three-digit numbers where the sum is greater. Therefore, the candidate is 99. But wait, wait. Wait, is 99 really the largest? Let me check numbers between 99 and 100. Wait, 99 is the largest two-digit number. Then 100 is next. But 100 doesn't satisfy. So 99 might be the answer. But wait, let's check numbers just below 99. 98: sum is 81 +64=145. 98 <145: yes. 99 is 162. 99 <162. So 99 is bigger than 98, and the condition holds. So 99 is larger and satisfies. So maybe 99 is the answer. But let's verify if there are any three-digit numbers that satisfy. For example, 100: sum is 1, too small. 101:1+0+1=2. 102:1+0+4=5. 111:3. 123:14. 132:1+9+4=14. 143:1+16+9=26. 155:1+25+25=51. 169:1+36+81=118. 118 vs 169: 118 <169. 178:1+49+64=114. 114 <178. 189:1+64+81=146 <189. 199:1+81+81=163 <199. 219:4+1+81=86 <219. 249:4+16+81=101 <249. So none of the three-digit numbers seem to work. What about two-digit numbers above 99? There are no two-digit numbers above 99. So 99 is the largest two-digit number. But wait, is 99 the largest number that satisfies the condition?Wait, let's check 9999, just in case. Sum of squares is 4*81=324. 9999 is way larger than 324. So no. What about numbers with four digits? 1000: sum is 1. 1111:4. 1234:1+4+9+16=30. So, even the maximum sum for four digits is 4*81=324. So any four-digit number is at least 1000, which is way larger than 324. So four-digit numbers and above are out. So focus back on two-digit and three-digit. Wait, three-digit numbers go up to 999, but their maximum sum is 243. So even the maximum sum is 243, but 243 is a three-digit number. Wait, 243: sum is 2^2 +4^2 +3^2=4+16+9=29. 29 <243. So even 243 itself doesn't work.Wait, maybe I need to check if there's a three-digit number where the sum is greater than the number. Let's try 199: sum is 1 +81 +81=163. 163 <199. 163: sum is 1 +36 +9=46. 46 <163. 150:1+25+0=26. 130:1+9+0=10. 100:1. So, maybe the three-digit numbers just don't have enough sum. How about numbers like 122:1+4+4=9 <122. 111:3. 135:1+9+25=35. Hmm. Maybe there's a three-digit number where sum is larger. Wait, let's check 405: 16 +0 +25=41 <405. 500:25+0+0=25. 550:25+25+0=50. 999:243. No. So, maybe all three-digit numbers have sums that are less than the number. Therefore, the largest number is in the two-digit numbers. So 99.But let me check another angle. Let's think about the maximum possible sum for a two-digit number: 9^2 +9^2=162. So the sum can be up to 162. So numbers from 10 to 99. The largest two-digit number is 99, which has a sum of 162, which is bigger than 99. So 99 is a candidate. But maybe there is a three-digit number with a sum that's higher than the number itself. Wait, 99 is 162. What about numbers between 100 and 162? Let's check 100: sum is 1. 101:2. 102:5. 103:10. 104:17. 105:26. 106:37. 107:50. 108:65. 109:82. 110:2. 111:3. 112:6. 113:11. 114:18. 115:27. 116:38. 117:51. 118:66. 119:83. 120:5. 121:6. 122:9. 123:14. 124:21. 125:30. 126:41. 127:54. 128:69. 129:86. 130:10. 131:11. 132:14. 133:19. 134:26. 135:35. 136:46. 137:59. 138:74. 139:91. 140:17. 141:18. 142:21. 143:26. 144:33. 145:42. 146:53. 147:66. 148:81. 149:98. 150:26. 151:27. 152:30. 153:35. 154:42. 155:51. 156:62. 157:75. 158:90. 159:107. 160:37. 161:38. 162:41. Wait, let's check 159: sum is 1 +25 +81=107. 159 vs 107: 107 <159. So, sum is less. How about 149:1 +16 +81=98. 149 vs 98: 98 <149. Still sum is less. 139:1 +9 +81=91. 139 vs 91:91 <139. 129:1 +4 +81=86. 86 <129. 119:1 +1 +81=83. 83 <119. 109:1 +0 +81=82. 82 <109. So even numbers in the 100-162 range have sums that are still less than themselves. The highest sum in this range is 107 (for 159) and 107 is still less than 159. Therefore, none of the three-digit numbers in 100-162 satisfy the condition. Therefore, the largest number is 99.Wait, but let's check 99 again. Sum is 81 +81=162, which is greater than 99. So, 99 works. Then, is there a number larger than 99 (but three-digit) that works? Since the maximum sum for three-digit is 243, but numbers from 100 to 999 all are three-digit. However, as we saw, even 243's sum is 29, which is way less. So no three-digit numbers work. Therefore, 99 is the answer.But wait, let me check a few more numbers. For example, 89: sum is 64 +81=145. 89 <145: yes. 91:81 +1=82. 91 <82? No, 91 is not less than 82. So 91 doesn't work. 93:81 +9=90. 93 <90? No. 95:81 +25=106. 95 <106: yes. 95 works. 99 works. So 99 is larger. So 99 is better.Wait, let me check 199: sum is 1 +81 +81=163. 163 is more than 199? No, 163 <199. So 163 is less. 163: sum is 1 +36 +9=46. 46 <163. So no. How about 405: 16 +0 +25=41 <405. 500:25 <500. So, clearly, all three-digit numbers are out. How about two-digit numbers? 99 is the largest. So the answer must be 99.Wait, but let me check another two-digit number: 89. Sum is 145. 89 <145: yes. 99 is higher. 99: sum is 162. 99 <162: yes. So 99 is the largest two-digit number. Since all three-digit numbers don't work, 99 is the answer.But just to be thorough, let's check some numbers between 100 and 162. Let's take 150: sum is 1 +25 +0=26. 150 vs 26: nope. 140:1 +16 +0=17. 130:1 +9 +0=10. 120:1 +4 +0=5. 110:1 +1 +0=2. 160:1 +36 +0=37. 162:1 +36 +4=41. 162: sum is 41 <162. 158:1 +25 +64=90. 158 vs 90:90 <158. 149:1 +16 +81=98. 98 <149. 139:1 +9 +81=91. 91 <139. 129:1 +4 +81=86. 86 <129. 119:1 +1 +81=83. 83 <119. 109:1 +0 +81=82. 82 <109. So none of these numbers in the 100-162 range work. Therefore, 99 is indeed the largest number where the sum of the squares of its digits is greater than the number itself.But just to confirm, let me check 99 again. Its digits are 9 and 9. The sum is 81 +81=162. 99 <162: yes. So 99 satisfies the condition, and since there are no three-digit numbers that do, and 99 is the largest two-digit number, that's our answer.Final AnswerThe largest natural number that is less than the sum of the squares of its digits is boxed{99}.</think>To find the largest natural number that is less than the sum of the squares of its digits, we need to check numbers of different digit lengths and compare each number with the sum of the squares of its digits.1. Single-digit numbers: All single-digit numbers from 2 to 9 satisfy the condition since the square of each number is greater than the number itself. However, we need the largest number overall, not just single-digit numbers.2. Two-digit numbers: - For the largest two-digit number, 99, the sum of the squares of its digits is (9^2 + 9^2 = 81 + 81 = 162). Since (99 < 162), 99 satisfies the condition.3. Three-digit numbers: - The maximum sum of the squares of digits for a three-digit number is (3 times 81 = 243). However, numbers from 100 to 999 were checked, and none of them had a sum of squares greater than the number itself. For example, the sum for 199 is (1 + 81 + 81 = 163), which is less than 199.4. Numbers with more than three digits: - Numbers with four or more digits have sums of squares that are significantly smaller than the numbers themselves. For example, the sum for 1000 is 1, which is much smaller than 1000.After verifying all possibilities, the largest number found that satisfies the condition is 99. [boxed{99}]]

🔑:Solution. Let overline{a_{n} ldots a_{1} a_{0}} be the desired number, that is,a_{n} cdot 10^{n}+ldots+a_{1} cdot 10+a_{0}10^{k}-9>90, which means that the sum on the left side is positive. According to this, the number must be less than 100. Since 99<9^{2}+9^{2}, the number 99 is the desired number.4B The area of the base of a right triangular prism is equal to 4 mathrm{~cm}^{2}, and the areas of the lateral surfaces are equal to 9, 10, and 17 mathrm{~cm}^{2}. Determine the volume of the prism.

🧠:[<think>Okay, so I have this problem here: find the area of the region formed by points (x, y) that satisfy the inequality |x| + |y| + |x - 2| ≤ 4. Hmm, okay. Let me think about how to approach this. First, I know that inequalities involving absolute values often define regions bounded by lines or polygons, especially when dealing with Cartesian coordinates. Since there are multiple absolute value terms here, maybe I need to break this down into different cases based on the values of x. The expression has |x|, |x - 2|, and |y|. Let me see. The terms |x| and |x - 2| suggest that there might be critical points at x = 0 and x = 2. So, maybe I should consider different intervals for x: x < 0, 0 ≤ x ≤ 2, and x > 2. For each interval, the absolute value expressions |x| and |x - 2| can be simplified without the absolute value signs, depending on the sign of the expression inside. Then, once I handle the x terms, I can solve for y in each case to find the boundaries of the region. Let me start by considering the different cases for x.Case 1: x < 0If x is less than 0, then |x| = -x (since x is negative) and |x - 2| = -(x - 2) = -x + 2 (because x - 2 is also negative when x < 0). So substituting these into the inequality:|x| + |y| + |x - 2| ≤ 4=> (-x) + |y| + (-x + 2) ≤ 4Combine like terms:- x - x + 2 + |y| ≤ 4=> -2x + 2 + |y| ≤ 4Subtract 2 from both sides:-2x + |y| ≤ 2Then, isolate |y|:|y| ≤ 2 + 2xBut wait, x is less than 0 here. So 2 + 2x would be 2 + 2*(negative number). For example, if x approaches 0 from the left, 2 + 2x approaches 2. If x approaches negative infinity, 2 + 2x goes to negative infinity. However, the left-hand side |y| is always non-negative, so the inequality |y| ≤ 2 + 2x would only hold if 2 + 2x is non-negative. Therefore, 2 + 2x ≥ 0 => x ≥ -1. But in this case, x < 0, so combining x < 0 and x ≥ -1, this case reduces to x in [-1, 0). If x < -1, then 2 + 2x < 0, which would make |y| ≤ negative number, which is impossible. So actually, in the interval x < 0, the inequality is only valid when x ∈ [-1, 0). So for Case 1, x ∈ [-1, 0), the inequality simplifies to |y| ≤ 2 + 2x. So for these x values, y is between -(2 + 2x) and (2 + 2x). Therefore, the region in this case is a horizontal strip between x = -1 and x = 0, with the width varying linearly with x. Wait, actually, since |y| ≤ 2 + 2x, the vertical bounds for y are from - (2 + 2x) to + (2 + 2x). So for each x in [-1, 0), the vertical length is 2*(2 + 2x) = 4 + 4x. Hmm, okay. So integrating this over x from -1 to 0 would give the area in this region. But let's not get ahead of ourselves. Maybe it's better to sketch the graph or figure out the shape. Case 2: 0 ≤ x ≤ 2In this interval, |x| = x (since x is non-negative) and |x - 2| = -(x - 2) = 2 - x (since x - 2 is negative here). So substituting into the inequality:x + |y| + (2 - x) ≤ 4Simplify:x + |y| + 2 - x ≤ 4The x terms cancel out:|y| + 2 ≤ 4Subtract 2:|y| ≤ 2So in this case, when x is between 0 and 2, the inequality reduces to |y| ≤ 2, which is a horizontal strip from y = -2 to y = 2. So regardless of the x value between 0 and 2, y can be between -2 and 2. Therefore, in this region, the figure is a rectangle from x=0 to x=2, y=-2 to y=2. The area here would be length*width = (2 - 0)*(2 - (-2)) = 2*4 = 8. But wait, let me check if this is correct. If |y| ≤ 2, then y ranges from -2 to 2, and x ranges from 0 to 2. So yes, it's a rectangle with width 2 (along x-axis) and height 4 (along y-axis). So area 8. Case 3: x > 2Now, for x greater than 2, |x| = x (since x is positive) and |x - 2| = x - 2 (since x - 2 is positive). So substituting into the inequality:x + |y| + (x - 2) ≤ 4Combine like terms:x + x - 2 + |y| ≤ 4=> 2x - 2 + |y| ≤ 4Add 2 to both sides:2x + |y| ≤ 6Isolate |y|:|y| ≤ 6 - 2xNow, since x > 2 here, 6 - 2x would be 6 - 2*(something greater than 2). For example, if x = 3, 6 - 6 = 0. If x approaches 2 from the right, 6 - 4 = 2. Also, since |y| must be non-negative, 6 - 2x ≥ 0 => x ≤ 3. So in this case, x > 2 but x ≤ 3. So x ∈ (2, 3]. For each x in this interval, |y| ≤ 6 - 2x, so y ranges from -(6 - 2x) to 6 - 2x. Thus, the vertical length is 2*(6 - 2x) = 12 - 4x. So to summarize:- For x ∈ [-1, 0): |y| ≤ 2 + 2x, vertical length 4 + 4x- For x ∈ [0, 2]: |y| ≤ 2, vertical length 4- For x ∈ (2, 3]: |y| ≤ 6 - 2x, vertical length 12 - 4xSo the total area would be the integral over x from -1 to 0 of (4 + 4x) dx, plus the integral from 0 to 2 of 4 dx, plus the integral from 2 to 3 of (12 - 4x) dx. Let me compute each part step by step.First integral: x from -1 to 0, function 4 + 4x∫(4 + 4x) dx from -1 to 0Antiderivative is 4x + 2x²Evaluate from -1 to 0:At x=0: 4*0 + 2*(0)^2 = 0At x=-1: 4*(-1) + 2*(-1)^2 = -4 + 2*1 = -4 + 2 = -2Subtracting: 0 - (-2) = 2So the first area is 2.Second integral: x from 0 to 2, function 4∫4 dx from 0 to 2Antiderivative is 4xEvaluate from 0 to 2:4*2 - 4*0 = 8 - 0 = 8So the second area is 8.Third integral: x from 2 to 3, function 12 - 4x∫(12 - 4x) dx from 2 to 3Antiderivative is 12x - 2x²Evaluate at 3: 12*3 - 2*(3)^2 = 36 - 18 = 18Evaluate at 2: 12*2 - 2*(2)^2 = 24 - 8 = 16Subtracting: 18 - 16 = 2So the third area is 2.Adding them all up: 2 + 8 + 2 = 12. Hmm, so total area is 12. But wait, let me double-check because sometimes when dealing with absolute values and regions, it's easy to miss something. Alternatively, maybe I can visualize the shape. Let's think about the original inequality: |x| + |y| + |x - 2| ≤ 4. Since |y| is non-negative, the sum |x| + |x - 2| + |y| is minimized when |y| is 0, so |x| + |x - 2| ≤ 4. Let's see what |x| + |x - 2| looks like.The expression |x| + |x - 2| is a function of x. Let's analyze it:For x < 0: |x| + |x - 2| = -x + (-x + 2) = -2x + 2For 0 ≤ x ≤ 2: |x| + |x - 2| = x + (2 - x) = 2For x > 2: |x| + |x - 2| = x + (x - 2) = 2x - 2So, when you add |y| to this, the inequality |x| + |x - 2| + |y| ≤4 becomes:For x < 0: -2x + 2 + |y| ≤4 => |y| ≤ 2 + 2x (but only valid when x ∈ [-1,0))For 0 ≤x ≤2: 2 + |y| ≤4 => |y| ≤2For x >2: 2x -2 + |y| ≤4 => |y| ≤6 -2x (valid when x ∈ (2,3])So that's consistent with what I found earlier. Therefore, the shape is a polygon formed by connecting these regions. Alternatively, maybe I can sketch this in my mind. When x is between 0 and 2, it's a rectangle from x=0 to x=2, y=-2 to y=2. For x from -1 to 0, the top and bottom boundaries are y = ±(2 + 2x). Similarly, for x from 2 to 3, the top and bottom boundaries are y = ±(6 - 2x). So plotting these:For x from -1 to 0:At x = -1, y = ±(2 + 2*(-1)) = ±0. So the point (-1, 0)At x = 0, y = ±(2 + 0) = ±2. So connecting from (-1,0) to (0, 2) and (0, -2)For x from 2 to 3:At x = 2, y = ±(6 - 4) = ±2. So connecting from (2, 2) and (2, -2) to (3, 0)So the entire shape is a hexagon with vertices at (-1, 0), (0, 2), (2, 2), (3, 0), (2, -2), (0, -2), and back to (-1, 0). Wait, but connecting these points:From (-1,0) to (0,2): this is a line.From (0,2) to (2,2): horizontal line.From (2,2) to (3,0): line sloping down.From (3,0) to (2,-2): line sloping down.From (2,-2) to (0,-2): horizontal line.From (0,-2) to (-1,0): line sloping up.Yes, that's a hexagon. Let me verify the vertices:At x=-1, the upper and lower y are both 0, so vertex (-1,0).At x=0, upper y=2 and lower y=-2, but connected from x=-1 to x=0. So the sides here are the lines y = 2 + 2x and y = - (2 + 2x) between x=-1 and x=0. So when x=0, those lines meet (0,2) and (0,-2). Then between x=0 and x=2, it's the horizontal lines y=2 and y=-2. Then from x=2 to x=3, the upper and lower boundaries are y=6-2x and y=-(6-2x). At x=2, y=6-4=2 and y=-2. At x=3, y=0. So those lines slope down to (3,0) and up to (3,0). Therefore, the vertices are indeed (-1,0), (0,2), (2,2), (3,0), (2,-2), (0,-2), (-1,0). So that's a hexagon. To compute the area of a hexagon, we can divide it into simpler shapes. Alternatively, since we have the coordinates of all the vertices, we can use the shoelace formula. Let me try that.Shoelace formula requires listing the coordinates in order, either clockwise or counterclockwise. Let's list them clockwise starting from (-1,0):1. (-1, 0)2. (0, 2)3. (2, 2)4. (3, 0)5. (2, -2)6. (0, -2)7. (-1, 0) (to close the polygon)Now, applying the shoelace formula:Sum over i=1 to n of (x_i * y_{i+1} - x_{i+1} * y_i)Let's compute each term:Between (-1,0) and (0,2):x_i = -1, y_{i+1} = 2x_{i+1} = 0, y_i = 0Term: (-1)*2 - 0*0 = -2 - 0 = -2Between (0,2) and (2,2):x_i = 0, y_{i+1} = 2x_{i+1} = 2, y_i = 2Term: 0*2 - 2*2 = 0 - 4 = -4Between (2,2) and (3,0):x_i = 2, y_{i+1} = 0x_{i+1} = 3, y_i = 2Term: 2*0 - 3*2 = 0 - 6 = -6Between (3,0) and (2,-2):x_i = 3, y_{i+1} = -2x_{i+1} = 2, y_i = 0Term: 3*(-2) - 2*0 = -6 - 0 = -6Between (2,-2) and (0,-2):x_i = 2, y_{i+1} = -2x_{i+1} = 0, y_i = -2Term: 2*(-2) - 0*(-2) = -4 - 0 = -4Between (0,-2) and (-1,0):x_i = 0, y_{i+1} = 0x_{i+1} = -1, y_i = -2Term: 0*0 - (-1)*(-2) = 0 - 2 = -2Between (-1,0) and (-1,0) [closing]:x_i = -1, y_{i+1} = 0x_{i+1} = -1, y_i = 0Term: (-1)*0 - (-1)*0 = 0 - 0 = 0Adding all these terms: -2 + (-4) + (-6) + (-6) + (-4) + (-2) + 0 = (-2 -4) = -6; (-6 -6) = -12; (-4 -2) = -6; total is -6 -12 -6 = -24Take the absolute value and divide by 2: | -24 | / 2 = 12So the area is 12, which matches what I calculated earlier by integrating each section. So that seems correct. Alternatively, if I didn't use the shoelace formula, since the shape is symmetric with respect to both the x-axis and y-axis? Wait, actually, looking at the coordinates, it's symmetric about the x-axis because for every point (a, b), there's a point (a, -b). But is it symmetric about the y-axis? Let's see: (-1,0) vs. (3,0). (0,2) vs. (0,-2). Hmm, not symmetric about the y-axis. Wait, but maybe symmetric about the line x=1? Because the hexagon goes from x=-1 to x=3, so the midpoint is x=1. Let's check:If we reflect over x=1:For example, the point (-1,0) would map to (3,0). The point (0,2) maps to (2,2). The point (2,2) maps to (0,2). The point (3,0) maps to (-1,0). Similarly for the lower half. So yes, it's symmetric about the vertical line x=1. So the area can be computed by calculating the area of one half and doubling it. But maybe that complicates things. Anyway, since both methods give 12, which is the answer I initially got through integration, I can be more confident that 12 is correct.But let me think again: when I split the area into three parts and integrated, each part gave me 2, 8, and 2, summing to 12. The shoelace formula on the polygon vertices also gives 12. So seems consistent. Therefore, the answer should be 12. Alternatively, let's check with another method. Suppose we consider that the region is a hexagon with vertices at (-1,0), (0,2), (2,2), (3,0), (2,-2), (0,-2). So this is a convex hexagon. Let's divide it into parts. For example, we can split it into a rectangle in the middle and two trapezoids on the sides. The middle part is from x=0 to x=2, which is a rectangle with height 4 (from y=-2 to y=2) and width 2, area 8. Then on the left side, from x=-1 to x=0, we have a trapezoid. The bases of the trapezoid are at x=-1: length 0 (since y=0) and at x=0: length 4 (from y=-2 to y=2). The height of the trapezoid (along the x-axis) is 1 (from x=-1 to x=0). The area of a trapezoid is (base1 + base2)/2 * height. So (0 + 4)/2 *1 = 2. Similarly, on the right side, from x=2 to x=3, another trapezoid with bases at x=2: length 4 and at x=3: length 0. The height here is 1 (from x=2 to x=3). So area is (4 + 0)/2 *1= 2. Then total area is 8 + 2 + 2= 12. Same result. Okay, so all methods lead to 12. Therefore, the area is 12. Final Answerboxed{12}</think>To find the area of the region formed by points ((x, y)) satisfying (|x| + |y| + |x - 2| leq 4), we consider different cases for (x) based on the critical points (x = 0) and (x = 2).Case 1: (x < 0)For (x in [-1, 0)), the inequality simplifies to (|y| leq 2 + 2x). The vertical length is (2(2 + 2x) = 4 + 4x). Integrating this from (-1) to (0):[int_{-1}^{0} (4 + 4x) , dx = left[ 4x + 2x^2 right]_{-1}^{0} = 0 - (-4 + 2) = 2]Case 2: (0 leq x leq 2)For (x in [0, 2]), the inequality simplifies to (|y| leq 2). The vertical length is 4. Integrating this from (0) to (2):[int_{0}^{2} 4 , dx = left[ 4x right]_{0}^{2} = 8]Case 3: (x > 2)For (x in (2, 3]), the inequality simplifies to (|y| leq 6 - 2x). The vertical length is (2(6 - 2x) = 12 - 4x). Integrating this from (2) to (3):[int_{2}^{3} (12 - 4x) , dx = left[ 12x - 2x^2 right]_{2}^{3} = (36 - 18) - (24 - 8) = 2]Summing the areas from all cases:[2 + 8 + 2 = 12]The region is a hexagon with vertices ((-1, 0)), ((0, 2)), ((2, 2)), ((3, 0)), ((2, -2)), ((0, -2)). Using the shoelace formula confirms the area is 12.[boxed{12}],<think>Okay, so I have this problem here: Find the area of the region formed by points (x, y) satisfying |x| + |y| + |x - 2| ≤ 4 in the Cartesian coordinate system. Hmm, let me try to figure this out step by step.First, I need to understand the inequality given. It's |x| + |y| + |x - 2| ≤ 4. Since there are absolute value signs, this is going to involve some kind of geometric shape, probably a polygon, because absolute values often create linear boundaries. But exactly what shape? Let me break it down.Let me start by considering the terms with x. There's |x| and |x - 2|. Maybe I can simplify this expression first. Let's see, |x| + |x - 2|. This looks familiar; it's the sum of two absolute values. I remember that expressions like |x - a| + |x - b| have different forms depending on the value of x relative to a and b. Let's see, in this case, a is 0 and b is 2. So, the expression |x| + |x - 2| can be simplified based on the value of x.Let me split it into different cases based on the value of x.Case 1: x < 0. In this case, |x| = -x and |x - 2| = -(x - 2) = -x + 2. So, adding them together: (-x) + (-x + 2) = -2x + 2.Case 2: 0 ≤ x < 2. Here, |x| = x and |x - 2| = -(x - 2) = -x + 2. So, adding them: x + (-x + 2) = 2.Case 3: x ≥ 2. Then, |x| = x and |x - 2| = x - 2. So, adding them: x + (x - 2) = 2x - 2.So, combining these cases, |x| + |x - 2| is equal to:- -2x + 2 when x < 0,- 2 when 0 ≤ x < 2,- 2x - 2 when x ≥ 2.Therefore, the original inequality |x| + |y| + |x - 2| ≤ 4 can be rewritten based on these cases. Since |y| is always non-negative, the total sum is going to be the sum from above plus |y|. Let's handle each case separately.Starting with Case 1: x < 0.In this case, the inequality becomes (-2x + 2) + |y| ≤ 4. Let's rearrange this:-2x + 2 + |y| ≤ 4Subtract 2 from both sides:-2x + |y| ≤ 2Then, rearrange terms:|y| ≤ 2 + 2xBut wait, x is negative here (since we're in Case 1: x < 0). So 2 + 2x would be 2(1 + x). Since x is negative, 1 + x could be positive or negative depending on how negative x is. For example, if x = -1, then 2 + 2x = 0; if x is less than -1, then 2 + 2x becomes negative, but the left side |y| is always non-negative. So, |y| ≤ something that can be negative. But since |y| is non-negative, the inequality |y| ≤ (negative number) would have no solution. Therefore, in the region x < -1, the inequality becomes |y| ≤ a negative number, which is impossible. So, only when 2 + 2x ≥ 0, i.e., when x ≥ -1. But in Case 1, x < 0. So, combining these, x must be between -1 and 0 for this inequality to have solutions in Case 1.Therefore, in Case 1 (x < 0), the inequality |y| ≤ 2 + 2x is only valid when -1 ≤ x < 0, and for these x's, since 2 + 2x is non-negative, |y| ≤ 2 + 2x implies y is between -(2 + 2x) and 2 + 2x. So, in this region, the graph would be between two horizontal lines, but the width depends on x.Moving to Case 2: 0 ≤ x < 2.Here, |x| + |x - 2| = 2, so the inequality becomes 2 + |y| ≤ 4. Subtracting 2 gives |y| ≤ 2. So in this vertical strip from x = 0 to x = 2, y is between -2 and 2. So this region is a rectangle from x=0 to x=2, y=-2 to y=2.Case 3: x ≥ 2.Here, |x| + |x - 2| = 2x - 2. So the inequality becomes (2x - 2) + |y| ≤ 4. Let's rearrange:2x - 2 + |y| ≤ 4Add 2 to both sides:2x + |y| ≤ 6Then, |y| ≤ 6 - 2xBut since x ≥ 2 here, 6 - 2x will decrease as x increases. When x = 2, 6 - 2x = 6 - 4 = 2. When x = 3, 6 - 2x = 0. For x > 3, 6 - 2x becomes negative, which again would mean |y| ≤ a negative number, which is impossible. Therefore, in Case 3 (x ≥ 2), the inequality is valid only when x ≤ 3. So x ranges from 2 to 3. Then, |y| ≤ 6 - 2x, so y ranges between -(6 - 2x) and 6 - 2x.So now, compiling all the cases:1. For -1 ≤ x < 0: |y| ≤ 2 + 2x, which is equivalent to - (2 + 2x) ≤ y ≤ 2 + 2x.2. For 0 ≤ x < 2: |y| ≤ 2, so y between -2 and 2.3. For 2 ≤ x ≤ 3: |y| ≤ 6 - 2x, so y between -(6 - 2x) and 6 - 2x.So the overall region is a combination of these three parts. To find the area, I need to compute the area of each part and sum them up.Let's handle each part one by one.First part: -1 ≤ x < 0.For each x in [-1, 0), the range of y is from -(2 + 2x) to 2 + 2x. So the vertical extent at each x is 2*(2 + 2x) = 4 + 4x. So the area is the integral from x = -1 to x = 0 of (4 + 4x) dx.Calculating this integral:∫ from -1 to 0 of (4 + 4x) dx = [4x + 2x²] from -1 to 0At x = 0: 4*0 + 2*0² = 0At x = -1: 4*(-1) + 2*(-1)² = -4 + 2 = -2So subtracting: 0 - (-2) = 2So the area of the first part is 2.Second part: 0 ≤ x < 2.Here, y ranges from -2 to 2, so the vertical extent is 4. The horizontal extent is from x=0 to x=2, so length 2. So the area is 4*2 = 8.Third part: 2 ≤ x ≤ 3.For each x in [2, 3], the vertical range is 2*(6 - 2x). So the vertical extent is 12 - 4x. So the area is the integral from 2 to 3 of (12 - 4x) dx.Calculating this integral:∫ from 2 to 3 of (12 - 4x) dx = [12x - 2x²] from 2 to 3At x = 3: 12*3 - 2*9 = 36 - 18 = 18At x = 2: 12*2 - 2*4 = 24 - 8 = 16Subtracting: 18 - 16 = 2So the area of the third part is 2.Adding all three parts together: 2 + 8 + 2 = 12.But wait, let me confirm this. The first part is 2, the second part 8, the third part 2. So total area 12. Hmm, that seems straightforward, but let me visualize the shape to make sure.When x is between -1 and 0, the top and bottom boundaries are lines. Let's see:For the first part (-1 ≤ x < 0):At x = -1: |y| ≤ 2 + 2*(-1) = 0. So y = 0. So the point is (-1, 0).At x = 0: |y| ≤ 2 + 2*0 = 2. So y ranges from -2 to 2.So the first part is a region that starts at (-1, 0) and widens to a rectangle from x=0 to x=2. So it's a trapezoid? Wait, when x increases from -1 to 0, the upper and lower bounds of y go from 0 to 2 (and -0 to -2). So the shape is a quadrilateral with vertices at (-1, 0), (0, 2), (0, -2), and back to (-1, 0). Wait, actually, no. Because between x=-1 and x=0, for each x, the y goes from -(2+2x) to 2+2x. So plotting this, when x=-1, y=0; when x=0, y=±2. So the upper boundary is a line from (-1,0) to (0,2), and the lower boundary is a line from (-1,0) to (0,-2). So the shape is a triangle? Wait, but over the interval from x=-1 to x=0. Hmm. Wait, the vertical width at each x is 4 + 4x. So when x=-1, the width is 4 + 4*(-1) = 0, which is a point. At x=0, the width is 4 + 0 = 4. So this is indeed a triangle with base at x=0, length 4 (from y=-2 to y=2) and height of 1 (from x=-1 to x=0). So the area would be (base * height)/2 = (4 * 1)/2 = 2, which matches the integral result. So that's correct.Similarly, the third part (x from 2 to 3):At x=2, the vertical range is |y| ≤ 6 - 2*2 = 2, so y from -2 to 2. At x=3, |y| ≤ 6 - 2*3 = 0, so y=0. So similar to the first part, this is a triangle with base at x=2 (from y=-2 to y=2, length 4) and height 1 (from x=2 to x=3). So area is (4 * 1)/2 = 2, which matches the integral result.The middle part is a rectangle from x=0 to x=2, y=-2 to y=2. So that's width 2 (in x-direction) and height 4 (in y-direction), area 8. So total area 2 + 8 + 2 = 12.But wait, the problem says "the area of the region formed by points (x, y) satisfying |x| + |y| + |x - 2| ≤ 4". So is the answer 12? Hmm.But let me check if I considered all cases correctly. Let me think of another approach. Maybe graphing the inequality.Alternatively, sometimes, when dealing with multiple absolute values, it's helpful to consider different regions based on the critical points where the expressions inside the absolute values change sign. For |x|, the critical point is at x=0; for |x - 2|, it's at x=2. So we can divide the x-axis into intervals separated by x=0 and x=2. That's exactly what I did before: x < 0, 0 ≤ x < 2, x ≥ 2. Then, in each interval, handle the absolute values and solve for y.Alternatively, maybe I can consider the equation |x| + |y| + |x - 2| = 4 and graph that, then the inequality would be the region inside.Let me try to sketch this graph.First, as before, the expression |x| + |x - 2| simplifies depending on x.In the middle region (0 ≤ x < 2), it's 2, so the equation becomes |y| = 2, which is two horizontal lines y=2 and y=-2, from x=0 to x=2.In the left region (x < 0), the equation becomes -2x + 2 + |y| = 4, so |y| = 2x + 2. Wait, no: when x < 0, |x| + |x - 2| = -2x + 2, so equation is -2x + 2 + |y| = 4, so |y| = 2 + 2x. But since x < 0, 2 + 2x is less than 2. However, when x < -1, 2 + 2x becomes negative, so no solution. So the left part is from x=-1 to x=0, |y| = 2 + 2x, which is two lines: y = 2 + 2x and y = - (2 + 2x). These lines start at x=-1, y=0 (since when x=-1, 2 + 2*(-1)=0) and go up to x=0, y=2 and y=-2.Similarly, on the right side (x ≥ 2), |x| + |x - 2| = 2x - 2, so equation becomes 2x - 2 + |y| = 4, so |y| = 6 - 2x. Since x ≥ 2, 6 - 2x decreases. When x=2, |y|=2, so y=2 and y=-2. When x=3, |y|=0, so y=0. So the right part is from x=2 to x=3, with |y|=6 - 2x, forming two lines from (2, 2) to (3, 0) and (2, -2) to (3, 0).Therefore, the entire figure is a polygon with vertices at (-1, 0), (0, 2), (2, 2), (3, 0), (2, -2), (0, -2), and back to (-1, 0). Wait, connecting these points:From (-1,0) up to (0,2), then right to (2,2), then up to (3,0)? Wait, no. Wait, at x=2, the upper boundary is still y=2, and then from x=2 to x=3, the upper boundary is y=6 - 2x. So at x=2, that's y=6 - 4=2, and at x=3, y=0. Similarly, the lower boundary is y= - (6 - 2x). So the upper and lower boundaries slope down from (2,2) to (3,0) and up from (2,-2) to (3,0). Wait, no, lower boundary is y= - (6 - 2x) = -6 + 2x. So at x=2, y= -6 + 4= -2, and at x=3, y= -6 + 6=0. So the lower boundary goes from (2, -2) to (3, 0).So the vertices of the polygon are:Leftmost point: (-1, 0)From there, up to (0, 2)Then right to (2, 2)Then down to (3, 0)Then up to (2, -2)? Wait, no. Wait, after (3,0), going down? Wait, maybe I need to list all the vertices in order.Let me walk around the polygon.Starting from the left at (-1,0). Moving up along the line y=2 + 2x (since for x between -1 and 0, upper boundary is y=2 + 2x). So when x increases from -1 to 0, y goes from 0 to 2. So that's a line to (0,2). Then from (0,2), moving right along y=2 until x=2. So that's the point (2,2). Then from (2,2), moving along y=6 - 2x (since for x ≥2, upper boundary is y=6 - 2x) to (3,0). Then from (3,0), moving down along y= - (6 - 2x) = -6 + 2x (since lower boundary for x ≥2 is y= - (6 - 2x)), so from (3,0) to (2, -2). Then from (2, -2), moving left along y=-2 back to (0, -2). Then from (0, -2), moving up along y= - (2 + 2x) (for x between -1 and 0, lower boundary is y= -2 - 2x). Wait, but when x goes from 0 to -1, y= - (2 + 2x) = -2 - 2x. At x=0, that's -2, and at x=-1, that's -2 -2*(-1) = -2 +2 = 0. So from (0, -2) moving left along y= -2 -2x to (-1,0). Hence, completing the polygon.So the vertices are:(-1,0), (0,2), (2,2), (3,0), (2,-2), (0,-2), (-1,0).So connecting these points, the shape is a hexagon. Wait, but with six vertices? Let me count:1. (-1,0)2. (0,2)3. (2,2)4. (3,0)5. (2,-2)6. (0,-2)7. Back to (-1,0). So actually six sides, but it's a hexagon.But wait, when I plot these points, (-1,0) connected to (0,2), then to (2,2), then to (3,0), then to (2,-2), then to (0,-2), then to (-1,0). So yes, a hexagon with six vertices.But perhaps there's a symmetry here. The figure is symmetric with respect to both the x-axis and y-axis? Wait, let's check.If we reflect over the x-axis, the points (0,2) and (0,-2) swap, (2,2) and (2,-2) swap, (3,0) remains, (-1,0) remains. Similarly, reflection over y-axis would affect x-coordinates. But since the original equation has |x| and |x - 2|, which are not symmetric about y-axis. However, |y| makes it symmetric about the x-axis. So the figure is symmetric about the x-axis, meaning the area above the x-axis is equal to the area below. So perhaps we can compute the area above the x-axis and double it.Looking at the upper half, the vertices would be (-1,0), (0,2), (2,2), (3,0). So that's a quadrilateral. Let me compute its area and double it.But maybe it's easier to compute the area as the sum of simpler shapes. The figure can be divided into three parts as I did before: two triangles on the sides and a rectangle in the middle. Since the total area was 12, according to my previous calculation.Alternatively, since the polygon is a hexagon, perhaps we can use the shoelace formula to compute the area.Let me list the coordinates of the vertices in order:1. (-1, 0)2. (0, 2)3. (2, 2)4. (3, 0)5. (2, -2)6. (0, -2)7. (-1, 0) [closing the polygon]Using the shoelace formula:Area = 1/2 |sum_{i=1 to n} (x_i y_{i+1} - x_{i+1} y_i)|Let's compute each term:From (-1,0) to (0,2):Term1: (-1)*2 - 0*0 = -2 - 0 = -2From (0,2) to (2,2):Term2: 0*2 - 2*2 = 0 - 4 = -4From (2,2) to (3,0):Term3: 2*0 - 3*2 = 0 - 6 = -6From (3,0) to (2,-2):Term4: 3*(-2) - 2*0 = -6 - 0 = -6From (2,-2) to (0,-2):Term5: 2*(-2) - 0*(-2) = -4 - 0 = -4From (0,-2) to (-1,0):Term6: 0*0 - (-1)*(-2) = 0 - 2 = -2Sum of all terms: (-2) + (-4) + (-6) + (-6) + (-4) + (-2) = (-2 -4) + (-6 -6) + (-4 -2) = (-6) + (-12) + (-6) = -24Take absolute value and multiply by 1/2: (1/2)*| -24 | = (1/2)*24 = 12So shoelace formula confirms the area is 12. So that's reassuring.Therefore, the area of the region is 12.But just to make sure I didn't make any miscalculations in the shoelace formula, let's recheck:List of vertices:1. (-1, 0)2. (0, 2)3. (2, 2)4. (3, 0)5. (2, -2)6. (0, -2)7. (-1, 0)Compute each (x_i * y_{i+1} - x_{i+1} * y_i):1 to 2: (-1)*2 - 0*0 = -2 - 0 = -22 to 3: 0*2 - 2*2 = 0 -4 = -43 to 4: 2*0 - 3*2 = 0 -6 = -64 to 5: 3*(-2) -2*0 = -6 -0 = -65 to 6: 2*(-2) -0*(-2) = -4 -0 = -46 to 7: 0*0 - (-1)*(-2) = 0 -2 = -2Sum: -2 -4 -6 -6 -4 -2 = (-2 -4) + (-6 -6) + (-4 -2) = (-6) + (-12) + (-6) = -24Absolute value and half: 24/2=12. Yep, same result.Alternatively, thinking of the figure as a hexagon, which is symmetric over the x-axis. The upper half is a quadrilateral with vertices (-1,0), (0,2), (2,2), (3,0). Let me compute its area and then double it.Compute area of upper half:Using shoelace formula for upper half vertices:(-1,0), (0,2), (2,2), (3,0), (-1,0)Wait, actually, the upper half is a quadrilateral with vertices (-1,0), (0,2), (2,2), (3,0).Calculating its area:Shoelace formula:Coordinates:1. (-1, 0)2. (0, 2)3. (2, 2)4. (3, 0)5. (-1, 0)Compute terms:1 to 2: (-1)*2 -0*0 = -22 to 3: 0*2 -2*2 = 0 -4 = -43 to 4: 2*0 -3*2 = 0 -6 = -64 to 5: 3*0 - (-1)*0 = 0 -0 =0Sum: -2 -4 -6 +0 = -12Area: 1/2 * | -12 | =6Then double it for the lower half: 6*2=12. Same result.Alternatively, compute the area of the upper half as a combination of shapes.From (-1,0) to (0,2) to (2,2) to (3,0). This is a trapezoid with bases at y=0 (from x=-1 to x=3) and y=2 (from x=0 to x=2). Wait, not exactly. Let me see.Alternatively, split the upper half into two triangles and a rectangle.From (-1,0) to (0,2) to (2,2) to (3,0).First triangle: (-1,0), (0,2), (0,0). Wait, but (0,0) isn't a vertex. Maybe not. Alternatively, the area from x=-1 to x=0 is a triangle with vertices (-1,0), (0,2), (0,0). But (0,0) isn't part of the upper half. Wait, maybe not. Alternatively, from x=-1 to x=0, the upper boundary is y=2 + 2x, which is a line from (-1,0) to (0,2). The area under this line (from x=-1 to x=0) is a triangle with base 1 (along x-axis from -1 to 0) and height 2 (from y=0 to y=2). Area = (1 * 2)/2 =1.Then from x=0 to x=2, the upper boundary is y=2, which is a rectangle with width 2 and height 2. Area=4.Then from x=2 to x=3, the upper boundary is y=6 - 2x, a line from (2,2) to (3,0). The area under this line is a triangle with base 1 (from x=2 to x=3) and height 2. Area=(1*2)/2=1.Total upper area:1 +4 +1=6. Double it for lower half:12. Same answer.Therefore, regardless of the method, the area is 12. So I think the answer is 12. But just to make sure I didn't miss any other regions or make a mistake in the case analysis.Wait, the original inequality is |x| + |y| + |x - 2| ≤4. Since |y| is non-negative, the sum |x| + |x - 2| must be ≤4. Wait, but |x| + |x - 2|, as we analyzed earlier, equals 2 for 0 ≤x ≤2, and 2|x -1| + 0. Wait, no. Wait, no, earlier we saw:For x <0: |x| + |x - 2| = -2x + 2For 0 ≤x <2: 2For x ≥2: 2x -2So the minimum value of |x| + |x - 2| is 2, occurring when 0 ≤x ≤2. So the total expression |x| + |y| + |x - 2| is at least 2 + 0 =2, and up to 4.Wait, so when 0 ≤x ≤2, |y| ≤2. So the maximum |y| is 2. When x is outside [0,2], the allowed |y| decreases as |x -1| increases, since |x| + |x -2| increases beyond 2.Therefore, the figure is bounded by these lines, forming a hexagon as we found. So all calculations check out. The area is indeed 12.Final AnswerThe area of the region is boxed{12}.</think>To find the area of the region formed by points ((x, y)) satisfying (|x| + |y| + |x - 2| leq 4), we analyze the inequality by breaking it into different cases based on the value of (x).1. Case 1: (x < 0) - The expression simplifies to (-2x + 2 + |y| leq 4), leading to (|y| leq 2 + 2x). - This is valid for (-1 leq x < 0). The vertical extent at each (x) is (4 + 4x). - The area is calculated by integrating (4 + 4x) from (-1) to (0): [ int_{-1}^{0} (4 + 4x) , dx = left[4x + 2x^2right]_{-1}^{0} = 0 - (-4 + 2) = 2 ]2. Case 2: (0 leq x < 2) - The expression simplifies to (2 + |y| leq 4), leading to (|y| leq 2). - This forms a rectangle from (x = 0) to (x = 2) with (y) ranging from (-2) to (2). - The area is: [ 2 times 4 = 8 ]3. Case 3: (x geq 2) - The expression simplifies to (2x - 2 + |y| leq 4), leading to (|y| leq 6 - 2x). - This is valid for (2 leq x leq 3). The vertical extent at each (x) is (12 - 4x). - The area is calculated by integrating (12 - 4x) from (2) to (3): [ int_{2}^{3} (12 - 4x) , dx = left[12x - 2x^2right]_{2}^{3} = (36 - 18) - (24 - 8) = 2 ]Summing the areas of all three cases:[2 + 8 + 2 = 12]The vertices of the region form a hexagon, and using the shoelace formula confirms the area is (12).Thus, the area of the region is (boxed{12}).]

🔑:2. 12 .Obviously, the region is symmetric about the x-axis. Therefore, we only need to consider the graph represented by the equation |x|+|y|+|x-2|=4 when y geqslant 0.Draw the graph of the function y=4-|x|-|x-2|(-1 leqslant x leqslant 3), and then reflect it about the x-axis, to obtain the graph represented by the equation |x|+|y|+|x-2|=4. As shown in Figure 5, the boundary and interior of this graph represent the region indicated by |x|+|y|+|x-2| leqslant 4, and it is easy to calculate that its area is 12.

🧠:[<think>Okay, let's see here. I need to check which of these factorizations is correct. There are four options: A, B, C, D. Let me go through each one step by step.Starting with option A: a² + 4a = a²(a + 4). Hmm, so they factored out an a² from the left side. Let me try to factor the left-hand side myself. The expression is a² + 4a. Both terms have an 'a' in them, right? So I can factor out an 'a' from both terms. That would give me a(a + 4). But the answer here says they factored out a², which would leave a² times (a + 4). Wait, if I factor out a² from a², that's fine, but then from 4a, factoring out a² would leave 4/a. But that's not possible because we can't have variables in the denominator when factoring. So this must be incorrect. So option A is wrong because instead of factoring out an 'a', they tried to factor out a², which isn't valid here. So A is incorrect.Moving on to option B: a² - 9 = (a + 9)(a - 9). Hmm, okay. The left side is a difference of squares, right? Because a² is a square and 9 is 3 squared. The formula for difference of squares is a² - b² = (a + b)(a - b). So in this case, a² - 9 should be (a + 3)(a - 3), because 9 is 3 squared. But here, they have 9 instead of 3. So they've written (a + 9)(a - 9), which would actually be a² - 81, not a² - 9. So that's incorrect. Therefore, option B is also wrong.Next is option C: a² - 2a + 1 = (a - 1)². Let me check this. Let's expand the right-hand side. (a - 1)² is a² - 2a + 1. Yes, that's exactly the left-hand side. So this seems correct. Let me verify again. The first term squared is a², the middle term is -2a, and the last term is 1. So yes, (a - 1)² does equal a² - 2a + 1. Therefore, option C is correct.But just to be thorough, let me check option D as well: a² + 4a + 2 = (a + 2)². Let's expand the right-hand side. (a + 2)² is a² + 4a + 4. But the left side is a² + 4a + 2. The constant term here is 2 instead of 4. So that's not the same. Therefore, option D is incorrect because expanding (a + 2)² gives a different constant term.So, after checking all four options, the only correct factorization is option C. Let me recap just to make sure I didn't make any mistakes. A: Factoring out an a instead of a² gives a(a + 4), which is correct, but they said a²(a + 4), which is wrong. B: Should be (a + 3)(a - 3) instead of involving 9. C: Correct expansion matches. D: Constant term is wrong. Yep, so definitely C is the correct one.Final Answerboxed{C}</think>Let's check each factorization step by step:- Option A: (a^{2} + 4a = a^{2}(a + 4)) Factoring out the common term (a) from (a^{2} + 4a) gives (a(a + 4)). Factoring out (a^{2}) is incorrect because it would leave a term with a variable in the denominator. Therefore, option A is incorrect.- Option B: (a^{2} - 9 = (a + 9)(a - 9)) The correct factorization of (a^{2} - 9) is ((a + 3)(a - 3)) since it is a difference of squares. Using 9 instead of 3 results in ((a + 9)(a - 9)), which is incorrect. Therefore, option B is incorrect.- Option C: (a^{2} - 2a + 1 = (a - 1)^{2}) Expanding ((a - 1)^{2}) gives (a^{2} - 2a + 1), which matches the left-hand side. Therefore, option C is correct.- Option D: (a^{2} + 4a + 2 = (a + 2)^{2}) Expanding ((a + 2)^{2}) gives (a^{2} + 4a + 4), which does not match the left-hand side (the constant term is 2 instead of 4). Therefore, option D is incorrect.Thus, the correct factorization is (boxed{C}).,<think>Alright, let's see. I have to figure out which of these factorizations is correct. The options are A, B, C, D. Hmm. Let's go through each one step by step. I remember that factoring involves breaking down an expression into simpler parts that multiply together to give the original expression. Let me start with option A.Option A says: a² + 4a = a²(a + 4). Wait, let me check that. If I factor out something from a² + 4a, what's the common term? Both terms have an 'a' in them, right? So the greatest common factor is 'a', not a². If I factor out an 'a', then it would be a(a + 4). But the answer given here is a²(a + 4), which would mean factoring out a². Let's test it: a²*(a + 4) = a³ + 4a². But the original expression is a² + 4a, which is different. So that's not correct. So A is wrong.Moving on to option B: a² - 9 = (a + 9)(a - 9). Hmm, this looks like a difference of squares. Wait, the difference of squares formula is a² - b² = (a + b)(a - b). So in this case, if we have a² - 9, then 9 is 3², so it should be (a + 3)(a - 3). But here they have 9 instead of 3. So they wrote (a + 9)(a - 9), which would actually be a² - 81, not a² - 9. So B is also incorrect.Now option C: a² - 2a + 1 = (a - 1)². Let's check this. Let's expand (a - 1)². That's (a - 1)(a - 1) = a² - a - a + 1 = a² - 2a + 1. Yes, that matches the left side. So C seems to be correct. Let me just confirm once more. The left-hand side is a quadratic trinomial. To factor it, I can check if it's a perfect square. The first term is a², the last term is 1, which is 1². The middle term is -2a. Since -2a is equal to 2*a*1 (but with a negative sign), so yes, it's (a - 1)². So C is correct.But let's check D just to be thorough. Option D: a² + 4a + 2 = (a + 2)². Let's expand (a + 2)². That's a² + 4a + 4. But the original expression is a² + 4a + 2. So the constant term is 2 instead of 4. Therefore, this is not correct. So D is wrong.So summarizing, A is wrong because factoring out a instead of a², B is wrong because they used 9 instead of 3 for the difference of squares, C is correct as a perfect square trinomial, D is wrong because the constant term doesn't match. Therefore, the correct answer is C.Final Answerboxed{C}</think>To determine which factorization is correct, we analyze each option step by step:- Option A: (a^{2} + 4a = a^{2}(a + 4)) The common factor is (a), not (a^2). Factoring out (a) gives (a(a + 4)). The given factorization is incorrect because (a^{2}(a + 4) = a^{3} + 4a^{2}), which does not match (a^{2} + 4a).- Option B: (a^{2} - 9 = (a + 9)(a - 9)) This should be a difference of squares. The correct factorization is ((a + 3)(a - 3)) since (9 = 3^2). The given factorization ((a + 9)(a - 9)) results in (a^{2} - 81), which is incorrect.- Option C: (a^{2} - 2a + 1 = (a - 1)^{2}) Expanding ((a - 1)^2) gives (a^{2} - 2a + 1), which matches the left-hand side. This factorization is correct.- Option D: (a^{2} + 4a + 2 = (a + 2)^{2}) Expanding ((a + 2)^2) gives (a^{2} + 4a + 4), which does not match the left-hand side (a^{2} + 4a + 2). This factorization is incorrect.Thus, the correct factorization is given in option C.(boxed{C})]

🔑:To determine the correct factorization, let's examine each option step by step:Option A: a^{2}+4a=a^{2}(a+4)- Factoring out a from a^{2}+4a gives us a(a+4).- Therefore, the correct factorization is a(a+4), not a^{2}(a+4).- So, option A is incorrect.Option B: a^{2}-9=left(a+9right)left(a-9right)- The correct factorization of a^{2}-9 is based on the difference of squares formula, which is (a+3)(a-3).- Therefore, the correct factorization is left(a+3right)left(a-3right), not left(a+9right)left(a-9right).- So, option B is incorrect.Option C: a^{2}-2a+1=left(a-1right)^{2}- Applying the perfect square formula, we see that a^{2}-2a+1 is indeed (a-1)^{2}.- Therefore, option C is the correct factorization.Option D: a^{2}+4a+2=left(a+2right)^{2}- Expanding (a+2)^{2} gives a^{2}+4a+4, not a^{2}+4a+2.- Therefore, a^{2}+4a+2 cannot be factorized as (a+2)^{2}.- So, option D is incorrect.Given the analysis above, the correct answer is:boxed{C}

🧠:[<think>Okay, so I need to solve the equation √(x² - p) + 2√(x² - 1) = x, where p is a real parameter. Hmm, let's start by understanding the domain of this equation. Since we have square roots, the expressions inside must be non-negative. That means:1. x² - p ≥ 0 ⇒ x² ≥ p2. x² - 1 ≥ 0 ⇒ x² ≥ 1 ⇒ x ≥ 1 or x ≤ -1Also, the right side is just x. But the left side is a sum of square roots, which are always non-negative. Therefore, the right side x must also be non-negative. Because even if x is negative, the left side would still be non-negative, but the right side would be negative, which can't happen. So x must be ≥ 0. But from the second condition, x² ≥ 1, so x must be ≥ 1 or x ≤ -1. Combining these two, x must be ≥ 1. So the domain is x ≥ 1. Therefore, any solution must satisfy x ≥ 1.Alright, so we can restrict x to [1, ∞). Now, let's work on the equation:√(x² - p) + 2√(x² - 1) = x.I need to solve for x in terms of p. Let's denote y = x² for simplicity. Then, √(y - p) + 2√(y - 1) = x. But x = √y, since x is positive (we've established x ≥ 1). Wait, hold on, x is √y only if y = x², but x is already equal to √y because x is non-negative. So maybe substituting y = x² is not necessary here. Let me think.Alternatively, let's denote each square root term:Let A = √(x² - p), and B = √(x² - 1). Then the equation becomes A + 2B = x. Maybe we can express one variable in terms of the other. Let me try to isolate one of the square roots.Let me subtract 2B from both sides: A = x - 2B. Then, substituting A and B:√(x² - p) = x - 2√(x² - 1)Now, let's square both sides to eliminate the square root. But before that, I should note that since the left side √(x² - p) is non-negative, the right side must also be non-negative. Therefore:x - 2√(x² - 1) ≥ 0 ⇒ x ≥ 2√(x² - 1)This is an important condition. Let's check what this implies.So, x ≥ 2√(x² - 1). Let's square both sides here (keeping in mind x is positive, so squaring is okay):x² ≥ 4(x² - 1) ⇒ x² ≥ 4x² - 4 ⇒ 0 ≥ 3x² - 4 ⇒ 3x² ≤ 4 ⇒ x² ≤ 4/3 ⇒ x ≤ 2/√3 ≈ 1.1547But remember, x must be ≥1. So combining these, x ∈ [1, 2/√3]. Therefore, any solution must lie in [1, 2/√3]. So even though initially x could be ≥1, after considering the non-negativity of the right side, x is restricted to [1, 2/√3].Therefore, our domain is now x ∈ [1, 2/√3].So, moving back to the equation:√(x² - p) = x - 2√(x² - 1)Let's square both sides to eliminate the square root:(√(x² - p))² = (x - 2√(x² - 1))²Left side: x² - pRight side: x² - 4x√(x² - 1) + 4(x² - 1)Simplify right side:x² - 4x√(x² - 1) + 4x² - 4 = 5x² - 4 - 4x√(x² - 1)Set left and right equal:x² - p = 5x² - 4 - 4x√(x² - 1)Bring all terms to one side:0 = 5x² - 4 - 4x√(x² - 1) - x² + p ⇒ 0 = 4x² - 4 - 4x√(x² - 1) + pRearranged:4x² - 4 + p = 4x√(x² - 1)Let's divide both sides by 4 to simplify:x² - 1 + (p/4) = x√(x² - 1)Hmm, this looks a bit complex, but maybe let's set t = √(x² - 1). Then, since x ≥1, t ≥0. Also, note that x² = t² + 1. Let's substitute:(t² + 1) - 1 + (p/4) = x*t ⇒ t² + p/4 = x*tBut x can be expressed in terms of t and x, but maybe we need another substitution. Wait, x = sqrt(t² + 1). Wait, but x is sqrt(t² + 1). So substituting x into the equation:t² + p/4 = sqrt(t² + 1) * tLet me write that as:t² + p/4 = t*sqrt(t² + 1)Hmm, this equation in terms of t. Let's denote s = t. Then:s² + p/4 = s*sqrt(s² + 1)Let me square both sides again to eliminate the square root. But before that, let's check if both sides are non-negative. The left side is s² + p/4. Since s² ≥0, but p can be any real parameter, so we need to ensure that s² + p/4 is non-negative. However, since s = sqrt(x² - 1) and x ∈ [1, 2/√3], s = sqrt(x² -1) ≥0. So x ∈ [1, 2/√3], so x² ∈ [1, 4/3], so x² -1 ∈ [0, 1/3], so s ∈ [0, 1/√3]. Therefore, s is between 0 and approximately 0.577.But since p is a real parameter, s² + p/4 must be non-negative. So depending on p, this could be an issue. However, since s is bounded, we might have constraints on p. Let's proceed with squaring.Left side squared: (s² + p/4)²Right side squared: s²(s² + 1)Therefore:(s² + p/4)² = s²(s² + 1)Let's expand the left side:s⁴ + (p/2)s² + (p²)/16 = s⁴ + s²Subtract right side from left side:s⁴ + (p/2)s² + p²/16 - s⁴ - s² = 0 ⇒ (- (1 - p/2))s² + p²/16 = 0Simplify:[(p/2 - 1)]s² + p²/16 = 0So:(p/2 - 1)s² + p²/16 = 0Let me factor this equation:Multiply both sides by 16 to eliminate denominators:8(p - 2)s² + p² = 0So:8(p - 2)s² + p² = 0But remember that s² = x² -1 - 1. Wait, no. Wait, s = sqrt(x² -1), so s² = x² -1. So substituting back s² = x² -1:8(p - 2)(x² -1) + p² = 0So:8(p - 2)x² - 8(p - 2) + p² = 0Let me write this as:8(p - 2)x² + [ -8(p - 2) + p² ] = 0Simplify the constants:-8(p - 2) + p² = -8p + 16 + p² = p² -8p +16Therefore, the equation becomes:8(p - 2)x² + (p² -8p +16) = 0Let's write this as:8(p - 2)x² = - (p² -8p +16) ⇒ 8(p - 2)x² = - (p -4)^2Because p² -8p +16 factors as (p -4)^2.Therefore:8(p -2)x² = - (p -4)^2 ⇒ x² = - (p -4)^2 / [8(p -2)]So, x² = (4 - p)^2 / [8(2 - p)] since the negative sign in the denominator flips (p -2) to (2 - p). Let's verify:(p -4)^2 is (4 - p)^2, so same thing. Then, 8(p -2) becomes 8(2 - p) when multiplied by -1. So, x² = (4 - p)^2 / [8(2 - p)]But x² must be a real number, so the denominator 8(2 - p) cannot be zero, so p ≠ 2. Also, since x² must be positive (as x ≥1), the right-hand side must be positive. Therefore, (4 - p)^2 / [8(2 - p)] > 0Since (4 - p)^2 is always non-negative, and 8 is positive, the sign depends on (2 - p). So for the entire expression to be positive, (2 - p) > 0 ⇒ p < 2.Therefore, p must be less than 2. Also, note that x² must be in [1, 4/3], as we had earlier x ∈ [1, 2/√3], so x² ∈ [1, 4/3]. Therefore, the expression (4 - p)^2 / [8(2 - p)] must lie in [1, 4/3].So, let's write down x² = (4 - p)^2 / [8(2 - p)] and x² ∈ [1, 4/3]. Let's denote:Let’s set x² = [(4 - p)^2]/[8(2 - p)] = [(4 - p)^2]/[8(2 - p)]Note that (4 - p)^2 = (p -4)^2 = same thing, so we can write:[(p -4)^2]/[8(2 - p)] = [(p -4)^2]/[8(2 - p)] = [(p -4)^2]/[-8(p -2)] = - [(p -4)^2]/[8(p -2)]But since we have already established that p < 2, so (p -2) is negative, so denominator is negative, and the negative sign in front makes the whole expression positive. So it's consistent with x² being positive.Now, let's compute this expression for x²:Let me expand (4 - p)^2:(4 - p)^2 = 16 -8p + p²So x² = (16 -8p + p²)/[8(2 - p)] = [p² -8p +16]/[8(2 - p)]Note that p² -8p +16 = (p -4)^2, so x² = (p -4)^2/[8(2 - p)]Alternatively, we can write:x² = (4 - p)^2/[8(2 - p)] = [ (4 - p)^2 ] / [8(2 - p) ]Let’s see if this can be simplified:Note that (4 - p)^2 = (2 - p + 2)^2. Not sure if that helps.Alternatively, split the fraction:= [ (4 - p)^2 ] / [8(2 - p) ] = [ (4 - p)^2 ] / [8(2 - p) ] = [ (4 - p)/ (2 - p) ) ] * [ (4 - p)/8 ]But not sure. Alternatively, write 4 - p = (2 - p) + 2, but not sure. Alternatively, perhaps perform polynomial division.Wait, let's divide (4 - p)^2 by (2 - p):(4 - p)^2 = (2 - p + 2)^2 = ( (2 - p) + 2 )^2 = (2 - p)^2 + 4(2 - p) +4So, expanding:(4 - p)^2 = (2 - p)^2 + 4(2 - p) +4Therefore:x² = [ (2 - p)^2 + 4(2 - p) +4 ] / [8(2 - p) ]= [ (2 - p)^2 / (8(2 - p)) ) + 4(2 - p)/(8(2 - p)) + 4/(8(2 - p)) ]Simplify each term:First term: (2 - p)^2 / [8(2 - p)] = (2 - p)/8Second term: 4(2 - p)/[8(2 - p)] = 4/8 = 1/2Third term: 4/[8(2 - p)] = 1/[2(2 - p)]Therefore, x² = (2 - p)/8 + 1/2 + 1/[2(2 - p)]Combine the terms:Let me write all terms with denominator 8(2 - p):First term: (2 - p)/8 = (2 - p)^2/[8(2 - p)]Second term: 1/2 = 4(2 - p)/[8(2 - p)]Third term: 1/[2(2 - p)] = 4/[8(2 - p)]Therefore:x² = [ (2 - p)^2 + 4(2 - p) +4 ] / [8(2 - p)]But this is the same as original expression. Maybe this approach is not helpful.Alternatively, let's consider x² = (4 - p)^2/[8(2 - p)] and set this equal to a variable. Let’s denote q = 2 - p. Since p < 2, q > 0.Then, x² = (4 - p)^2/(8q). But 4 - p = 4 - (2 - q) = 4 -2 + q = 2 + q. Therefore, x² = (2 + q)^2/(8q) = (4 +4q + q²)/(8q) = (q² +4q +4)/8q = (q + 2)^2/(8q)Hmm, not sure if this helps. Alternatively, maybe take derivative or find critical points? Wait, perhaps not necessary. Let's recall that x² must be in [1, 4/3]. So:1 ≤ (4 - p)^2/[8(2 - p)] ≤ 4/3Given that p < 2, so 2 - p > 0. Let's write this as:1 ≤ [(4 - p)^2]/[8(2 - p)] ≤ 4/3Multiply all parts by 8(2 - p). Since 8(2 - p) > 0 (because p <2), the inequality signs don't change.So:8(2 - p) ≤ (4 - p)^2 ≤ (4/3)*8(2 - p)Simplify:Left inequality: 8(2 - p) ≤ (4 - p)^2Right inequality: (4 - p)^2 ≤ (32/3)(2 - p)Let's solve left inequality first:8(2 - p) ≤ (4 - p)^2Expand the right side:16 -8p + p²So:16 -8p + p² ≥8(2 -p) ⇒ 16 -8p + p² -16 +8p ≥0 ⇒ p² ≥0Which is always true since p² is non-negative. So the left inequality holds for all real p.Now, the right inequality:(4 - p)^2 ≤ (32/3)(2 - p)Again, expand the left side:16 -8p + p² ≤ (32/3)(2 - p)Multiply both sides by 3 to eliminate the denominator:3(16 -8p + p²) ≤32(2 - p)Expand:48 -24p +3p² ≤64 -32pBring all terms to left side:48 -24p +3p² -64 +32p ≤0 ⇒ -16 +8p +3p² ≤0 ⇒3p² +8p -16 ≤0Now, solve the quadratic inequality 3p² +8p -16 ≤0First, find roots:p = [-8 ±√(64 +192)]/6 = [-8 ±√256]/6 = [-8 ±16]/6So,p = (-8 +16)/6 = 8/6 = 4/3 ≈1.333p = (-8 -16)/6 = -24/6 = -4Therefore, the quadratic 3p² +8p -16 is ≤0 between p = -4 and p =4/3. So the inequality holds for p ∈ [-4, 4/3].But remember that p <2 from previous condition, and here p must be in [-4,4/3] to satisfy the right inequality. Therefore, combining p <2 and p ∈ [-4,4/3], we get p ∈ [-4,4/3].Therefore, the original equation has solutions only when p ∈ [-4, 4/3]. Because for p <2 but not in [-4,4/3], the expression x² would be greater than 4/3, which is outside our domain.Therefore, the allowable values of p are p ∈ [-4, 4/3], and for these p, x² = (4 - p)^2/[8(2 - p)] must be in [1,4/3]. Let's verify this.But since we derived that inequality, and we found that x² is in [1,4/3] exactly when p ∈ [-4,4/3], then we can proceed.So, given that p ∈ [-4,4/3], then x² = (4 - p)^2/[8(2 - p)]. Then x = sqrt[(4 - p)^2/[8(2 - p)]]. But since x ≥1, we take the positive square root.But we also need to check that x is in [1,2/√3]. Let's see, for p ∈ [-4,4/3], does x calculated as sqrt[(4 - p)^2/[8(2 - p)]] lie in [1,2/√3]?Alternatively, since we already derived x² ∈ [1,4/3], so x ∈ [1,2/√3], as needed.Therefore, the solutions for x are x = sqrt[(4 - p)^2/[8(2 - p)]] = |4 - p| / sqrt[8(2 - p)].But since p ∈ [-4,4/3], 4 - p is positive (since p ≤4/3 <4), so |4 - p|=4 -p. Therefore, x = (4 - p)/sqrt[8(2 - p)]Simplify sqrt(8(2 - p)) = 2*sqrt(2(2 - p))Therefore, x = (4 - p)/[2*sqrt(2(2 - p))] = [ (4 - p) ] / [2*sqrt{2(2 - p)} ]We can rationalize or simplify further. Let's see:Multiply numerator and denominator by sqrt{2(2 - p)}:x = [ (4 - p) * sqrt{2(2 - p)} ] / [2 * 2(2 - p) ] = [ (4 - p) * sqrt{2(2 - p)} ] / [4(2 - p) ]Simplify (4 - p)/(4(2 - p)) = (4 - p)/(4(2 - p)) = [ (4 - p) / (2 - p) ] /4But 4 - p = (2 - p) + 2, so:= [ (2 - p +2)/(2 - p) ] /4 = [1 + 2/(2 - p) ] /4Not sure if helpful.Alternatively, leave it as x = (4 - p)/[2*sqrt{2(2 - p)} ]Alternatively, factor numerator and denominator:Note that 4 - p = 4 - p, denominator is 2*sqrt{2(2 - p)}. Let me write sqrt{2(2 - p)} as sqrt{2}*sqrt{2 - p}Therefore, x = (4 - p)/(2*sqrt{2}*sqrt{2 - p}) = (4 - p)/(2*sqrt{2}*sqrt{2 - p}) = [ (4 - p)/sqrt{2 - p} ] / (2*sqrt{2})Which is [ sqrt{(4 - p)^2/(2 - p)} ] / (2*sqrt{2})But since (4 - p)^2/(2 - p) =8x² from the previous equation, but maybe not helpful.Alternatively, leave it as x = (4 - p)/[2√{2(2 - p)} ]But perhaps we can write it as x = sqrt{(4 - p)^2/[8(2 - p)]} = sqrt{ [ (4 - p)^2 ] / [8(2 - p) ] } which is the same as previous expression.Therefore, the solution is x = sqrt{ (4 - p)^2 / [8(2 - p) ] } = |4 - p| / sqrt{8(2 - p)} = (4 - p)/[2√{2(2 - p)} ]But we need to check if this solution satisfies the original equation. Remember that when we squared both sides, we might have introduced extraneous solutions. So we need to verify the solution.So, let's verify. Let's take x = (4 - p)/[2√{2(2 - p)} ] and check if it satisfies the original equation:√(x² - p) + 2√(x² -1) = xFirst, compute x²:x² = (4 - p)^2/[8(2 - p)]Therefore, x² - p = (4 - p)^2/[8(2 - p)] - p= [ (16 -8p + p²) -8p(2 - p) ] / [8(2 - p)]Compute numerator:16 -8p + p² -16p +8p² =16 -8p + p² -16p +8p² =16 -24p +9p²So x² -p = (9p² -24p +16)/[8(2 - p)]Similarly, x² -1 = (4 - p)^2/[8(2 - p)] -1= [ (4 - p)^2 -8(2 - p) ] / [8(2 - p)]Compute numerator:16 -8p +p² -16 +8p = p²Therefore, x² -1 = p²/[8(2 - p)]Therefore, the left side:√(x² - p) + 2√(x² -1) = sqrt( (9p² -24p +16)/[8(2 - p)] ) + 2*sqrt( p²/[8(2 - p)] )Simplify sqrt terms:sqrt( (9p² -24p +16)/[8(2 - p)] ) = sqrt( (9p² -24p +16) ) / sqrt(8(2 - p)) )Similarly, sqrt( p²/[8(2 - p)] ) = |p| / sqrt(8(2 - p)) )But since p ∈ [-4,4/3], and the equation requires x ≥1, but p can be negative. However, x is positive.Wait, but when p is negative, x² must still be ≥p and ≥1. Since p ∈ [-4,4/3], so if p is negative, x² ≥1 is automatically satisfied.But in the term sqrt(x² - p), when p is negative, x² - p is x² + |p|, which is okay.But back to verifying the solution:Compute sqrt(9p² -24p +16). Let's see:9p² -24p +16 = (3p -4)^2Yes, because (3p -4)^2 =9p² -24p +16. So sqrt(9p² -24p +16) = |3p -4|But since p ∈ [-4,4/3], let's check the sign of 3p -4:When p ∈ [-4,4/3], 3p ranges from -12 to 4. Therefore, 3p -4 ranges from -16 to 0. Therefore, |3p -4| = 4 -3p.Therefore, sqrt(9p² -24p +16) =4 -3p.Similarly, sqrt(p²) = |p|. But p can be negative. However, in the original equation, when we derived the solution, we squared both sides, so we need to check if the solution satisfies the original equation.Therefore, substituting back:Left side: [4 -3p]/sqrt(8(2 - p)) + 2*[ |p| / sqrt(8(2 - p)) ]Right side: x = (4 -p)/[2*sqrt(2(2 - p)) ] = (4 -p)/[2*sqrt(2)*sqrt(2 - p) ] = (4 -p)/[2*sqrt(2(2 - p)) ]But sqrt(8(2 - p)) = sqrt(4*2(2 - p)) = 2*sqrt(2(2 - p))Therefore, Left side:[4 -3p]/[2*sqrt(2(2 - p))] + 2*[ |p| ]/[sqrt(8(2 - p))] = [4 -3p]/[2*sqrt(2(2 - p))] + 2*|p|/[2*sqrt(2(2 - p))]Simplify:= [4 -3p + 2|p| ]/[2*sqrt(2(2 - p))]Compare to right side x = (4 -p)/[2*sqrt(2(2 - p))]Therefore, for the left side to equal the right side, we need:4 -3p +2|p| =4 -pSo:4 -3p +2|p| =4 -p ⇒ -3p +2|p| = -p ⇒ 2|p| =2p ⇒ |p|=pWhich holds if p ≥0.But p ∈ [-4,4/3]. So if p ≥0, then |p|=p, and the equality holds. If p <0, then |p|=-p, so 2|p|= -2p, so:-3p -2p = -p ⇒ -5p = -p ⇒ -5p +p =0 ⇒-4p=0⇒p=0, but p <0 in this case, so no solution.Therefore, the equality 4 -3p +2|p| =4 -p holds only when p ≥0.Therefore, when p <0, the left side becomes [4 -3p -2p]/[2*sqrt(2(2 - p))] = [4 -5p]/[2*sqrt(2(2 - p))], which is not equal to (4 -p)/[2*sqrt(2(2 - p))] unless p=0.Therefore, our earlier solution x = (4 -p)/[2√{2(2 -p)} ] is valid only when p ≥0, because for p <0, the verification fails.Therefore, we need to split into cases: p ≥0 and p <0.But wait, the original problem states that p is a real parameter. So we need to check for p ∈ [-4,4/3], but split into p ≥0 and p <0.Case 1: p ≥0.In this case, |p|=p, so the verification holds, so x = (4 -p)/[2√{2(2 -p)} ] is a valid solution.Case 2: p <0.In this case, the verification does not hold, meaning that the solution x = (4 -p)/[2√{2(2 -p)} ] does not satisfy the original equation. Therefore, for p <0, there is no solution.Wait, but that seems conflicting with our previous result where p ∈ [-4,4/3]. But we need to check whether there are solutions for p <0.Wait, perhaps when p <0, even though x² = (4 -p)^2/[8(2 -p)] is in [1,4/3], the actual x does not satisfy the original equation because of the absolute value introduced when squaring.Therefore, for p <0, the equation has no solution, even though x² is in [1,4/3]. Hence, the solution x is valid only when p ∈ [0,4/3].Wait, but how is that possible? Let's take a specific example. Let’s choose p = -1 (which is in [-4,4/3]). Let's compute x from the formula:x = (4 - (-1))/[2*sqrt{2(2 - (-1))}] =5/[2*sqrt(2*3)] =5/[2*sqrt(6)] ≈5/(2*2.449)≈5/4.898≈1.02.Now, check the original equation:√(x² - p) + 2√(x² -1) =x.Compute x≈1.02, x²≈1.0404.Compute √(x² - (-1)) =√(1.0404 +1)=√2.0404≈1.428.Compute 2√(x² -1)=2√(1.0404 -1)=2√0.0404≈2*0.201≈0.402Sum:1.428 +0.402≈1.83, but x≈1.02. Not equal. Therefore, the solution doesn't satisfy the original equation when p is negative. Therefore, even though algebraically we derived a solution, it does not satisfy the original equation when p <0.Therefore, we must have p ∈ [0,4/3] for the solution to be valid.Therefore, the final solution is x = (4 -p)/[2√{2(2 -p)} ] for p ∈ [0,4/3].But we need to confirm this with another example. Let's take p=0.Then x=(4 -0)/[2√{2(2 -0)} ]=4/[2*√4]=4/[2*2]=1.Check original equation: √(1² -0) +2√(1² -1)=√1 +2√0=1+0=1=x. Correct.Take p=4/3.x=(4 -4/3)/[2√{2(2 -4/3)} ]=(8/3)/[2√{2*(2/3)} ]=(8/3)/[2*(√(4/3))]=(8/3)/[2*(2/√3)]=(8/3)/(4/√3)=(8/3)*(√3/4)=2√3/3≈1.1547, which is 2/√3≈1.1547. So x=2/√3, which is the upper limit. Check original equation:√(x² -4/3) +2√(x² -1)=xx=2/√3, x²=4/3.√(4/3 -4/3)=√0=0.2√(4/3 -1)=2√(1/3)=2*(1/√3)=2/√3.Sum:0 +2/√3=2/√3=x. Correct.Another example, p=1.x=(4 -1)/[2√{2(2 -1)} ]=3/[2√2]≈3/2.828≈1.0607.Check original equation:√(x² -1) +2√(x² -1)=x.Wait, x²= (3)^2/[8(2 -1)]=9/8≈1.125.Wait, compute x²= (4 -1)^2/[8(2 -1)]=9/8.So x=3/(2√2)=3/(2*1.414)=~1.06.Compute left side:√(9/8 -1)=√(1/8)=1/(2√2)≈0.3536.2√(9/8 -1)=2*(1/(2√2))=1/√2≈0.7071.Sum:0.3536 +0.7071≈1.0607≈x. Correct.Therefore, when p ∈ [0,4/3], the solution is valid. When p ∈ [-4,0), even though the algebra gives a solution, it does not satisfy the original equation. Hence, there are no solutions for p ∈ [-4,0).But wait, let's check p=-4. If p=-4, according to our previous result, x²=(4 -(-4))²/[8(2 -(-4))]=(8)^2/[8*6]=64/48=4/3. So x=2/√3≈1.1547. Let's check the original equation:√(x² - (-4)) +2√(x² -1)=√(4/3 +4)+2√(4/3 -1)=√(16/3)+2√(1/3)=4/√3 +2/√3=6/√3=2√3≈3.464. But x=2/√3≈1.1547. Not equal. So even at p=-4, the solution does not hold. Hence, p must be in [0,4/3].But earlier, when solving the inequality, we found p ∈ [-4,4/3]. But in reality, only p ∈ [0,4/3] gives valid solutions. Therefore, the correct domain for p is [0,4/3].Therefore, the solution is x=(4 -p)/[2√{2(2 -p)} ] for p ∈ [0,4/3].Hence, the answer is x= (4 - p)/(2√[2(2 - p)]) for 0 ≤ p ≤4/3. To write this neatly, we can rationalize or write in different forms, but probably simplest as:x = frac{4 - p}{2 sqrt{2(2 - p)}}And p must be between 0 and 4/3 inclusive.Therefore, all values of x that satisfy the equation are x = (4 - p)/(2√[2(2 - p)]) for 0 ≤ p ≤ 4/3.Final AnswerThe solution is boxed{dfrac{4 - p}{2 sqrt{2(2 - p)}}} for (0 leq p leq dfrac{4}{3}).</think>To solve the equation (sqrt{x^{2} - p} + 2 sqrt{x^{2} - 1} = x) where (p) is a real parameter, we start by determining the domain. The expressions inside the square roots must be non-negative, leading to (x geq 1). The right side (x) must be non-negative, confirming (x geq 1).We rewrite the equation as:[sqrt{x^2 - p} = x - 2sqrt{x^2 - 1}]Squaring both sides and simplifying, we get:[x^2 - p = x^2 - 4xsqrt{x^2 - 1} + 4(x^2 - 1)]Simplifying further:[4x^2 - 4 + p = 4xsqrt{x^2 - 1}]Dividing by 4 and substituting (t = sqrt{x^2 - 1}), we get:[t^2 + frac{p}{4} = tsqrt{t^2 + 1}]Squaring both sides again and simplifying, we find:[8(p - 2)t^2 + p^2 = 0]Substituting (t^2 = x^2 - 1), we solve for (x^2):[x^2 = frac{(4 - p)^2}{8(2 - p)}]We check the domain constraints and verify that (x geq 1). This leads to (0 leq p leq frac{4}{3}). For (p) in this interval, the solution for (x) is:[x = frac{4 - p}{2 sqrt{2(2 - p)}}]Thus, the values of (x) that satisfy the equation are:[boxed{dfrac{4 - p}{2 sqrt{2(2 - p)}}}]for (0 leq p leq dfrac{4}{3}).,<think>Okay, so I need to solve the equation √(x² - p) + 2√(x² - 1) = x, where p is a real parameter. Hmm, let me start by understanding the domain of this equation. Since we have square roots, the expressions inside must be non-negative. That means x² - p ≥ 0 and x² - 1 ≥ 0. Also, the right side is x, so x must be ≥ 0 because the left side is a sum of square roots which are non-negative. So combining these, x has to be ≥ max{√p, 1} and x ≥ 0. But since √p is non-negative (as p is real, but we need x² ≥ p and x² ≥1). So the domain is x ≥ max{√p, 1} if p ≥0. If p is negative, then x² ≥1 because x² - p would automatically be positive. Wait, p is a real parameter, so p can be any real number, but for the square root to be real, x² - p must be ≥0. So, if p is negative, then x² - p is always positive for all real x, but we still have the other square root requiring x² ≥1, so domain is x ≥1. If p is non-negative, then x² ≥ p and x² ≥1, so x ≥ max{√p, 1}. But since √p could be greater than 1 if p >1, or less than 1 if p <1. So the domain is x ≥ √p when p ≥1, and x ≥1 when p ≤1. Also, x must be ≥0, but since x ≥1 or x ≥√p ≥0, the domain is covered.Now, the equation is √(x² - p) + 2√(x² - 1) = x. Let me think how to approach this. Maybe isolate one of the square roots and square both sides? Let's try.First, let's denote A = √(x² - p) and B = 2√(x² - 1). The equation becomes A + B = x. Let me rearrange it as A = x - B, then square both sides:A² = (x - B)²Which gives:x² - p = x² - 2xB + B²Simplify:x² - p = x² - 2xB + B²Subtract x² from both sides:-p = -2xB + B²Multiply both sides by -1:p = 2xB - B²But B is 2√(x² -1), so substitute back:p = 2x*(2√(x² -1)) - [2√(x² -1)]²Compute each term:First term: 2x*(2√(x² -1)) = 4x√(x² -1)Second term: [2√(x² -1)]² = 4(x² -1)So, p = 4x√(x² -1) - 4(x² -1)Let's factor out 4:p = 4[ x√(x² -1) - x² +1 ]Hmm, that's one equation. Maybe we can express this in terms of x. But this seems complicated. Alternatively, perhaps another substitution. Let me consider substituting t = x². Then √(t - p) + 2√(t -1) = √t. Wait, but x is in the equation. Wait, x is positive because we concluded x ≥1. So x = √t. Hmm, but x is sqrt(t), so original equation becomes √(t - p) + 2√(t -1) = √t. Maybe this substitution could help? Let's try.Let t = x², then equation is √(t - p) + 2√(t -1) = √t.But then, perhaps this is another equation to solve for t, but with p as a parameter. Let me attempt solving this.Let me denote √(t - p) = √t - 2√(t -1). Then square both sides:t - p = t - 4√t√(t -1) + 4(t -1)Simplify RHS:t -4√[t(t -1)] + 4t -4 = 5t -4 -4√[t(t -1)]Thus:t - p =5t -4 -4√[t(t -1)]Bring t terms to the left:-p =4t -4 -4√[t(t -1)]Multiply both sides by -1:p = -4t +4 +4√[t(t -1)]Hmm, so p =4√[t(t -1)] -4t +4.Not sure if that helps. Let me see if I can write this equation as:4√[t(t -1)] =4t -4 +pDivide both sides by 4:√[t(t -1)] = t -1 + p/4Now square both sides again:t(t -1) = [t -1 + p/4]^2Expand RHS:(t -1)^2 + 2(t -1)(p/4) + (p/4)^2Which is:t² -2t +1 + (p/2)(t -1) + p²/16Thus, equation is:t² - t = t² -2t +1 + (p/2)(t -1) + p²/16Subtract t² from both sides:-t = -2t +1 + (p/2)(t -1) + p²/16Bring all terms to left:-t +2t -1 - (p/2)(t -1) - p²/16 =0Simplify:t -1 - (p/2)(t -1) - p²/16=0Factor out (t -1):[1 - p/2](t -1) - p²/16 =0Hmm, so:[ (2 - p)/2 ] (t -1) - p²/16 =0Multiply both sides by 16 to eliminate denominators:8(2 - p)(t -1) - p² =0Expand:16(2 - p)(t -1) - p² =0? Wait, no, wait:Wait, [ (2 - p)/2 ] (t -1) - p²/16 =0Multiply by 16:8(2 - p)(t -1) - p² =0Yes. So:8(2 - p)(t -1) = p²Divide both sides by 8(2 - p):t -1 = p² / [8(2 - p)]Thus:t =1 + [p²]/[8(2 - p)]But t =x², so:x² =1 + [p²]/[8(2 - p)]Therefore, x= sqrt[1 + p²/(8(2 - p))]But wait, this seems like a possible solution, but we need to check for validity since we squared twice, which can introduce extraneous solutions. Also, we need to consider the domain for t. Remember t =x² ≥1, and t ≥ p. So, the expression under the square root must be valid, and x must be ≥1. Let's analyze.First, the expression 1 + p²/(8(2 - p)) must be ≥1, because x² must be ≥1. Let's check:1 + [p²]/[8(2 - p)] ≥1Which implies [p²]/[8(2 - p)] ≥0So the denominator 8(2 - p) must be positive, because numerator p² is always non-negative. So 8(2 - p) >0 => 2 -p >0 => p <2.If p <2, then denominator is positive, and numerator p² ≥0, so [p²]/[8(2 - p)] ≥0. Hence, x² ≥1. So that's okay.If p >2, then denominator 8(2 - p) <0, and numerator p² >0, so [p²]/[8(2 - p)] is negative. Then 1 + negative number would be less than 1, which would make x² <1, which contradicts the domain x ≥1. Therefore, for p >2, this solution would not be valid.If p=2, then denominator is zero, so expression is undefined.So possible solutions when p <2. Let's note that.So, for p <2, we have x= sqrt[1 + p²/(8(2 - p))]But we need to check if this solution satisfies the original equation and the domain conditions.Additionally, when we squared the equation twice, we might have introduced extraneous roots, so we need to verify.Also, remember that when we started, we had the domain x ≥ max{√p,1}. If p <2, then depending on p, √p could be less than 1 or greater. For p ≤1, √p ≤1, so domain is x ≥1. For 1 <p <2, √p is between 1 and √2 ≈1.414, so domain is x ≥√p. But our solution is x= sqrt[1 + p²/(8(2 - p))]. Let's check if this is ≥ max{√p,1}.Let me compute sqrt[1 + p²/(8(2 - p))] and compare with max{√p,1}.First, consider p ≤1. Then max{√p,1}=1. Let's see if sqrt[1 + p²/(8(2 - p))] ≥1. Since p²/(8(2 - p)) is non-negative (as p ≤1 <2, so denominator positive, numerator p² ≥0), so the sqrt is ≥1. So okay.For 1 <p <2, we need sqrt[1 + p²/(8(2 - p))] ≥√p. Let's square both sides:1 + p²/(8(2 - p)) ≥pMultiply both sides by 8(2 - p) (which is positive since p <2):8(2 - p) + p² ≥8p(2 - p)Compute LHS: 16 -8p + p²RHS:16p -8p²Bring all terms to left:16 -8p + p² -16p +8p² ≥0Combine like terms:9p² -24p +16 ≥0Factor:Let me compute discriminant: 576 - 4*9*16=576 -576=0. So it's a perfect square.Thus, (3p -4)^2 ≥0. Which is always true. Equality when 3p -4=0 => p=4/3. So for all real p, (3p -4)^2 ≥0. Thus, the inequality holds, so sqrt[1 + p²/(8(2 - p))] ≥√p for 1 <p <2. Therefore, the solution x= sqrt[1 + p²/(8(2 - p))] is in the domain for p <2.Now, let's check when p ≥2. For p ≥2, the expression 1 + p²/(8(2 - p)) becomes 1 minus a positive number (since denominator is negative), so if p²/(8(2 - p)) is negative, then x²=1 + negative number, which may be less than 1, which is invalid. But we can check if maybe the equation still has solutions for p ≥2?Wait, when p ≥2, the domain requires x² ≥ p and x² ≥1, so x ≥√p. But original equation is √(x² - p) + 2√(x² -1) =x. Let's suppose p=2. Then equation becomes √(x² -2) +2√(x² -1)=x. Is there a solution here? Let's test x=√p=√2 ≈1.414. Then √(2 -2)=0, 2√(2 -1)=2*1=2. Left side=0 +2=2. Right side=√2≈1.414. Not equal. So x=√2 is not a solution. Let's see if there's a larger x. Suppose x=2. Then left side: √(4 -2)=√2≈1.414, 2√(4 -1)=2√3≈3.464. Total≈4.878. Right side=2. Not equal. So maybe no solution for p=2.For p>2, let's pick p=3. Then x must be ≥√3≈1.732. Let's test x=2. Left side: √(4 -3)=1, 2√(4 -1)=2√3≈3.464. Total≈4.464. Right side=2. Not equal. Try x=3. Left side: √(9 -3)=√6≈2.45, 2√(9 -1)=2*2.828≈5.656. Total≈8.106 vs right side=3. Not equal. So maybe no solutions for p ≥2.Alternatively, maybe for p ≥2, there's no solution.But let's check. Suppose p=0. Then equation becomes √(x²) +2√(x² -1)=x. Which is |x| +2√(x² -1)=x. But since x ≥1 (domain), so |x|=x. So equation is x +2√(x² -1)=x. Then subtract x: 2√(x² -1)=0. Then √(x² -1)=0. So x² -1=0 =>x=1. But x=1 is in the domain (p=0, so max{√0,1}=1). Check original equation: √(1 -0) +2√(1 -1)=1 +0=1=x. So x=1 is a solution when p=0. But according to our previous formula, x= sqrt[1 +0/(8(2 -0))]=sqrt[1 +0]=1. So correct.Similarly, let's check for p=1. Then x= sqrt[1 +1/(8(2 -1))]=sqrt[1 +1/8]=sqrt[9/8]= (3)/(2√2)≈1.06. Wait, but p=1, so domain is x≥1. So x≈1.06 is valid. Let's check in original equation: √(x² -1) +2√(x² -1)=3√(x² -1). So 3√(x² -1)=x. Let x=3/(2√2). Compute x²=9/(8). Then x² -1=1/8. So √(1/8)=1/(2√2). Then 3*(1/(2√2))=3/(2√2)=x. Which matches. So correct.Another test case: p=1.5. Then x= sqrt[1 + (2.25)/(8*(0.5))]=sqrt[1 +2.25/4]=sqrt[1 +0.5625]=sqrt[1.5625]=1.25. Check domain: p=1.5, so √p≈1.2247. So x=1.25 ≥1.2247, so okay. Now check original equation: √(1.5625 -1.5)=√0.0625=0.25, and 2√(1.5625 -1)=2√0.5625=2*0.75=1.5. Sum:0.25 +1.5=1.75. But x=1.25. Wait, 1.75≠1.25. Wait, contradiction. So something's wrong here. Wait, what's happening.Wait, p=1.5, so according to our formula, x= sqrt[1 + (p²)/(8(2 -p))] = sqrt[1 + (2.25)/(8*(0.5))] = sqrt[1 +2.25/4] = sqrt[1 +0.5625] = sqrt[1.5625] =1.25. Let's plug into original equation:Left side: sqrt(x² - p) +2 sqrt(x² -1)=sqrt(1.5625 -1.5)+2 sqrt(1.5625 -1)=sqrt(0.0625)+2 sqrt(0.5625)=0.25 +2*0.75=0.25+1.5=1.75. Right side: x=1.25. 1.75≠1.25. So the solution doesn't satisfy the equation. Which means our formula gives an extraneous solution. So this is a problem.Therefore, even though algebraically we derived x= sqrt[1 + p²/(8(2 - p))], in reality, when we plug in p=1.5, it doesn't satisfy. So our solution is invalid. That means we made a mistake in the process.Wait, so we need to check the solutions after squaring. So when we square equations, we can get extraneous solutions, so need to verify.So perhaps even though we derived x= sqrt[1 + p²/(8(2 - p))], only some p values make this a valid solution.Wait, let's check when p=1.5, x=1.25. Let's see the domain: x=1.25, which is greater than √1.5≈1.2247, so okay. But the original equation gives 1.75=1.25, which is false. Therefore, the solution is extraneous. Therefore, there must be a mistake in the steps.Let me go back through the steps.Original equation:√(x² - p) +2√(x² -1)=x.First step: move √(x² -p) to the other side? Wait, no, actually, when I first rearranged, I set A=√(x² -p), B=2√(x² -1), then wrote A =x - B. Then squared both sides.But when I squared, I might have introduced extraneous solutions. So, perhaps even though algebraically we derived x= sqrt[1 + p²/(8(2 - p))], in reality, only certain p satisfy the original equation.Alternatively, maybe there's a restriction on p. Let's check p=1.5. Our solution x=1.25 gives left side=1.75, right side=1.25. Not equal. So the solution is extraneous. So need to figure out why.Wait, maybe when solving, after squaring, we need additional constraints.Let me re-examine the steps.Starting with:√(x² - p) +2√(x² -1) =x.Let me isolate one radical:√(x² - p) =x -2√(x² -1)Now, before squaring, both sides must be non-negative. The left side is a square root, so non-negative. Therefore, the right side must also be non-negative: x -2√(x² -1) ≥0.So x ≥2√(x² -1). Let's square this inequality (keeping in mind x ≥0):x² ≥4(x² -1)x² ≥4x² -40 ≥3x² -43x² ≤4x² ≤4/3x ≤2/√3≈1.1547But from the domain, x ≥max{√p,1}. If p=1.5, then x≥√1.5≈1.2247. But our derived solution x=1.25 is greater than 1.2247 but the inequality x ≥2√(x² -1) requires x ≤2/√3≈1.1547. Hence, for p=1.5, the solution x=1.25 does not satisfy x ≤1.1547, which is a necessary condition for the initial equation rearrangement. Therefore, the solution is extraneous.Therefore, in order for the equation √(x² - p) =x -2√(x² -1) to hold, the right side must be non-negative, so x ≥2√(x² -1). So this inequality must be satisfied by the solution. Therefore, combining this with the domain x ≥max{√p,1}, we can find the valid solutions.Therefore, after deriving x= sqrt[1 + p²/(8(2 - p))], we must check two things:1. x ≥max{√p,1}2. x ≥2√(x² -1)But since x= sqrt[1 + p²/(8(2 - p))], perhaps we can express the second condition in terms of p.Condition 2: x ≥2√(x² -1)Let me substitute x²=1 + p²/(8(2 - p)).Then x= sqrt(1 + p²/(8(2 - p)))So √(x² -1)=sqrt(p²/(8(2 -p)))=|p|/(2√[2(2 -p)]). Since p is a real parameter, but we are considering p <2 (since for p ≥2, solution is invalid). For p <2, 2 -p >0. Also, since x is positive, p can be negative?Wait, p can be negative. Let me check if p can be negative. Let's suppose p=-1. Then the equation becomes √(x² +1)+2√(x² -1)=x. The domain here is x ≥1. Let's check if the formula gives a solution.For p=-1, x= sqrt[1 + (1)/(8(2 -(-1)))]=sqrt[1 +1/(24)]=sqrt[25/24]≈1.0206. Then check condition x ≥2√(x² -1). Compute RHS:2√(25/24 -1)=2√(1/24)=2/(2√6)=1/√6≈0.408. So x≈1.0206 ≥0.408. True. Also, check the original equation. √(x² -p)=√(25/24 +1)=√(49/24)=7/(2√6)≈1.429. 2√(x² -1)=2*(1/(2√6))=1/√6≈0.408. Sum≈1.429+0.408≈1.837. But x≈1.0206. Not equal. So again, the solution is extraneous.Wait, so even for p negative, the formula gives a solution that doesn't satisfy the original equation. Therefore, there must be a mistake in the process.Alternatively, perhaps there are no solutions for p <2 except for specific cases.Wait, when p=0, x=1. We checked that x=1 works. Let's see if p=0, the solution x=1 satisfies condition 2: x ≥2√(x² -1). x=1, then 1 ≥2√(1 -1)=0. True. So that's okay.But when p=1, x=3/(2√2)≈1.06. Check condition 2: x ≥2√(x² -1). Compute RHS: 2√(9/8 -1)=2√(1/8)=2*(1/(2√2))=1/√2≈0.707. So 1.06 ≥0.707, true. Then check original equation: √(x² -1) +2√(x² -1)=3√(x² -1)=3*(1/(2√2))≈1.06. Which equals x≈1.06. Correct.So for p=1, the solution works. So why for p=1.5 it doesn't? Because in the case p=1, x=3/(2√2)≈1.06, which is less than 2/√3≈1.1547. Wait, but in the case p=1.5, x=1.25 which is greater than 2/√3≈1.1547. Wait, but condition 2 is x ≥2√(x² -1). Let's compute 2√(x² -1) for x=1.25:2√(1.5625 -1)=2√0.5625=2*0.75=1.5. So x=1.25 vs 1.5. So 1.25 is not ≥1.5. Therefore, the condition x ≥2√(x² -1) is not satisfied. Hence, the solution is invalid. Therefore, even though x= sqrt[1 + p²/(8(2 - p))] is in the domain, it fails the second condition. Therefore, we need to enforce that x ≥2√(x² -1).Let me express this condition in terms of p.Given x= sqrt[1 + p²/(8(2 - p))], we need x ≥2√(x² -1)Let me substitute x²=1 + p²/(8(2 - p)) into √(x² -1):√(x² -1)=sqrt(p²/(8(2 - p)))= |p|/(2√[2(2 - p)])Since p <2 (as we're considering p <2 for the solution to exist), and x is positive, so:x ≥2√(x² -1) => sqrt[1 + p²/(8(2 - p))] ≥2*(|p|/(2√[2(2 - p)])) )Simplify RHS: (|p|)/√[2(2 - p)]Therefore:sqrt[1 + p²/(8(2 - p))] ≥ |p|/√[2(2 - p)]Square both sides (both sides are non-negative):1 + p²/(8(2 - p)) ≥ p²/(2(2 - p))Multiply both sides by 8(2 - p):8(2 - p) + p² ≥4p²Simplify:16 -8p +p² ≥4p²Bring all terms to left:16 -8p +p² -4p² ≥0Simplify:-3p² -8p +16 ≥0Multiply by -1 (reverse inequality):3p² +8p -16 ≤0Solve the quadratic inequality 3p² +8p -16 ≤0.Find roots:p = [-8 ±√(64 +192)]/(6) = [-8 ±√256]/6 = [-8 ±16]/6So p=(8)/6=4/3≈1.333 or p=(-24)/6=-4.The quadratic opens upwards (coefficient 3>0), so the inequality 3p² +8p -16 ≤0 holds between the roots: p ∈ [-4, 4/3]Therefore, the condition x ≥2√(x² -1) is satisfied only when p ∈ [-4,4/3]Therefore, the solution x= sqrt[1 + p²/(8(2 - p))] is valid only when p ∈ (-∞, 2) and p ∈ [-4,4/3]. So combining these, p ∈ [-4,4/3]Therefore, the original equation has solutions only when p ∈ [-4,4/3], and the solution is x= sqrt[1 + p²/(8(2 - p))]Wait, but when p=4/3, then the inequality becomes equality. Let's check p=4/3.p=4/3: x= sqrt[1 + (16/9)/(8*(2 -4/3))] = sqrt[1 + (16/9)/(8*(2/3))] = sqrt[1 + (16/9)/(16/3)] = sqrt[1 + (16/9)*(3/16)] = sqrt[1 + 1/3] = sqrt(4/3)=2/√3≈1.1547Check original equation: √(x² -4/3)+2√(x² -1)=x.x=2/√3, x²=4/3. Then √(4/3 -4/3)=0, 2√(4/3 -1)=2√(1/3)=2/√3. So total left side=0 +2/√3=2/√3. Right side=2/√3. So equality holds. Hence, p=4/3 is valid.Similarly, p=-4. Let's check p=-4.x= sqrt[1 + (16)/(8*(2 -(-4)))]=sqrt[1 +16/(8*6)]=sqrt[1 +16/48]=sqrt[1 +1/3]=sqrt(4/3)=2/√3≈1.1547Check original equation: √(x² -(-4))+2√(x² -1)=√(4/3 +4)+2√(4/3 -1)=√(16/3)+2√(1/3)=4/√3 + 2/√3=6/√3=2√3≈3.464. But x=2/√3≈1.1547. Not equal. So even though p=-4 is in the interval, the solution is extraneous. Contradiction. So something is wrong.Wait, perhaps when p is negative, even if the algebraic conditions are satisfied, the solution may not satisfy the original equation.Let me check p=-4. x=2/√3≈1.1547. Compute left side:√(x² -p)=√( (4/3) -(-4))=√(4/3 +4)=√(16/3)=4/√3≈2.30942√(x² -1)=2√(4/3 -1)=2√(1/3)=2/√3≈1.1547Sum≈2.3094+1.1547≈3.464. Right side=x≈1.1547. Not equal. Therefore, even though p=-4 is in the interval, the solution doesn't hold. Hence, our conclusion is wrong.Therefore, the quadratic inequality 3p² +8p -16 ≤0 gives p ∈ [-4,4/3], but for p negative, even if this is satisfied, the solution x doesn't satisfy the original equation.Wait, but why? Because when p is negative, the term √(x² -p) becomes √(x² +|p|), which is larger, but the equation requires this to add to 2√(x² -1) to be equal to x. However, when p is negative, the left side becomes larger, making it harder to equal x.Therefore, perhaps even within p ∈ [-4,4/3], we need additional constraints. Let's check for p=0, which worked. p=1, worked. p=4/3, worked. But p=-4 didn't work. Maybe the actual valid interval is p ∈ [0,4/3].Let's check p= -1, which we tried earlier. The solution x≈1.02 doesn't satisfy original equation. Therefore, perhaps only for p ∈ [0,4/3], the solution is valid.Wait, but why does the quadratic inequality suggest p ∈ [-4,4/3], but in reality, even within this interval, negative p don't work. Maybe another condition we missed.Wait, when we derived the condition 3p² +8p -16 ≤0, it was based on the premise that x ≥2√(x² -1). But when p is negative, we have another condition from the domain: x ≥1. However, when p is negative, the domain is x ≥1. But the solution x= sqrt[1 + p²/(8(2 - p))] must also satisfy x ≥1. For p negative, let's compute x:For p negative, p² is positive, 8(2 -p) is 8*(2 -p)=16 -8p. Since p is negative, denominator is greater than16. So p²/(8(2 -p)) is positive, but divided by a larger number. So x= sqrt[1 + positive], which is greater than1. So x ≥1 is satisfied.But even so, when p is negative, the solution doesn't satisfy the original equation, as seen in p=-1 and p=-4.Wait, perhaps the problem is that when we squared the equation, we lost the information about the sign. Wait, let's see.Original equation after first rearrangement:√(x² - p) =x -2√(x² -1)When p is negative, √(x² -p)=√(x² +|p|) is greater than x. But the right side is x -2√(x² -1). Since x ≥1, 2√(x² -1) is at least 0, so x -2√(x² -1) could be positive or negative. But from earlier, we required x -2√(x² -1) ≥0, leading to x ≥2√(x² -1). Which we translated into p ∈ [-4,4/3].But for p negative, even if x ≥2√(x² -1), the left side √(x² -p) is greater than x, and the right side is x minus something. So their equality might not hold.Wait, take p negative, say p=-k, k>0. Then equation becomes √(x² +k) +2√(x² -1)=x. Since √(x² +k) >x, the left side is greater than x +2√(x² -1). But x +2√(x² -1) is greater than x. So left side is greater than x, but right side is x. Contradiction. Therefore, for p negative, there are no solutions. So even though algebraically we can derive solutions, they don't actually satisfy the original equation.Therefore, the valid interval for p is actually p ∈ [0,4/3]. Because for p negative, no solutions exist, despite the algebraic derivation. Because when p is negative, left side >x, right side=x, so no solution.But in our earlier test, p=0 gives x=1, which works. p=1 gives x=3/(2√2)≈1.06, which works. p=4/3 gives x=2/√3≈1.1547, which works. So perhaps the actual valid p is [0,4/3]. Let me check for p=0.5.p=0.5: x= sqrt[1 +0.25/(8*(2 -0.5))]=sqrt[1 +0.25/(12)]≈sqrt[1 +0.0208]≈1.010. Check original equation:√(1.0201 -0.5)=√0.5201≈0.721, 2√(1.0201 -1)=2√0.0201≈2*0.142≈0.284. Sum≈0.721+0.284≈1.005≈x≈1.010. Close, but not exact. Due to approximation, let's compute exactly.x²=1 + (0.25)/(8*1.5)=1 +0.25/12=1 +1/48=49/48. So x=7/√48=7/(4√3)≈7/6.928≈1.010.Left side:√(49/48 -0.5)=√(49/48 -24/48)=√(25/48)=5/(4√3)≈0.721. 2√(49/48 -1)=2√(1/48)=2/(4√3)=1/(2√3)≈0.288. Sum≈0.721+0.288≈1.009≈x≈1.010. Close enough, considering rounding errors. So it works.Hence, p=0.5 is okay. Therefore, conclusion: the solution x= sqrt[1 +p²/(8(2 -p))] is valid only when p ∈ [0,4/3]. For p <0, even though algebraically the solution exists, it doesn't satisfy the original equation. For p ∈ [0,4/3], it works. For p >4/3, the solution doesn't satisfy x ≥2√(x² -1), so invalid. For p=4/3, it works. For p>4/3 up to p=2, the solution is invalid. For p≥2, no solution.Therefore, the final answer is that the equation has a solution x= sqrt[1 +p²/(8(2 -p))] if and only if p ∈ [0,4/3], and x= sqrt[1 +p²/(8(2 -p))].But we need to confirm this with p=4/3.p=4/3: x= sqrt[1 + (16/9)/(8*(2 -4/3))]=sqrt[1 + (16/9)/(8*(2/3))]=sqrt[1 + (16/9)/(16/3)]=sqrt[1 + (16/9)*(3/16)]=sqrt[1 +1/3]=sqrt(4/3)=2/√3. Check original equation:√(4/3 -4/3) +2√(4/3 -1)=0 +2√(1/3)=2/√3. Which equals x=2/√3. Correct.For p=0: x=1. Correct.For p=1: x=3/(2√2). Correct.For p=0.5: approximately correct.For p=-1: solution exists but doesn't satisfy equation. Hence, only p ∈ [0,4/3].But why does the quadratic inequality 3p² +8p -16 ≤0 include p ∈ [-4,4/3], but in reality, for p negative, solutions are invalid. Therefore, the correct interval is the intersection of p ∈ [-4,4/3] and p ≥0, giving p ∈ [0,4/3].Therefore, the equation has solutions only when p is in [0,4/3], and the solution is x= sqrt(1 + p²/(8(2 -p))).Hence, the answer is x= sqrt( (8(2 -p) +p²)/[8(2 -p)] ) = sqrt( (16 -8p +p²)/[8(2 -p)] ). This simplifies to sqrt( (p² -8p +16)/[8(2 -p)] )=sqrt( ( (p -4)^2 )/[8(2 -p)] ). Since p ≤4/3 <4, so (p -4)^2=(4 -p)^2. Therefore:x= (4 -p)/sqrt(8(2 -p)) = (4 -p)/(2√[2(2 -p)])But maybe simplifying differently:x= sqrt(1 +p²/(8(2 -p)))=sqrt( [8(2 -p) +p²]/[8(2 -p)] ) numerator=16 -8p +p²=(p -4)^2. So:x= sqrt( (p -4)^2 / [8(2 -p)] ) = |p -4| / sqrt(8(2 -p)).Since p ≤4/3 <4, so |p -4|=4 -p. Hence:x=(4 -p)/sqrt(8(2 -p)) = (4 -p)/(2√[2(2 -p)] )This can be further simplified as:x=(4 -p)/(2√[2(2 -p)] ) = [ (4 -p) ] / [2√{2(2 -p)} ] = [ (4 -p) ] / [2√{2}√(2 -p)} ] = [ (4 -p) ] / [2√2√(2 -p)} ] = [ (4 -p) ] / [2√{2(2 -p)} } ]Alternatively, rationalizing:x= (4 -p)/(2√[2(2 -p)] )But it's probably simplest to leave it as x= sqrt(1 +p²/(8(2 -p))).Thus, the solution exists only when p ∈ [0,4/3], and x= sqrt(1 +p²/(8(2 -p))).Final AnswerThe values of ( x ) that satisfy the equation are (boxed{sqrt{1 + dfrac{p^2}{8(2 - p)}}}) for ( 0 leq p leq dfrac{4}{3} ).</think>To solve the equation (sqrt{x^{2} - p} + 2 sqrt{x^{2} - 1} = x) where (p) is a real parameter, we start by considering the domain of the equation. The expressions inside the square roots must be non-negative, so (x^2 geq max{p, 1}). Additionally, the right side (x) must be non-negative, so (x geq 0). Therefore, the domain is (x geq max{sqrt{p}, 1}).We isolate one of the square roots and square both sides:[sqrt{x^2 - p} = x - 2sqrt{x^2 - 1}]Squaring both sides, we get:[x^2 - p = x^2 - 4xsqrt{x^2 - 1} + 4(x^2 - 1)]Simplifying, we find:[-p = -4xsqrt{x^2 - 1} + 4x^2 - 4]Rearranging terms and solving for (p), we get:[p = 4xsqrt{x^2 - 1} - 4(x^2 - 1)]Substituting (t = x^2) and solving, we find:[p = 4sqrt{t(t - 1)} - 4t + 4]Further simplification leads to:[p = 8(2 - p)(t - 1)]Solving for (t), we get:[t = 1 + frac{p^2}{8(2 - p)}]Thus, (x = sqrt{1 + frac{p^2}{8(2 - p)}}).We must check the validity of this solution. The right side (x - 2sqrt{x^2 - 1}) must be non-negative, leading to the condition (x geq 2sqrt{x^2 - 1}). This translates to solving the quadratic inequality (3p^2 + 8p - 16 leq 0), which gives (p in [-4, 4/3]). However, for (p < 0), the solution does not satisfy the original equation. Therefore, the valid interval for (p) is ([0, 4/3]).Thus, the values of (x) that satisfy the equation are:[boxed{sqrt{1 + dfrac{p^2}{8(2 - p)}}}]for (0 leq p leq dfrac{4}{3}).]

🔑:[Solve] sqrt{x^{2}-p}+2 sqrt{x^{2}-1}=xRearrange and square both sides to getx^{2}-p=x^{2}+4left(x^{2}-1right)-4 x sqrt{x^{2}-1}which simplifies to 4left(x^{2}-1right)+p=4 x sqrt{x^{2}-1}.Square both sides again to get16left(x^{2}-1right)^{2}+8 pleft(x^{2}-1right)+p^{2}=16 x^{2}left(x^{2}-1right)Rearranging gives quad 8(2-p) x^{2}=(p-4)^{2}.Now, we discuss the parameter p.(1) When p=2, the equation (2) becomes0 cdot x^{2}=4Clearly, there is no solution. Thus, the original equation has no solution.(2) When p>2, for any real number x, we have8(2-p) x^{2} leqslant 0while the right side of (2) is quad(p-4)^{2} geqslant 0.Thus, only when p=4, equation (2) has a solution: x=0.However, substituting x=0 into the original equation (1), the left side is undefined, so x=0 is not a solution to the original equation. Therefore, when p>2, the original equation has no solution.(3) When p<2, we have 8(2-p) x^{2} > 0, sox=frac{4-p}{2 sqrt{2(2-p)}}Substituting (3) into the original equation (1) for verification, we getfrac{|3 p-4|}{2 sqrt{2(2-p)}}+frac{2|p|}{2 sqrt{2(2-p)}}=frac{4-p}{2 sqrt{2(2-p)}}which simplifies to|3 p-4|+2|p|=4-pWe discuss p in different cases.When p>frac{4}{3}, the original equation has no solution.When 0 leqslant p leqslant frac{4}{3}, the original equation has a unique solution:x=frac{4-p}{2 sqrt{2(2-p)}}.

🧠:[<think>Okay, so I need to figure out which of these sets can be the lengths of the sides of a right-angled triangle. The options are A, B, C, D. Let me recall that for a triangle to be right-angled, the Pythagorean theorem must hold. That is, the sum of the squares of the two shorter sides should equal the square of the longest side. The longest side is the hypotenuse.Let me start with option A: 7, 12, 13. First, I should check if these numbers can form a triangle. The triangle inequality states that the sum of any two sides should be greater than the third side. Let's verify that. 7 + 12 = 19, which is greater than 13. 7 + 13 = 20, which is greater than 12. 12 + 13 = 25, which is greater than 7. So, the triangle inequality holds. Now, check if it's a right-angled triangle. The longest side is 13, so let's see if 7² + 12² equals 13².Calculating: 7² is 49, 12² is 144. Adding them: 49 + 144 = 193. Now, 13² is 169. Hmm, 193 ≠ 169. So, that doesn't work. Therefore, option A is not a right-angled triangle.Moving on to option B: 3, 4, 7. First, check the triangle inequality. 3 + 4 = 7. Wait, but that equals the third side, which is 7. But the triangle inequality requires the sum to be greater than the third side. Since 3 + 4 is exactly 7, that would form a degenerate triangle, which is basically a straight line. So, this can't be a triangle at all. Therefore, option B is invalid as a triangle, so it can't be a right-angled triangle.Option C: 8, 15, 17. Check triangle inequality first. 8 + 15 = 23, which is greater than 17. 8 + 17 = 25, which is greater than 15. 15 + 17 = 32, which is greater than 8. So, triangle inequality holds. Now check Pythagorean theorem. The longest side is 17. Compute 8² + 15²: 64 + 225 = 289. 17² is 289. So, 64 + 225 = 289, which equals 17². Therefore, this satisfies the Pythagorean theorem. So, option C is a right-angled triangle.Option D: 1.6, 2, 2.5. Let's check triangle inequality. 1.6 + 2 = 3.6, which is greater than 2.5. 1.6 + 2.5 = 4.1, which is greater than 2. 2 + 2.5 = 4.5, which is greater than 1.6. So, triangle inequality holds. Now, check if it's right-angled. The longest side is 2.5. Let's compute 1.6² + 2². 1.6 squared is 2.56. 2 squared is 4. Adding them: 2.56 + 4 = 6.56. Now, 2.5 squared is 6.25. Wait, 6.56 ≠ 6.25. Hmm. So, that's not equal. But maybe I made a mistake in calculations? Let me check again.1.6 squared: 1.6 * 1.6. 1.6 * 1 is 1.6, 1.6 * 0.6 is 0.96. So, total is 1.6 + 0.96 = 2.56. Correct. 2 squared is 4. 2.56 + 4 is indeed 6.56. 2.5 squared is 6.25. So, they aren't equal. Therefore, D does not satisfy the Pythagorean theorem. So, not a right-angled triangle.Wait, but maybe the sides are in a different order? Like, perhaps 2 is the hypotenuse? Let me check. If we consider 1.6 and 2.5 as the legs, and 2 as the hypotenuse? Let's see: 1.6² + 2.5² should equal 2². But 2.56 + 6.25 = 8.81, and 2² is 4. So, that's even worse. Alternatively, 1.6 and 2 as legs, hypotenuse 2.5. Which we checked earlier. So, no.Alternatively, maybe the order is different. Wait, the sides are given as 1.6, 2, 2.5. The hypotenuse must be the largest, which is 2.5. So, the legs should be 1.6 and 2. But as we saw, 1.6² + 2² = 2.56 + 4 = 6.56, which is not equal to 2.5² = 6.25. Therefore, regardless of the combination, it doesn't satisfy. So, D is not a right-angled triangle.So, recapping:A: Not a right triangle.B: Not a triangle.C: Right triangle.D: Not a right triangle.Therefore, the correct answer is option C.Final Answerboxed{C}</think>To determine which set of numbers can be the lengths of the sides of a right-angled triangle, we use the Pythagorean theorem, which states that the sum of the squares of the two shorter sides must equal the square of the longest side (hypotenuse).Option A: 7, 12, 13- Check triangle inequality: Valid.- Check Pythagorean theorem: (7^2 + 12^2 = 49 + 144 = 193), (13^2 = 169). Not equal. Not a right-angled triangle.Option B: 3, 4, 7- Check triangle inequality: (3 + 4 = 7) (degenerate triangle). Not a valid triangle.Option C: 8, 15, 17- Check triangle inequality: Valid.- Check Pythagorean theorem: (8^2 + 15^2 = 64 + 225 = 289), (17^2 = 289). Equal. Right-angled triangle.Option D: 1.6, 2, 2.5- Check triangle inequality: Valid.- Check Pythagorean theorem: (1.6^2 + 2^2 = 2.56 + 4 = 6.56), (2.5^2 = 6.25). Not equal. Not a right-angled triangle.Thus, the correct answer is (boxed{C}).,<think>Okay, let's see. I need to figure out which of these sets of numbers can be the lengths of the sides of a right-angled triangle. The options are A, B, C, D. Hmm, right-angled triangles follow the Pythagorean theorem, right? So, for any triangle with sides a, b, and c, where c is the hypotenuse (the longest side), it should satisfy a² + b² = c². Let me start with option A: 7, 12, 13. First, I need to identify the longest side here. Looking at the numbers, 13 is the largest. So, let's check if 7² + 12² equals 13². Calculating 7 squared is 49, 12 squared is 144. Adding those together: 49 + 144 = 193. Now, 13 squared is 169. Wait, 193 isn't equal to 169. So, that doesn't work. Therefore, option A is not a right-angled triangle. Hmm, maybe I did the math wrong? Let me double-check. 7² is definitely 49, 12² is 144. 49 + 144 is 193. 13² is 169. Yep, that's correct. So A is out.Moving on to option B: 3, 4, 7. Again, the longest side here is 7. Let's check if 3² + 4² equals 7². 3 squared is 9, 4 squared is 16. Adding those: 9 + 16 = 25. 7 squared is 49. 25 vs. 49 – nope, that's not equal either. So B isn't a right triangle. Wait, but 3,4,5 is a classic Pythagorean triple. Maybe there's a typo here? But no, the given numbers are 3,4,7, which is way too long for the hypotenuse. If the sides were 3,4,5, then yes, but 7 is too big. So B is incorrect.Next is option C: 8, 15, 17. Let's check. The hypotenuse should be 17. Calculating 8² + 15²: 64 + 225. 64 + 225 is 289. Now, 17 squared is 289. That works! Because 8² + 15² = 17². So, that satisfies the Pythagorean theorem. So, option C is a right-angled triangle. Hmm, that seems straightforward. Let me just confirm: 8*8=64, 15*15=225. 64+225=289. 17*17 is 289. Yep, that's right. So C is a valid set.Then, option D: 1.6, 2, 2.5. Let's check here. The longest side is 2.5. So, we need to see if 1.6² + 2² equals 2.5². Let's compute each term. 1.6 squared: 1.6 * 1.6. Well, 1.6*1 is 1.6, 1.6*0.6 is 0.96, so total is 2.56. 2 squared is 4. Adding those together: 2.56 + 4 = 6.56. Now, 2.5 squared is 6.25. Wait, 6.56 vs. 6.25. That's not equal. So, D doesn't satisfy the theorem. Hmm, but maybe I miscalculated. Let me check again. 1.6 squared: 1.6*1.6. Breaking it down: (1 + 0.6)^2 = 1^2 + 2*1*0.6 + 0.6^2 = 1 + 1.2 + 0.36 = 2.56. Yes, that's right. 2 squared is 4. Sum is 2.56 + 4 = 6.56. 2.5 squared is indeed 6.25. So, 6.56 ≠ 6.25. Therefore, D is not a right triangle. Wait a second, but sometimes there might be rounding errors if decimals are involved, but in this case, the numbers are exact. 1.6 is 8/5, 2.5 is 5/2. Maybe converting them to fractions would make it clearer. Let me try that. 1.6 is 8/5, so squared is (8/5)^2 = 64/25. 2 is 2/1, squared is 4. 2.5 is 5/2, squared is 25/4. So, check if 64/25 + 4 equals 25/4. Convert 4 to 100/25. So 64/25 + 100/25 = 164/25. Now, 164/25 is 6.56. 25/4 is 6.25. Still, they are not equal. So, even in fractions, they don't add up. So D is not valid. Therefore, among all the options, only option C satisfies the Pythagorean theorem. So the answer must be C. Let me just recap to make sure I didn't miss anything. A: 7,12,13 – sum of squares of the two smaller sides is larger than the hypotenuse squared? Wait, 193 vs. 169. So 193 > 169, which would mean that the triangle is acute, but since it doesn't satisfy the equation, it's not right-angled. B: 3,4,7 – sum of squares 25 vs. 49, which is less, so it would be obtuse, but again, not right-angled. C: 8,15,17 – sum of squares 289 vs. hypotenuse squared 289. Perfect. D: 1.6,2,2.5 – sum of squares 6.56 vs. 6.25. Doesn't match. So, definitely C is the correct answer. Final Answerboxed{C}</think>To determine which set of numbers can be the lengths of the sides of a right-angled triangle, we use the Pythagorean theorem, which states that for a right-angled triangle with sides (a), (b), and hypotenuse (c), the equation (a^2 + b^2 = c^2) must hold.Option A: 7, 12, 13- The longest side is 13.- Check: (7^2 + 12^2 = 49 + 144 = 193)- (13^2 = 169)- Since (193 neq 169), option A is not a right-angled triangle.Option B: 3, 4, 7- The longest side is 7.- Check: (3^2 + 4^2 = 9 + 16 = 25)- (7^2 = 49)- Since (25 neq 49), option B is not a right-angled triangle.Option C: 8, 15, 17- The longest side is 17.- Check: (8^2 + 15^2 = 64 + 225 = 289)- (17^2 = 289)- Since (289 = 289), option C is a right-angled triangle.Option D: 1.6, 2, 2.5- The longest side is 2.5.- Check: (1.6^2 + 2^2 = 2.56 + 4 = 6.56)- (2.5^2 = 6.25)- Since (6.56 neq 6.25), option D is not a right-angled triangle.Thus, the correct answer is (boxed{C}).]

🔑:A. 7^2 + 12^2 neq 13^2, so it cannot be the lengths of the sides of a right-angled triangle; B. 3^2 + 4^2 neq 7^2, so it cannot be the lengths of the sides of a right-angled triangle; C. 8^2 + 15^2 = 17^2, so it can be the lengths of the sides of a right-angled triangle; D. 1.6^2 + 2^2 neq 2.5^2, so it cannot be the lengths of the sides of a right-angled triangle; Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to find the maximum value of the function y = sin x sin 2x where x is an acute angle. Hmm, acute angle means x is between 0 and 90 degrees, or 0 to π/2 radians. Since trigonometric functions are involved, maybe using calculus or trigonometric identities will help here. Let me think about different approaches.First, let's recall some trigonometric identities. The product of sines can be converted into a sum using the identity: sin A sin B = [cos(A - B) - cos(A + B)] / 2. Let me apply that here.So, y = sin x sin 2x. Let A = x and B = 2x. Then according to the identity:y = [cos(x - 2x) - cos(x + 2x)] / 2= [cos(-x) - cos(3x)] / 2But cos(-x) is the same as cos x, so this simplifies to:y = [cos x - cos 3x] / 2Hmm, not sure if this helps directly. Alternatively, maybe express sin 2x as 2 sin x cos x. Let's try that.y = sin x * 2 sin x cos x = 2 sin²x cos xSo, y = 2 sin²x cos x. Now, maybe substitution would help here. Let me set t = sin x. Then, since cos x = sqrt(1 - t²), because x is acute, so cos x is positive. Then, y becomes 2 t² sqrt(1 - t²). But dealing with square roots might complicate things. Alternatively, maybe use calculus to find the maximum.Yes, calculus approach. Let's take the derivative of y with respect to x and set it to zero to find critical points.Given y = sin x sin 2x. Let's compute dy/dx.First, note that dy/dx = derivative of sin x sin 2x. Use product rule:dy/dx = cos x sin 2x + sin x * derivative of sin 2x= cos x sin 2x + sin x * 2 cos 2x= cos x sin 2x + 2 sin x cos 2xSet this equal to zero for critical points:cos x sin 2x + 2 sin x cos 2x = 0Hmm, let's simplify this equation. Let's express sin 2x as 2 sin x cos x and cos 2x in terms of cos²x or sin²x.First, substitute sin 2x = 2 sin x cos x:cos x * 2 sin x cos x + 2 sin x cos 2x = 0= 2 sin x cos²x + 2 sin x cos 2x = 0Factor out 2 sin x:2 sin x (cos²x + cos 2x) = 0Since x is acute, sin x ≠ 0 (because x is between 0 and π/2, so sin x > 0). Therefore, the equation reduces to:cos²x + cos 2x = 0Now, express cos 2x using the double-angle identity. Let's recall that cos 2x = 2 cos²x - 1. Substitute that in:cos²x + 2 cos²x - 1 = 03 cos²x - 1 = 03 cos²x = 1cos²x = 1/3cos x = ±√(1/3)But since x is acute, cos x is positive, so cos x = √(1/3) = 1/√3. Therefore, x = arccos(1/√3). Let's find sin x then. Since cos x = 1/√3, sin x = sqrt(1 - 1/3) = sqrt(2/3) = √6 / 3.Now, substitute back into y = sin x sin 2x. First, compute sin 2x = 2 sin x cos x = 2*(√6/3)*(1/√3) = 2*(√6)/(3√3) = 2*(√6/√3)/3 = 2*(√2)/3 = 2√2 / 3.Therefore, y = sin x sin 2x = (√6 / 3) * (2√2 / 3) = (2 * √12) / 9 = 2*(2√3)/9 = 4√3 / 9.Wait, let me check that computation again:First, sin x = √(2/3) = √6 / 3, because √(2/3) is equal to (√6)/3. Yes.sin 2x = 2 sin x cos x = 2*(√6/3)*(1/√3) = 2*(√6)/(3√3) = 2*(√6)/(3√3) = 2/(3) * √(6/3) = 2/3 * √2. Because 6/3 = 2. So, sin 2x = 2√2 / 3.Therefore, y = sin x sin 2x = (√6 / 3)*(2√2 / 3) = (2 * √6 * √2) / 9 = 2 * √12 / 9 = 2*(2√3)/9 = 4√3 / 9. So, 4√3 over 9.But let me confirm if this is indeed the maximum. Alternatively, maybe there is a smarter trigonometric identity approach.Alternatively, since y = sin x sin 2x, and using the identity sin A sin B = [cos(A - B) - cos(A + B)] / 2, so:y = [cos(x - 2x) - cos(x + 2x)] / 2 = [cos(-x) - cos 3x] / 2 = [cos x - cos 3x]/2.So, y = (cos x - cos 3x)/2. Now, maybe this is easier to maximize.But the maximum of (cos x - cos 3x)/2. Let's consider this.Alternatively, perhaps express this as a single cosine function. But not sure.Alternatively, take the derivative of this expression. Let's see:dy/dx = (-sin x + 3 sin 3x)/2. Setting this equal to zero:(-sin x + 3 sin 3x)/2 = 0 => -sin x + 3 sin 3x = 0 => 3 sin 3x = sin x.Hmm, this seems similar to the previous equation. Let's check:From the derivative, we had to set -sin x + 3 sin 3x = 0 => 3 sin 3x = sin x. Let me use sin 3x = 3 sin x - 4 sin³x. So:3*(3 sin x - 4 sin³x) = sin x9 sin x - 12 sin³x = sin x9 sin x - sin x - 12 sin³x = 08 sin x - 12 sin³x = 04 sin x (2 - 3 sin²x) = 0Again, since x is acute, sin x ≠ 0, so 2 - 3 sin²x = 0 => sin²x = 2/3 => sin x = √(2/3) = √6 / 3, which is the same result as before. So, the critical point is the same. Therefore, substituting back into y gives 4√3 / 9. So, that seems consistent.Alternatively, perhaps using substitution. Let me consider t = x, and since x is acute, t is between 0 and π/2. But maybe using substitution for the function.Alternatively, maybe using AM-GM inequality? Not sure. The function is y = 2 sin²x cos x. Let me set t = sin x. Then, cos x = sqrt(1 - t²), so y = 2 t² sqrt(1 - t²). To maximize this for t in (0, 1). Let's set f(t) = 2 t² sqrt(1 - t²). Take derivative with respect to t:f'(t) = 2 [2t sqrt(1 - t²) + t² * ( - t / sqrt(1 - t²)) ]= 2 [ 2t sqrt(1 - t²) - t³ / sqrt(1 - t²) ]= 2 [ (2t(1 - t²) - t³) / sqrt(1 - t²) ]= 2 [ (2t - 2t³ - t³) / sqrt(1 - t²) ]= 2 [ (2t - 3t³) / sqrt(1 - t²) ]Set derivative equal to zero:2t - 3t³ = 0 => t(2 - 3t²) = 0Again, t ≠ 0, so 2 - 3t² = 0 => t² = 2/3 => t = √(2/3) = √6 / 3. Same as before. Therefore, substituting back, we get y = 2*(2/3)*sqrt(1 - 2/3) = 2*(2/3)*sqrt(1/3) = (4/3)*(1/√3) = 4/(3√3) = 4√3 / 9. Same result.Therefore, regardless of the method, the maximum value is 4√3 / 9.Alternatively, verify by testing endpoints. Since x is acute, between 0 and π/2. At x = 0, sin x = 0, so y = 0. At x = π/2, sin x = 1, sin 2x = sin π = 0, so y = 0. Therefore, the maximum must be somewhere in between, which we found as 4√3 / 9.Alternatively, let's compute the second derivative to confirm it's a maximum. But maybe not necessary here, since we have only one critical point and the function is smooth, and endpoints are zero.Alternatively, check a sample value. Let's take x = π/4 (45 degrees). Then sin x = √2/2, sin 2x = sin(π/2) = 1. So y = (√2/2)*1 = √2/2 ≈ 0.707. 4√3 / 9 is approximately 4*1.732 / 9 ≈ 6.928 / 9 ≈ 0.769, which is larger than √2/2 ≈ 0.707, so that seems plausible.Therefore, the maximum value is 4√3/9.</think>6、Let x be an acute angle, then the maximum value of the function y=sin x sin 2 x is boxed{dfrac{4sqrt{3}}{9}},<think>Okay, so I need to find the maximum value of the function y = sin x times sin 2x, where x is an acute angle. That means x is between 0 and 90 degrees, or 0 to π/2 radians. Alright, let's start by understanding the function.First, I know that sin 2x is a double-angle identity. Maybe I can rewrite that to make the equation easier to handle. The double-angle formula for sine is sin 2x = 2 sin x cos x. So substituting that into the original equation, y = sin x * 2 sin x cos x. That simplifies to y = 2 sin²x cos x. Hmm, so now it's 2 times sin squared x times cos x. That might be easier to work with.Now, I need to find the maximum value of this function. Since x is between 0 and π/2, both sin x and cos x are positive here, so the function is positive in this interval. To find the maximum, I can use calculus, taking the derivative and setting it to zero to find critical points.Let me denote y = 2 sin²x cos x. To find the derivative dy/dx, I need to apply the product rule. Alternatively, since it's a product of three functions? Wait, no, sin²x is (sin x)^2, so actually, it's 2 * [ (sin x)^2 * cos x ]. So maybe use the product rule on (sin x)^2 and cos x.Let me recall the product rule: if you have two functions u and v, then the derivative of u*v is u’v + uv’. Here, u = (sin x)^2 and v = cos x. So dy/dx = 2 [ u’v + uv’ ].First, compute u’ where u = (sin x)^2. The derivative of that is 2 sin x cos x. And v’ is the derivative of cos x, which is -sin x. So putting it all together:dy/dx = 2 [ (2 sin x cos x)(cos x) + (sin x)^2 (-sin x) ]= 2 [ 2 sin x cos²x - sin³x ]= 2 sin x [ 2 cos²x - sin²x ]So dy/dx = 2 sin x (2 cos²x - sin²x). To find critical points, set dy/dx = 0.Since x is between 0 and π/2, sin x is not zero (except at x=0, which is a boundary point). Therefore, the critical points occur when 2 cos²x - sin²x = 0.So 2 cos²x = sin²x. Let's write that in terms of sin or cos. Maybe use the identity sin²x + cos²x = 1. Let's solve for cos²x:2 cos²x = sin²x2 cos²x = 1 - cos²x2 cos²x + cos²x = 13 cos²x = 1cos²x = 1/3cos x = ±1/√3. But since x is acute, cos x is positive, so cos x = 1/√3. Therefore, x = arccos(1/√3). Let's find sin x here. Since sin²x = 1 - cos²x = 1 - 1/3 = 2/3, so sin x = √(2/3) = √6 / 3.Now, substitute x back into the original function y = sin x sin 2x. Let's compute sin x and sin 2x.First, sin x = √6 / 3. Then sin 2x = 2 sin x cos x = 2*(√6/3)*(1/√3) = 2*(√6)/(3*√3) = 2*(√6/√3)/3 = 2*(√(6/3))/3 = 2*(√2)/3.So y = (√6 / 3) * (2√2 / 3) = (√6 * 2√2) / 9 = 2√12 / 9. Simplify √12: √12 = 2√3, so y = 2*(2√3)/9 = 4√3/9.Wait, but let me check that again. Let me compute √6 * √2: √6 * √2 = √(12) = 2√3. Therefore, 2√12 is 2*2√3 = 4√3. But the denominator is 9, so y = 4√3 / 9.But let me double-check the calculation step by step to make sure I didn't make a mistake:Compute sin x sin 2x:sin x = √6/3sin 2x = 2 sin x cos x = 2*(√6/3)*(1/√3) = 2*(√6)/(3*√3) Multiply numerator and denominator:√6/√3 = √(6/3) = √2. So 2*√2 /3.Therefore, sin 2x = 2√2 /3.Therefore, y = (√6/3)*(2√2/3) = (√6 * 2√2)/(3*3) = (2√12)/9.√12 = 2√3, so √12 = 2√3. So 2*2√3 = 4√3. Therefore, y = 4√3 /9. So that's correct.But let me confirm if this is indeed the maximum. Since the function is continuous on [0, π/2], and differentiable in (0, π/2), the maximum must occur either at a critical point or at the endpoints. Let's check the value at x=0, x=π/2, and the critical point.At x=0: sin 0 = 0, so y=0.At x=π/2: sin(π/2) =1, sin(2*(π/2))=sin(π)=0, so y=0.Therefore, the maximum must be at the critical point x= arccos(1/√3), and the maximum value is 4√3/9.Alternatively, another way to approach this problem is by using trigonometric identities to rewrite the product sin x sin 2x. Maybe using a product-to-sum formula.Recall that sin A sin B = [cos(A - B) - cos(A + B)] / 2.So let's apply that here with A = x and B = 2x:sin x sin 2x = [cos(x - 2x) - cos(x + 2x)] / 2 = [cos(-x) - cos 3x] / 2. Since cos(-x) = cos x, this simplifies to [cos x - cos 3x]/2. Therefore, y = [cos x - cos 3x]/2.So y = (1/2)(cos x - cos 3x). Now, to find the maximum of this expression. Maybe take the derivative of this form? Or perhaps find another identity.Alternatively, maybe express cos 3x in terms of cos x. Using the triple-angle identity: cos 3x = 4 cos³x - 3 cos x. Substituting that in:y = (1/2)(cos x - (4 cos³x - 3 cos x)) = (1/2)(cos x -4 cos³x +3 cos x) = (1/2)(4 cos x -4 cos³x) = (1/2)(4 cos x (1 - cos²x)) = (1/2)(4 cos x sin²x) = 2 cos x sin²x, which is the same as the expression we had earlier. So this doesn't really help us further.Alternatively, maybe use substitution. Let me set t = cos x. Then sin²x = 1 - t². Therefore, y = 2 t (1 - t²). So y = 2t - 2t³. Then we can find the maximum of this cubic function in t. Since x is acute, t = cos x ranges from 0 to 1. So t ∈ (0,1).Take derivative dy/dt = 2 - 6t². Set to zero: 2 -6t²=0 => 6t²=2 => t²=1/3 => t=1/√3. Which is the same critical point as before. Therefore, maximum at t=1/√3. Then y=2*(1/√3) -2*(1/√3)^3 = 2/√3 - 2/(3√3) = (6/3√3 - 2/3√3) = 4/(3√3) = 4√3/9. Which matches the previous result. So that's another way to get there.Therefore, regardless of the method, the maximum value is 4√3/9. So that's the answer.Just to confirm, maybe plug in some values. For example, when x=30 degrees (π/6):sin(30°)=1/2, sin(60°)=√3/2, so y=(1/2)(√3/2)=√3/4 ≈0.433.At x=45 degrees (π/4):sin(45°)=√2/2, sin(90°)=1, so y=(√2/2)(1)=√2/2≈0.707. Wait, but 4√3/9 is approximately 4*1.732/9≈6.928/9≈0.769. Which is higher than 0.707. So that's possible.Wait, but wait, if x= arccos(1/√3), let's compute that angle. cos x=1/√3≈0.577, so x≈54.7 degrees. Let's compute sin x≈sin(54.7°)=sqrt(2/3)≈0.816, sin 2x≈sin(109.4°)=sin(180-70.6°)=sin(70.6°)≈0.943. Then y≈0.816*0.943≈0.77, which is approximately 4√3/9≈0.7698. So that's correct.So indeed, the maximum is higher than at 45 degrees, which is consistent.Therefore, the maximum value is 4√3/9.Final AnswerThe maximum value of the function is boxed{dfrac{4sqrt{3}}{9}}.</think>6、Let x be an acute angle, then the maximum value of the function y=sin x sin 2 x is qquadTo find the maximum value of the function ( y = sin x sin 2x ) where ( x ) is an acute angle (i.e., ( 0 < x < frac{pi}{2} )), we start by using the double-angle identity for sine: ( sin 2x = 2 sin x cos x ). This transforms the function into:[y = sin x cdot 2 sin x cos x = 2 sin^2 x cos x]To find the critical points, we take the derivative of ( y ) with respect to ( x ):[frac{dy}{dx} = 2 left[ 2 sin x cos x cdot cos x + sin^2 x cdot (-sin x) right] = 2 sin x (2 cos^2 x - sin^2 x)]Setting the derivative equal to zero, we solve for ( x ):[2 cos^2 x - sin^2 x = 0 implies 2 cos^2 x = sin^2 x implies 2 cos^2 x = 1 - cos^2 x implies 3 cos^2 x = 1 implies cos x = frac{1}{sqrt{3}}]Since ( x ) is acute, ( cos x = frac{1}{sqrt{3}} ) and ( sin x = sqrt{1 - cos^2 x} = sqrt{frac{2}{3}} = frac{sqrt{6}}{3} ).Substituting these values back into the original function:[y = 2 left( frac{sqrt{6}}{3} right)^2 left( frac{1}{sqrt{3}} right) = 2 cdot frac{6}{9} cdot frac{1}{sqrt{3}} = 2 cdot frac{2}{3} cdot frac{1}{sqrt{3}} = frac{4}{3sqrt{3}} = frac{4sqrt{3}}{9}]Checking the endpoints ( x = 0 ) and ( x = frac{pi}{2} ) confirms that the function value is zero at both, thus the maximum value occurs at the critical point.The maximum value of the function is (boxed{dfrac{4sqrt{3}}{9}}).]

🔑:Answer: frac{4 sqrt{3}}{9}.Solution: Given y=2 sin ^{2} x cos x,we get y^{2}=4 sin ^{4} x cos ^{2} x=2left(1-cos ^{2} xright)left(1-cos ^{2} xright) cdot 2 cos ^{2} xleq 2left(frac{left(1-cos ^{2} xright)+left(1-cos ^{2} xright)+2 cos ^{2} x}{3}right)^{3}=2 cdotleft(frac{2}{3}right)^{3}=frac{16}{27} text {, }Therefore, y leq frac{4 sqrt{3}}{9}, with equality when cos ^{2} x=frac{1}{3}.