🧠:[<think>Okay, so I need to figure out how many six-digit numbers are "simply cool." The definition says an odd six-digit number that uses only prime digits, and no two identical digits are next to each other. Let me break this down step by step.First, let me recall which digits are prime numbers. The single-digit primes are 2, 3, 5, and 7. So the digits allowed are 2, 3, 5, 7. Wait, but the number has to be odd. So the last digit must be an odd digit. The prime digits that are odd are 3, 5, and 7. So the last digit can only be 3, 5, or 7. The other digits (positions 1 to 5) can be any of the prime digits (2, 3, 5, 7), but with the restriction that no two identical digits are adjacent.So, this seems like a permutation problem with restrictions. Let me structure it as such.The six-digit number: D1 D2 D3 D4 D5 D6Constraints:1. D1 cannot be 0, but since all prime digits are non-zero (2,3,5,7), this is already satisfied.2. D6 must be 3, 5, or 7 (to be odd).3. All digits D1-D6 must be in {2,3,5,7}.4. For i from 1 to 5: D_i ≠ D_{i+1}So, the problem reduces to counting the number of sequences of length 6 where:- The first five digits can be 2,3,5,7 with no two consecutive digits the same.- The sixth digit must be 3,5,7 and not equal to the fifth digit.This seems like a problem that can be approached using recursion or combinatorics with multiplication principle. Let me try the multiplication principle approach.Let me consider the number of choices for each digit, considering the constraints.Starting with D1: Since it's the first digit, there are 4 choices (2,3,5,7).For D2: It can be any prime digit except D1. So 3 choices.For D3: Any prime digit except D2. 3 choices.Similarly, D4: 3 choices (not D3)D5: 3 choices (not D4)D6: Must be 3,5,7 (since it's odd) and not equal to D5.So here, the number of choices for D6 depends on what D5 was. If D5 is a 2, then D6 can be 3,5,7 (3 choices). If D5 is 3,5, or 7, then D6 has 2 choices (excluding D5).Therefore, the total number of such numbers would be equal to the sum over all possible D5 values of (number of sequences from D1 to D5 where D5 is a specific digit) multiplied by (number of choices for D6 given D5).Alternatively, maybe we can model this as a recurrence relation where we track the number of sequences ending with each digit at each position.Let me define:For each position i (from 1 to 6), and for each digit d in {2,3,5,7}, let’s define C(i, d) as the number of sequences of length i ending with digit d, satisfying the constraints.Then, the total number of sequences of length 6 would be the sum over d in {3,5,7} of C(6, d).Now, let's build this up step by step.Starting with position 1 (D1):C(1, 2) = 1C(1, 3) = 1C(1, 5) = 1C(1, 7) = 1But wait, since there are 4 choices for D1, each digit is possible, so each C(1, d) = 1? Wait, actually, no. If we are counting the number of sequences, then for each digit, there is 1 way to have a sequence of length 1 ending with that digit. So total sequences of length 1: 4.But when we get to position 6, we only sum over the odd digits. So this seems manageable.For position 2:For each digit d, C(2, d) = sum of C(1, d') for all d' ≠ d. Since the previous digit can't be the same as current.Therefore, for each d in {2,3,5,7}:C(2, d) = C(1, 2) + C(1, 3) + C(1, 5) + C(1, 7) - C(1, d)But since all C(1, d) = 1, this becomes 4 - 1 = 3. So each C(2, d) = 3.Therefore, total sequences of length 2: 4 * 3 = 12.Similarly, for position 3:For each d, C(3, d) = sum of C(2, d') for all d' ≠ d. Since each C(2, d') = 3, sum over d' ≠ d is 3 * 3 = 9. Wait, let me check:Total sum over all d' of C(2, d') is 4 * 3 = 12. Then subtract C(2, d) to get sum over d' ≠ d: 12 - 3 = 9. So yes, C(3, d) = 9 for each d.Similarly, C(4, d) = sum over d' ≠ d of C(3, d') = 12 - 9 = 3? Wait, wait, no. Wait, if each C(3, d') = 9, then sum over all d' ≠ d would be 9 * 3 = 27. Wait, but that can't be. Wait, no, maybe my approach is wrong here.Wait, let's clarify. For position 3:C(3, d) = sum of C(2, d') for all d' ≠ d. Since in position 2, each C(2, d') = 3. So for each d, how many terms are in the sum? 3 terms (since d' can be 4 digits, excluding d). Therefore, C(3, d) = 3 + 3 + 3 = 9. Yes, because each of the three other digits contributes 3 sequences. So yes, C(3, d) = 9 for each d.Similarly, for position 4:C(4, d) = sum of C(3, d') for d' ≠ d. Each C(3, d') is 9, so sum over three d' gives 9*3 = 27. Therefore, C(4, d) = 27 for each d.Similarly, position 5:C(5, d) = sum of C(4, d') for d' ≠ d. Each C(4, d') = 27, so sum over three d' gives 27*3 = 81. So C(5, d) = 81 for each d.Now, moving to position 6. Here, D6 must be an odd digit: 3, 5, or 7. So for each odd digit d in {3,5,7}, C(6, d) = sum of C(5, d') for d' ≠ d. However, note that d' can be any digit except d, but d' can be 2,3,5,7 (but d is in {3,5,7} here). So for each odd digit d at position 6, the previous digit d' can be 2, or the other two odd digits. So sum over d' in {2,3,5,7} excluding d.Therefore, for each d in {3,5,7}:C(6, d) = sum_{d' ≠ d} C(5, d') = sum_{d' ≠ d} 81 = 81 * 3 = 243. Wait, but hold on. Wait, C(5, d') for d' ≠ d. Since each C(5, d') is 81 regardless of d', then sum over three d' (since d is one of the four digits, so excluding d, there are 3 digits left). Wait, but the digits are four: 2,3,5,7. If d is 3, then d' can be 2,5,7. If d is 5, d' can be 2,3,7. If d is 7, d' can be 2,3,5. So in all cases, there are three digits d' allowed for each d at position 6.But in position 5, each digit (2,3,5,7) has 81 possibilities. So sum over three digits (excluding d) would be 81 + 81 + 81 = 243. Therefore, each C(6, d) = 243. Since there are three such digits (3,5,7), total number is 243 * 3 = 729.Wait, but this seems too straightforward. Let me check this step again.Wait, if we use the recurrence relation:At position 5, each digit (2,3,5,7) has 81 numbers ending with that digit. Then, for position 6, which must be an odd digit (3,5,7), the number of numbers ending with, say, 3 would be the sum of numbers ending with 2,5,7 at position 5. Similarly for 5 and 7.Therefore, for each odd digit d at position 6:C(6, d) = C(5, 2) + C(5, other odd digits not equal to d)Since C(5,2) = 81, and for the other two odd digits (since d is one of 3,5,7), there are two digits, each with C(5, d') = 81. So total C(6, d) = 81 + 81 + 81 = 243. Wait, but that's including 2 and the other two odd digits. Wait, if d is 3, then allowed d' are 2,5,7. So yes, 2,5,7 each contribute 81. So 3*81=243. Similarly for d=5 and d=7.Therefore, each C(6,3), C(6,5), C(6,7) is 243, so total is 243*3=729.But wait, this would mean the total number of simply cool numbers is 729. But let me verify this with another approach to make sure.Alternative approach: Let's consider the first five digits. Each digit can be 2,3,5,7, with no two consecutive digits the same. The number of such sequences is 4 * 3^4. Because the first digit has 4 choices, each subsequent digit has 3 choices (can't be the same as the previous). So 4 * 3^4 = 4 * 81 = 324.Then, the sixth digit must be 3,5,7, and not equal to the fifth digit. So, depending on what the fifth digit is, the number of choices for the sixth digit changes.If the fifth digit is 2, then the sixth digit can be 3,5,7 (3 choices). If the fifth digit is 3,5, or 7, then the sixth digit can be the other two odd digits (2 choices).Therefore, total number of sequences is sum over all possible fifth digits:If fifth digit is 2: number of sequences where fifth digit is 2 multiplied by 3.If fifth digit is 3,5,7: number of sequences where fifth digit is each of these multiplied by 2 each.So let's compute the number of sequences where the fifth digit is 2, and those where it's 3,5,7.First, compute the number of sequences of length 5 where the fifth digit is 2. Let's denote this as N5(2).Similarly, N5(3), N5(5), N5(7).Given that each position from 1 to 5, the digits are 2,3,5,7 with no two consecutive same digits.This is similar to the earlier recurrence. Let me compute N5(2), N5(3), N5(5), N5(7).From the previous recurrence:At position 1: N1(d) = 1 for d in {2,3,5,7}At position 2: N2(d) = sum of N1(d') for d' ≠ d. So N2(d) = 3 for all d.Similarly, position 3: N3(d) = sum of N2(d') for d' ≠ d = 3*3 = 9 for each d.Position 4: N4(d) = sum of N3(d') for d' ≠ d = 3*9 = 27 for each d.Position 5: N5(d) = sum of N4(d') for d' ≠ d = 3*27 = 81 for each d.Wait, so N5(2) = 81, N5(3)=81, N5(5)=81, N5(7)=81.But that can't be right. Wait, according to this recurrence, each position's count is the same for each digit. So for position 5, each digit has 81 sequences ending with it. So total sequences of length 5: 4 * 81 = 324, which matches the earlier 4*3^4=324.But then, going back to the sixth digit:If the fifth digit is 2 (which occurs in 81 sequences), then sixth digit has 3 choices (3,5,7). So these contribute 81 * 3.If the fifth digit is 3,5,7 (each occurs in 81 sequences), then sixth digit has 2 choices each. So each contributes 81 * 2, and there are three such digits. Wait, no. Wait, fifth digit can be 3,5,7: each of these has 81 sequences, and for each, sixth digit has 2 choices. So total for these three digits: 3 * 81 * 2.Therefore, total sequences:81 * 3 + 3 * 81 * 2 = 81 * (3 + 6) = 81 * 9 = 729.Which matches the previous result. So total simply cool numbers are 729.But let me check again for errors. Maybe there's a miscalculation here.Wait, the fifth digit can be 2,3,5,7. If it's 2, sixth digit has 3 options. If it's 3, fifth digit is 3, sixth digit can be 5,7. Similarly for 5 and 7.Therefore, the total number is:Number of sequences where fifth digit is 2: 81. Each can be extended by 3,5,7. So 81 * 3.Number of sequences where fifth digit is 3: 81. Each can be extended by 5,7. So 81 * 2.Same for fifth digit 5: 81 * 2.Same for fifth digit 7: 81 * 2.Wait, wait a second. Wait, fifth digit is 2: 81 * 3.Fifth digits 3,5,7: each 81 * 2. So total:81 * 3 + 81 * 2 + 81 * 2 + 81 * 2. But that would be 81*(3 + 2 + 2 + 2) = 81*9=729.But wait, but fifth digit cannot be all four digits. Wait, fifth digit is one of 2,3,5,7. So when you split into fifth digit is 2 vs 3,5,7, then:If fifth digit is 2: 81 sequences, sixth digit: 3 options.If fifth digit is 3,5,7: each of the three digits has 81 sequences, sixth digit: 2 options each.Therefore, total is 81*3 + 3*81*2 = 81*3 + 81*6 = 81*9=729. Same as before.So both approaches agree. Therefore, the total number is 729.But let me check with a smaller case to ensure that the logic holds.Suppose we have a 2-digit simply cool number. Then, it must be odd, so the second digit is 3,5,7. First digit is prime digit (2,3,5,7) and not equal to the second digit.So total number: For each second digit (3,5,7), first digit can be 2,3,5,7 except the second digit. So 3 choices for first digit (if second digit is 3, then first digit can be 2,5,7). So for each second digit (3 choices), 3 first digits. So total 3*3=9.But according to our formula: For 2-digit numbers:First digit: 4 choices. Second digit: 3 choices (must be 3,5,7 and not equal to first digit). So if first digit is 2: second digit has 3 choices. If first digit is 3,5,7: second digit has 2 choices. So total number is:Number of first digits which are 2: 1 (i.e., digit 2). Second digits: 3.Number of first digits which are 3,5,7: 3. For each, second digits: 2.Total: 1*3 + 3*2 = 3 + 6 = 9. Which matches.Using our previous recurrence:For position 1: C(1, d) =1 for each d.For position 2: C(2, d) where d is 3,5,7.C(2, d) = sum of C(1, d') for d' ≠ d.C(1, d') =1 for each d', so sum is 3 if d is 3,5,7 (since d' can be 2 and the other two odd digits). Wait, for d=3, sum over d'=2,5,7. So 3 terms. Each C(1, d')=1. So C(2,3)=3. Similarly for 5 and 7. So total C(2,3)+C(2,5)+C(2,7)=3+3+3=9. Which matches.Therefore, the recurrence works for n=2.Another test: n=3.Total simply cool numbers with 3 digits. Let's compute using both methods.First method:First digit: 4 choices.Second digit: 3 choices (not equal to first).Third digit: must be 3,5,7 and not equal to second digit.So:Case 1: Third digit is 3. Then second digit can't be 3. So first digit can be 2,5,7 (if second digit is 3, but second digit can't be 3 if third digit is 3). Wait, no. Wait, third digit is 3. Then second digit can be 2,5,7.Similarly, first digit can be anything except second digit.Alternatively, let's compute using the recurrence.C(3,3) = sum of C(2, d') for d' ≠3. C(2, d') for d'=2,5,7. Each C(2,2)=3 (wait, no, in the previous recurrence, at position 2, all C(2,d)=3, regardless of d. So for position 3, C(3,3)= C(2,2)+C(2,5)+C(2,7)=3+3+3=9. Similarly, C(3,5)=9, C(3,7)=9. Total simply cool numbers: 9+9+9=27.Alternatively, using direct counting:First digit: 4 choices.Second digit: 3 choices (not first).Third digit: if second digit is 2, then third digit can be 3,5,7 (3 choices). If second digit is 3,5,7, then third digit can be the other two odd digits and 2? Wait, no. Wait, the third digit must be odd (3,5,7) and not equal to the second digit.So:Third digit is 3,5,7. So for each third digit, the second digit must not be equal to it.But second digit can be 2,3,5,7, not equal to first digit.This is getting complex, but let's see:Total third digits: 3 choices (3,5,7). For each third digit d3:Number of sequences where third digit is d3 is equal to number of sequences where second digit is not d3, and first digit is not equal to second digit.So for each d3, the number is:Sum over d2 ≠ d3 of [number of d1 ≠ d2].So for each d3:If d2 is 2: then d1 can be 3,5,7 (3 choices)If d2 is 3,5,7 but ≠ d3: for each such d2 (there are 2 of them), d1 can be 2,3,5,7 except d2, so 3 choices.Therefore, for each d3:Number of sequences = (number of d2=2 sequences) + (number of d2≠2,d2≠d3 sequences)= [1 (d2=2) * 3 (d1)] + [2 (d2= other two odd digits) * 3 (d1)]= 3 + 6 = 9 per d3.Since there are 3 choices for d3, total 27. Which matches the recurrence result.Therefore, the recurrence seems valid.Therefore, returning to the original problem, the total number is 729. But let me just check once more.Wait, if we have six-digit numbers, with digits 1-5 being any prime digits with no two consecutive same, and digit 6 being an odd prime digit not equal to digit 5.The count is:First five digits: 4 * 3^4 = 324.For the sixth digit:If fifth digit is 2: 3 choices.If fifth digit is 3,5,7: 2 choices.How many sequences have fifth digit 2? Let's compute.Number of sequences where fifth digit is 2: For the first four digits, they form a sequence of four prime digits with no two consecutive same, ending with a digit not equal to 2 (since fifth digit is 2, so fourth digit ≠2).Wait, no. Wait, fifth digit is 2, so fourth digit can be any except 2.Wait, the number of sequences where the fifth digit is 2 is equal to the number of sequences of length 4 (digits 1-4) ending with a digit ≠2, multiplied by 1 (digit 2). Wait, not quite.Wait, to compute the number of sequences of length 5 ending with 2, it's equal to the number of sequences of length 4 ending with digits ≠2, each extended by 2.So N5(2) = sum of N4(d') for d' ≠2.From the recurrence earlier, N4(d') for d' ≠2: There are three digits (3,5,7) each with N4(d')=27. So sum is 3*27=81. Therefore, N5(2)=81, which matches our previous result.Therefore, the number of sequences where fifth digit is 2 is 81, each contributing 3 choices for the sixth digit (3,5,7). So 81*3=243.Similarly, sequences where fifth digit is 3,5,7: each has N5(d)=81. For each of these, the sixth digit can't be d, so 2 choices. So for each of the three digits, 81*2=162. Total for all three: 162*3=486.Total sequences: 243 + 486=729.Therefore, the answer is 729. Hence, the number of simply cool numbers is 729.Final Answerboxed{729}</think>To determine the number of "simply cool" six-digit numbers, we need to consider the following constraints:1. The number must be odd, so the last digit must be 3, 5, or 7.2. All digits must be prime numbers (2, 3, 5, 7).3. No two identical digits can be adjacent.We use a recurrence relation approach to count the number of valid sequences. Let ( C(i, d) ) represent the number of sequences of length ( i ) ending with digit ( d ).1. Initialization (position 1): [ C(1, 2) = 1, quad C(1, 3) = 1, quad C(1, 5) = 1, quad C(1, 7) = 1 ]2. Recurrence relation for positions 2 to 5: For each position ( i ) from 2 to 5, the number of sequences ending with digit ( d ) is the sum of sequences ending with all other digits: [ C(i, d) = sum_{d' neq d} C(i-1, d') ] This results in: - ( C(2, d) = 3 ) for all ( d ) - ( C(3, d) = 9 ) for all ( d ) - ( C(4, d) = 27 ) for all ( d ) - ( C(5, d) = 81 ) for all ( d )3. Position 6 (last digit): The sixth digit must be 3, 5, or 7 and not equal to the fifth digit. For each odd digit ( d ) (3, 5, 7): [ C(6, d) = sum_{d' neq d} C(5, d') ] Since each ( C(5, d') = 81 ), we sum over three digits (excluding ( d )): [ C(6, d) = 81 times 3 = 243 ]4. Total number of "simply cool" numbers: Sum the counts for each valid last digit: [ C(6, 3) + C(6, 5) + C(6, 7) = 243 + 243 + 243 = 729 ]Thus, the number of "simply cool" six-digit numbers is (boxed{729}).,<think>Okay, so I need to figure out how many six-digit numbers are "simply cool." Let me start by breaking down the problem. The number has to be odd, six digits long, made up of prime digits, and no two identical digits next to each other. Alright, let's take this step by step.First, let's recall which digits are prime. The single-digit prime numbers are 2, 3, 5, and 7. So the digits allowed are 2, 3, 5, 7. That's four digits in total. But since the number has to be odd, the last digit must be an odd prime digit. Among the primes we have, the odd ones are 3, 5, 7. So the last digit can be 3, 5, or 7. The first digit, since it's a six-digit number, can't be 0, but since all our digits are 2, 3, 5, 7, 0 isn't an option anyway. So the first digit can be any of the four primes, except we need to make sure that consecutive digits aren't the same.So, to structure this, maybe I can model it as a permutation problem with restrictions. Let's consider each digit position from the first to the sixth. The key constraints are:1. The sixth digit must be 3, 5, or 7 (to be odd).2. No two adjacent digits can be the same.3. All digits must be in {2, 3, 5, 7}.Let me approach this using the multiplication principle. For each position, we need to count the number of choices we have, considering the previous digit.Let's start with the first digit. Since there are no restrictions except that it can't be 0 (which is already satisfied), there are 4 choices: 2, 3, 5, 7.For the second digit, it can't be the same as the first digit. So regardless of what the first digit was, there are 3 choices left.Similarly, the third digit can't be the same as the second, so again 3 choices.Fourth digit: same logic, 3 choices.Fifth digit: 3 choices.Now, the sixth digit must be odd (3, 5, 7) and also different from the fifth digit. So here, the number of choices depends on what the fifth digit was.If the fifth digit is one of the odd primes (3, 5, 7), then the sixth digit has 2 remaining choices. If the fifth digit is 2 (which is even), then the sixth digit can be any of 3, 5, 7, so 3 choices.Therefore, the total number of possibilities isn't just straightforward multiplication because the sixth digit's options depend on the fifth digit. Hmm, so maybe I need to break this into cases based on whether the fifth digit is 2 or an odd prime.Alternatively, perhaps I can model this recursively or use dynamic programming. Let me think.Let’s denote two cases for each position up to the fifth digit:- The current digit is 2.- The current digit is an odd prime (3, 5, 7).But since the fifth digit could be 2 or an odd prime, affecting the sixth digit's choices, perhaps tracking the number of sequences ending with 2 and ending with an odd prime at each position would help.Let’s formalize this:Let’s define:- ( a_n ) as the number of n-digit numbers where the last digit is 2.- ( b_n ) as the number of n-digit numbers where the last digit is an odd prime (3, 5, 7).Then, the total number of n-digit numbers is ( a_n + b_n ).Our goal is to compute ( b_6 ), since the sixth digit must be an odd prime. Wait, no. Wait, actually, the total number of six-digit numbers that satisfy the conditions is equal to the total number of sequences where the sixth digit is 3, 5, or 7 and no two adjacent digits are the same. So maybe we can model this as:At each position ( i ), we can track the number of sequences ending with 2 or ending with an odd prime.For n=1 (first digit):- ( a_1 = 1 ) (only number 2)- ( b_1 = 3 ) (numbers 3, 5, 7)Wait, but actually, the first digit can be any of the four primes. So:Wait, no. Wait, ( a_n ) is the number of n-digit numbers ending with 2, and ( b_n ) is the number ending with an odd prime. So for n=1:- ( a_1 = 1 ) (just 2)- ( b_1 = 3 ) (3, 5, 7)Total numbers for n=1: 4, which is correct.For n=2:To compute ( a_2 ), the number of two-digit numbers ending with 2. The previous digit (first digit) must be an odd prime (since we can't have two 2s in a row). So, the number is ( b_1 times 1 ) (since after an odd prime, we can choose 2). So ( a_2 = b_1 times 1 = 3 times 1 = 3 ).Similarly, ( b_2 ): number of two-digit numbers ending with an odd prime. The previous digit can be either 2 or another odd prime. If the previous digit was 2, then we can choose any of the three odd primes. If the previous digit was an odd prime, we can choose any of the remaining two odd primes (since they can't be the same). Wait, no. Wait, if the previous digit is an odd prime, the next digit can be any odd prime except the one just used. So:Actually, when building ( b_n ), the previous digit can be 2 or an odd prime.If the previous digit is 2, then there are 3 choices for the current digit (3, 5, 7). If the previous digit is an odd prime, then there are 2 choices left (since we can't repeat the same digit). So ( b_n = a_{n-1} times 3 + b_{n-1} times 2 ).Similarly, ( a_n ): To end with 2, the previous digit must be an odd prime (can't be 2). So ( a_n = b_{n-1} times 1 ).So we have the recursive relations:- ( a_n = b_{n-1} )- ( b_n = 3a_{n-1} + 2b_{n-1} )With base cases:- ( a_1 = 1 )- ( b_1 = 3 )Let me verify this with n=2:( a_2 = b_1 = 3 )( b_2 = 3a_1 + 2b_1 = 3*1 + 2*3 = 3 + 6 = 9 )Total numbers for n=2: 3 + 9 = 12. Let's check manually. For two-digit numbers with digits from {2,3,5,7}, no two same digits. First digit: 4 choices, second digit: 3 choices. Total 4*3=12. Correct.So the recursion works here.Similarly, n=3:( a_3 = b_2 = 9 )( b_3 = 3a_2 + 2b_2 = 3*3 + 2*9 = 9 + 18 = 27 )Total for n=3: 9 + 27 = 36. Which is 4*3*3 = 36. Correct.Continuing, but since we need to compute up to n=6, let's compute each step:n=1:- a1=1- b1=3Total=4n=2:- a2= b1=3- b2=3*1 + 2*3=3+6=9Total=12n=3:- a3= b2=9- b3=3*a2 +2*b2=3*3 +2*9=9+18=27Total=36n=4:- a4= b3=27- b4=3*a3 +2*b3=3*9 +2*27=27 +54=81Total=108n=5:- a5= b4=81- b5=3*a4 +2*b4=3*27 +2*81=81 +162=243Total=324n=6:- a6= b5=243- b6=3*a5 +2*b5=3*81 +2*243=243 +486=729Total=972But wait, the problem says that the number must be odd, so the sixth digit must be an odd prime. So actually, the total number of "simply cool" numbers is b6, which is 729. But let me check that.Wait, according to the recursion, b_n counts the number of n-digit numbers ending with an odd prime, and a_n counts those ending with 2. So the total number of six-digit numbers that are simply cool is b6, since they must end with an odd prime (to be odd). So the answer should be 729.But let's verify with smaller n. For example, n=1: numbers are 2,3,5,7. The odd ones are 3,5,7, which is 3, which is b1=3. Correct.For n=2: total numbers are 12. How many are odd? The second digit must be odd. For each first digit (4 choices), second digit (3 choices, but only 3 if first digit is even (2), or 2 if first digit is odd). Wait, maybe alternative count:First digit: 2, then second digit: 3,5,7 (3 choices). So numbers starting with 2 and ending with odd: 3.First digit: 3,5,7 (3 choices), then second digit: 3 primes excluding the first digit. Since first digit is odd, second digit can be 2 or the other two odd primes. Wait, no. Wait, if first digit is 3, second digit can be 2,5,7 (three choices). Wait, but no two same digits. If first digit is 3, second digit can be 2,5,7 (3 choices). Similarly for first digit 5 or 7. So total numbers with first digit odd (3 options) and second digit can be 3 choices (since they can't be the same as the first digit). So total numbers ending with odd: first digit even (2) and second digit odd: 1*3=3. First digit odd and second digit odd: 3*2=6 (since after an odd digit, second digit can be 2 or two other odds). Wait, no. Wait, if first digit is odd (3,5,7), the second digit can be 2, or any of the other two odd digits. So 3 choices for second digit. So total numbers ending with odd digit: 1*3 (first digit 2) + 3*3 (first digit 3,5,7) = 3 +9=12? Wait, but total numbers for n=2 is 12, but how many are odd? All numbers ending with odd digit. Wait, but for n=2, numbers like 22 are invalid, but since digits can't be same, so all numbers are 4*3=12, and numbers ending with odd digits would be first digit can be any, second digit must be odd and different from first digit.Wait, actually, let's compute it manually for n=2. The numbers are:First digit 2:23, 25, 27 (3 numbers, all ending with odd)First digit 3:32, 35, 37 (3 numbers, all ending with 2,5,7. Among these, 35,37 are odd)Wait, wait, no. Wait, if first digit is 3, second digit can be 2,5,7. So numbers: 32 (ends with 2), 35 (ends with 5), 37 (ends with 7). So two of them are odd (35,37). Similarly for first digit 5 and 7.So first digit 3: 2 odd numbersFirst digit 5: 2 odd numbersFirst digit 7: 2 odd numbersFirst digit 2: 3 odd numbersTotal odd numbers for n=2: 3 + 2+2+2=9. But according to our earlier recursion, b2=9. Which matches. But according to the total numbers, there are 12 numbers, 9 ending with odd, 3 ending with even (2). So the recursion correctly counts b2=9. Therefore, in the recursion, b_n is the number of n-digit numbers ending with odd, which is exactly what we need. So for n=6, the answer is b6=729.Wait, but let me check n=3. For n=3, the total numbers are 36. The numbers ending with odd (b3=27) and ending with 2 (a3=9). Let's see if that makes sense. For the third digit to be odd, the second digit can be 2 or another odd. If second digit is 2, then third digit can be 3,5,7 (3 choices). If second digit is odd, third digit can be 2 or two other odds. Wait, no. Wait, the third digit being odd can be preceded by a 2 or an odd digit different from itself. Let's see:If we have a three-digit number ending with an odd digit. The second digit can be 2 or another odd digit. If the second digit is 2, then first digit can be 3,5,7 (3 choices), second digit is 2 (1 choice), third digit is 3,5,7 (3 choices). So 3*1*3=9.If the second digit is an odd digit (different from the third digit), then first digit can be 2, or another odd digit (different from the second digit). Wait, this is getting complicated. Alternatively, since the recursion gives b3=27, and we know total numbers are 36, then a3=9. So numbers ending with 2 are 9, which would be sequences where the third digit is 2. The second digit must be odd (since no two 2s in a row). The first digit can be anything except the second digit. Hmm, this is a bit involved. Maybe it's easier to trust the recursion since it worked for n=1, n=2, and n=3. Let's proceed.Thus, following the recursion:n=1: a1=1, b1=3n=2: a2=3, b2=9n=3: a3=9, b3=27n=4: a4=27, b4=81n=5: a5=81, b5=243n=6: a6=243, b6=729So according to this, the number of six-digit simply cool numbers is 729. But let me just verify with another approach.Alternative approach:The first five digits can be any primes (including 2), with no two adjacent digits the same. The sixth digit must be an odd prime different from the fifth digit.So, for the first five digits, how many possibilities are there? It's similar to arranging five digits with four choices each time, no two same in a row. The number is 4*3*3*3*3=4*3^4=4*81=324.Wait, but that would be if each subsequent digit has 3 choices. But in reality, the first digit has 4 choices, then each next digit has 3 choices (since they can't be the same as the previous one). So for five digits, it's 4*3^4=324.Then, the sixth digit must be an odd prime (3 choices) not equal to the fifth digit. So depending on what the fifth digit is:If the fifth digit is 2, then the sixth digit can be 3,5,7 (3 choices). If the fifth digit is an odd prime, then the sixth digit can be the other two odd primes (2 choices).So to compute the total, we need to know how many five-digit numbers end with 2 and how many end with an odd prime.Wait, this seems similar to the previous recursion. Let me denote:Let’s call the number of five-digit numbers ending with 2 as A, and ending with odd primes as B. Then total five-digit numbers is A + B = 324.Then, the number of six-digit numbers is A*3 (since if the fifth digit is 2, sixth digit can be 3,5,7) + B*2 (since if fifth digit is odd, sixth digit can be two other odds).But from our recursion earlier, A = a5 =81, B = b5=243.Therefore, total six-digit numbers =81*3 +243*2=243 +486=729. Which matches the previous result.Alternatively, if we didn't use recursion, how would we compute A and B?Given that for the five-digit numbers, the number ending with 2 (A) and ending with odd (B). Let's see:For the first digit: 4 choices.Each subsequent digit (positions 2-5) can't be the same as the previous one.We can model this as a Markov chain with two states: last digit is 2 or last digit is odd.At each step, the number of sequences ending with 2 is equal to the number of sequences ending with odd in the previous step (since you can only append a 2 if the previous ended with odd). Similarly, the number of sequences ending with odd is equal to (number of sequences ending with 2)*3 + (number of sequences ending with odd)*2. Wait, actually, that's exactly the recursion we used earlier.So, yeah, this seems consistent. Therefore, the total number is 729.But let's just do one more check with n=3.If we compute manually:For three-digit numbers, first digit: 4 choices. Second digit: 3 choices (not same as first). Third digit: if second digit was 2, third digit can be 3 choices (3,5,7). If second digit was odd, third digit can be 2 or two other odds, so 3 choices (but wait, if second digit was odd, third digit can't be same as second digit. So, if second digit is 3, third digit can be 2,5,7 (3 choices). Similarly, if second digit is 5 or7. So regardless of whether second digit was 2 or odd, third digit has 3 choices. Wait, but in reality:Wait, if second digit is 2: third digit can be 3,5,7 (3 choices).If second digit is odd (3,5,7): third digit can be 2, or the other two odd digits. So 1 + 2 = 3 choices. Therefore, regardless of second digit, third digit has 3 choices. Thus, total three-digit numbers: 4*3*3=36. Which matches.But how many of these end with odd? Let's compute:Case 1: third digit is odd. Then, second digit can be 2 or another odd digit.Number of sequences where third digit is odd:If second digit is 2: first digit can be any of 3,5,7 (3 choices), second digit is 2, third digit is 3,5,7 (3 choices). So 3*1*3=9.If second digit is odd: first digit can be 2 or another odd digit. Let's see:First digit: 4 choices.Second digit: odd, not equal to first digit.Wait, this is getting complicated, but if second digit is odd, then first digit can be 2 or an odd digit different from the second.Let me think. For sequences ending with odd digit:Third digit is odd. Second digit can be 2 or another odd digit.If second digit is 2:First digit: can be 3,5,7 (3 choices). Second digit: 2. Third digit: 3,5,7 (3 choices). So 3*1*3=9.If second digit is odd:First digit: can be 2 or an odd digit different from the second.Second digit: odd, not equal to third digit. Wait, third digit is also odd, and must be different from second digit.Wait, maybe better to compute:Total sequences where third digit is odd: total sequences where third digit is 3,5,7.For each choice of third digit (3 options):The second digit can be 2, or any odd digit except the third digit.So for each third digit:If second digit is 2: first digit can be any of the three other primes (excluding 2). Wait, no, first digit can be any prime except the second digit (which is 2). Since second digit is 2, first digit can be 3,5,7. So 3 choices.If second digit is odd (and not equal to third digit): second digit has 2 choices (since third digit is fixed as an odd, so two other odds). Then first digit can be any prime except the second digit. So first digit has 3 choices (since primes are 2,3,5,7; exclude the second digit).Therefore, for each third digit:Number of sequences ending with that digit:= (number with second digit 2) + (number with second digit odd)= 3 (first digits) *1 (second digit 2) + 3 (first digits) *2 (second digit odd)Wait, no:Wait, for each third digit:If second digit is 2: first digit can be 3,5,7 (3 choices). Second digit is 2. Third digit is fixed. So 3*1=3.If second digit is odd (excluding third digit): second digit has 2 choices. Then first digit can be any prime except second digit. So if second digit is 3 (and third digit is, say,5), first digit can be 2,5,7 (but 5 is excluded because third digit is 5?). Wait, no. Wait, third digit is fixed. Wait, no, the first digit only needs to be different from the second digit. Wait, the third digit is separate. Wait, confusion here.Let me clarify:Suppose we fix the third digit as 3. Then:Second digit can be 2,5,7.If second digit is 2: first digit can be 3,5,7 (but not 2). But since third digit is 3, which is separate, the first digit just needs to be different from the second digit (2). So first digit can be 3,5,7. So 3 choices.If second digit is 5: first digit can be 2,3,7 (since first digit can't be 5). 3 choices.If second digit is 7: first digit can be 2,3,5 (since first digit can't be 7). 3 choices.So total sequences ending with 3:Second digit=2: 3Second digit=5: 3Second digit=7:3Total: 9.Similarly, if third digit is 5 or7, same thing. So for each third digit (3 options), 9 sequences. So total sequences ending with odd:3*9=27. Which matches our recursion result b3=27.So this method works. Hence, trusting the recursion, for n=6, the total is 729.Therefore, the number of "simply cool" numbers is 729.Final Answerboxed{729}</think>To determine the number of "simply cool" six-digit numbers, we need to consider the following constraints:1. The number must be odd, so the last digit must be 3, 5, or 7.2. All digits must be prime numbers (2, 3, 5, 7).3. No two adjacent digits can be the same.We use a recursive approach to count the number of valid sequences ending with 2 or an odd prime at each position. Define:- ( a_n ) as the number of n-digit numbers ending with 2.- ( b_n ) as the number of n-digit numbers ending with an odd prime (3, 5, 7).The recursive relations are:- ( a_n = b_{n-1} ) (since a number ending with 2 must be preceded by an odd prime)- ( b_n = 3a_{n-1} + 2b_{n-1} ) (since a number ending with an odd prime can be preceded by 2 or another odd prime)Base cases:- ( a_1 = 1 ) (only the number 2)- ( b_1 = 3 ) (numbers 3, 5, 7)We compute the values up to ( n = 6 ):[begin{aligned}&n = 1: & a_1 = 1, quad b_1 = 3 &n = 2: & a_2 = b_1 = 3, quad b_2 = 3a_1 + 2b_1 = 3 cdot 1 + 2 cdot 3 = 9 &n = 3: & a_3 = b_2 = 9, quad b_3 = 3a_2 + 2b_2 = 3 cdot 3 + 2 cdot 9 = 27 &n = 4: & a_4 = b_3 = 27, quad b_4 = 3a_3 + 2b_3 = 3 cdot 9 + 2 cdot 27 = 81 &n = 5: & a_5 = b_4 = 81, quad b_5 = 3a_4 + 2b_4 = 3 cdot 27 + 2 cdot 81 = 243 &n = 6: & a_6 = b_5 = 243, quad b_6 = 3a_5 + 2b_5 = 3 cdot 81 + 2 cdot 243 = 729 end{aligned}]Thus, the number of "simply cool" six-digit numbers is ( b_6 ), which is (boxed{729}).]

🔑:Solution. There are four single-digit prime numbers in total - 2, 3, 5, 7. We will place the digits starting from the least significant digit. In the units place, three of them can stand -3, 5, 7. In the tens place, there are also three out of the four (all except the one placed in the units place). In the hundreds place, there are also three out of the four (all except the one placed in the tens place). And so on. As a result, the number of "prime cool" numbers is 3^{2}, i.e., 729.Answer. 729.Recommendations for checking. Only the correct answer - 1 point. All 4 numbers are included in the units place - 3 points. The number 1 is considered a prime number, and the subsequent solution is correct based on this - 5 points.

🧠:[<think>Okay, let's try to figure out how to solve this problem. So, we have a triangle ABC with sides AB = 10 cm, BC = 17 cm, and AC = 21 cm. There's a point M inside this triangle, and the distances from M to sides AC and BC are given as 2 cm and 4 cm, respectively. We need to find the distance from point M to the line AB.Hmm, first, let me recall some concepts. Since M is inside the triangle, and we know its distances to two sides, maybe we can use area relations here? Because the area of the triangle can be split into smaller areas from point M to each side. I remember something about the areas adding up to the total area of the triangle. Let me think.Yes, if a point is inside a triangle, the sum of the areas of the three smaller triangles formed by joining the point to the vertices should equal the area of the original triangle. Each smaller triangle's area can be calculated as (1/2)*base*height, where the base is a side of the original triangle, and the height is the distance from point M to that side.So, in this case, the total area of triangle ABC should be equal to the sum of the areas of triangles AMB, BMC, and AMC. Let's write that down:Area of ABC = Area of AMB + Area of BMC + Area of AMC.We need to find the distance from M to AB, which would be the height corresponding to base AB in triangle AMB. Let's denote this distance as h. The distances from M to AC and BC are given as 2 cm and 4 cm, so those are the heights for triangles AMC and BMC, respectively.First, let's compute the area of triangle ABC. Since we know all three sides, maybe we can use Heron's formula. Heron's formula states that the area of a triangle with sides a, b, c is sqrt[s(s - a)(s - b)(s - c)], where s is the semi-perimeter.Let's calculate the semi-perimeter (s):s = (AB + BC + AC)/2 = (10 + 17 + 21)/2 = (48)/2 = 24 cm.Now, the area is sqrt[24(24 - 10)(24 - 17)(24 - 21)] = sqrt[24*14*7*3].Let me compute that step by step:24*14 = 336336*7 = 23522352*3 = 7056So, the area is sqrt(7056). Hmm, what's the square root of 7056? Let me check:84^2 = 7056, since 80^2 = 6400 and 85^2 = 7225, so between 80 and 85. 84*84 = 7056. So the area is 84 cm².Okay, so the area of triangle ABC is 84 cm².Now, let's express the areas of the three smaller triangles.Area of AMB: This triangle has base AB = 10 cm and height h (the distance from M to AB). So, area is (1/2)*10*h = 5h.Area of BMC: This triangle has base BC = 17 cm and height 4 cm (distance from M to BC). So, area is (1/2)*17*4 = 34 cm².Area of AMC: This triangle has base AC = 21 cm and height 2 cm (distance from M to AC). So, area is (1/2)*21*2 = 21 cm².Adding these up: 5h + 34 + 21 = 5h + 55.But we know that this sum must equal the area of ABC, which is 84 cm². So:5h + 55 = 84Subtract 55 from both sides:5h = 84 - 55 = 29Divide both sides by 5:h = 29/5 = 5.8 cm.Wait, that seems straightforward, but let me verify if I did everything right.First, the semi-perimeter was calculated correctly: 10+17+21=48, divided by 2 is 24. Then Heron's formula gives sqrt(24*14*7*3). Let me check 24*14=336, 7*3=21, 336*21=7056. sqrt(7056)=84. Correct.Then, areas of the smaller triangles:AMB: 1/2 * AB * h = 5h.BMC: 1/2 * BC * 4 = 1/2*17*4=34. Correct.AMC: 1/2 * AC * 2 = 1/2*21*2=21. Correct.Adding them: 5h + 34 +21 =5h+55=84. Therefore, 5h=29, so h=29/5=5.8. So, 5.8 cm is 29/5 cm. To write this as a mixed number, it would be 5 and 4/5 cm. But the problem might prefer fractional form or decimal. The answer is 5.8 cm, which is 29/5 cm. Let me check if that's correct.Wait, but 29 divided by 5 is 5.8. Yes. So, the distance from M to AB is 29/5 cm, which is 5.8 cm.But wait, let me make sure there's no mistake here. Because sometimes in geometry problems, especially with triangles, you have to consider the orientation or whether the point is inside, but in this case, the problem says M is inside the triangle, so the areas should add up directly. I think this approach is correct.Alternatively, another way to check is to compute the area contributions. If the total area is 84, and BMC is 34, AMC is 21, then together they are 55. So, the remaining area, 84 -55=29, must be the area of AMB, which is 5h. So 5h=29, so h=29/5. Yep, same result.Therefore, I think the answer is 29/5 cm, which is 5.8 cm. So, the distance from point M to line AB is 29/5 cm.Final AnswerThe distance from point ( M ) to line ( AB ) is boxed{dfrac{29}{5}} cm.</think>To solve the problem, we start by calculating the area of triangle (ABC) using Heron's formula. The sides of the triangle are (AB = 10 , text{cm}), (BC = 17 , text{cm}), and (AC = 21 , text{cm}). First, we find the semi-perimeter (s):[s = frac{AB + BC + AC}{2} = frac{10 + 17 + 21}{2} = 24 , text{cm}]Using Heron's formula, the area (A) of triangle (ABC) is:[A = sqrt{s(s - AB)(s - BC)(s - AC)} = sqrt{24 times (24 - 10) times (24 - 17) times (24 - 21)} = sqrt{24 times 14 times 7 times 3} = sqrt{7056} = 84 , text{cm}^2]Next, we use the areas of the smaller triangles formed by point (M) inside triangle (ABC). The distances from (M) to sides (AC) and (BC) are 2 cm and 4 cm, respectively. Let (h) be the distance from (M) to (AB).The areas of the triangles (AMC), (BMC), and (AMB) are calculated as follows:- Area of (AMC): (frac{1}{2} times AC times 2 = frac{1}{2} times 21 times 2 = 21 , text{cm}^2)- Area of (BMC): (frac{1}{2} times BC times 4 = frac{1}{2} times 17 times 4 = 34 , text{cm}^2)- Area of (AMB): (frac{1}{2} times AB times h = frac{1}{2} times 10 times h = 5h , text{cm}^2)The sum of these areas must equal the area of triangle (ABC):[21 + 34 + 5h = 84]Solving for (h):[5h + 55 = 84 5h = 29 h = frac{29}{5} = 5.8 , text{cm}]Thus, the distance from point (M) to line (AB) is (boxed{dfrac{29}{5}}) cm.,<think>Okay, so I have this geometry problem here. Let me read it again to make sure I understand. The problem says: "The distances from a point M, lying inside triangle ABC, to its sides AC and BC are 2 cm and 4 cm, respectively. Calculate the distance from point M to line AB, if AB=10 cm, BC=17 cm, AC=21 cm."Hmm. So, point M is inside triangle ABC. We know its distances to two sides: AC and BC are 2 cm and 4 cm. We need to find its distance to AB. The triangle has sides AB=10, BC=17, AC=21. First thought: Maybe this has something to do with areas? Because if we know the distances from a point to the sides, those can be related to the areas of smaller triangles within ABC. Let me recall that the area of a triangle can be calculated as (base * height)/2. If point M is inside ABC, then the areas of the triangles AMC, BMC, and AMB should add up to the area of ABC. Also, the distance from M to a side would be the height of the triangle formed with that side as the base. So, maybe if I can compute the area of ABC, and then set up equations using the known distances from M to AC and BC, then solve for the unknown distance to AB?Let's try that approach step by step.First, compute the area of triangle ABC. Since we know all three sides (10, 17, 21), we can use Heron's formula. Heron's formula states that the area of a triangle with sides a, b, c is sqrt[s(s - a)(s - b)(s - c)], where s is the semi-perimeter: (a + b + c)/2.Let's compute s first:s = (AB + BC + AC)/2 = (10 + 17 + 21)/2 = (48)/2 = 24 cm.Now, the area is sqrt[s(s - AB)(s - BC)(s - AC)].So that's sqrt[24(24 - 10)(24 - 17)(24 - 21)].Compute each term:24 - 10 = 14,24 - 17 = 7,24 - 21 = 3.So, Area = sqrt[24 * 14 * 7 * 3].Let me compute that. First, multiply 24 and 3: 24*3=72.Then, 14*7=98.So, now multiply 72 and 98: 72*98. Let's compute that. 70*98=6860, and 2*98=196, so total 6860 + 196 = 7056.Therefore, the area is sqrt[7056]. What's sqrt(7056)? Let me see. 84^2 = 7056, right? Because 80^2=6400, 85^2=7225, so between 80 and 85. 84*84: 80*80=6400, 80*4=320, 4*80=320, 4*4=16. So, (80+4)^2=80^2 + 2*80*4 + 4^2=6400 + 640 + 16=7056. Yes, so sqrt(7056)=84. Therefore, the area of ABC is 84 cm².Okay, that's the area of the whole triangle. Now, the point M inside the triangle. Let's denote the distances from M to the three sides as follows:Distance to AC: 2 cm (given),Distance to BC: 4 cm (given),Distance to AB: let's call this h (unknown, what we need to find).Now, if we can relate these distances to the areas of the smaller triangles AMC, BMC, and AMB.Specifically, the area of triangle AMC would be (1/2)*AC*(distance from M to AC). Similarly for the others.But wait, triangle AMC: base is AC=21 cm, height is the distance from M to AC, which is 2 cm. So area of AMC is (1/2)*21*2 = 21 cm².Similarly, triangle BMC: base is BC=17 cm, height is 4 cm. So area of BMC is (1/2)*17*4 = 34 cm².Triangle AMB: base is AB=10 cm, height is h (distance from M to AB). So area of AMB is (1/2)*10*h = 5h cm².Since M is inside the triangle ABC, the sum of these three areas should equal the area of ABC. So:Area AMC + Area BMC + Area AMB = Area ABC21 + 34 + 5h = 84Let's compute that:21 + 34 = 5555 + 5h = 84Subtract 55 from both sides:5h = 84 - 55 = 29So, h = 29 / 5 = 5.8 cmWait, 29 divided by 5 is 5.8, which is 29/5. So, 29/5 cm is 5.8 cm. But in fractional form, it's 29/5. The problem doesn't specify the form, but probably as a decimal or fraction. But let me check if I did this correctly.Wait, the logic here is that the sum of the areas of the three smaller triangles (AMC, BMC, AMB) equals the area of ABC. Is that always true?Yes, because point M is inside the triangle, so the three smaller triangles partition ABC without overlapping. Therefore, their areas should sum up to the area of ABC. So, that seems correct.Let me double-check the calculations:Area AMC: (1/2)*21*2 = 21. Correct.Area BMC: (1/2)*17*4 = 34. Correct.Area AMB: (1/2)*10*h = 5h.Total: 21 + 34 + 5h = 55 + 5h. Set equal to 84.55 + 5h = 84 => 5h = 29 => h = 29/5 = 5.8. So 5.8 cm. Which is 29/5 cm.But the problem is presented with all sides as integers, but the distances are 2 and 4. So the answer might be fractional. So, 29/5 cm is 5 and 4/5 cm. So, 5.8 cm. So, maybe they want it as 29/5 or 5.8?But let me check once again if the approach is correct.Alternative way: Using coordinates. Maybe set up coordinates for triangle ABC, assign coordinates to point M based on the given distances, and solve for h. But that might be more complicated. But let me try.Let's assign coordinates. Let me place point C at the origin (0,0). Then, side AC is along the x-axis, so point A is at (21,0). Point B is somewhere in the plane. We need to find coordinates of B such that AB=10, BC=17, and AC=21. Wait, but AB is 10, BC is 17, AC is 21. Let me see if that triangle is possible.Wait, we already computed the area as 84, so it is possible.So, if we place C at (0,0), A at (21,0). Then, point B is at (x,y). The distances: BC=17, so from (0,0) to (x,y) is 17, so x² + y² = 289. AB=10, so from (21,0) to (x,y) is 10, so (x - 21)^2 + y² = 100. Let's solve these two equations.Equation 1: x² + y² = 289Equation 2: (x - 21)^2 + y² = 100Subtract equation 2 from equation 1:x² + y² - [(x - 21)^2 + y²] = 289 - 100x² - (x² - 42x + 441) = 189x² - x² + 42x - 441 = 18942x - 441 = 18942x = 189 + 441 = 630x = 630 / 42 = 15Then, from equation 1: x² + y² = 289 => 15² + y² = 289 => 225 + y² = 289 => y² = 64 => y = 8 or -8. Since it's a triangle, assuming it's above the x-axis, y=8. So, coordinates of B are (15,8).So, coordinates:C: (0,0)A: (21,0)B: (15,8)Now, point M is inside ABC, with distances to AC and BC being 2 cm and 4 cm. Let's find coordinates of M.First, distance from M to AC: since AC is along the x-axis from (0,0) to (21,0), the distance from a point to AC is simply the y-coordinate. Wait, no. Wait, AC is the x-axis from (0,0) to (21,0). The distance from a point (x,y) to AC is the vertical distance, which is y. But in the problem, the distance from M to AC is 2 cm. So, that would mean the y-coordinate of M is 2. But wait, but in coordinate system, if AC is along x-axis, then the distance from any point to AC is the y-coordinate. But in that case, if M is inside the triangle, then its y-coordinate should be between 0 and 8, since the height of the triangle is 8 (from earlier calculation, since point B is at (15,8)).But wait, in the problem, the distance from M to AC is 2 cm, so according to coordinates, that would be y=2. Similarly, distance from M to BC is 4 cm. So, we need to find a point inside the triangle with y=2 and distance to BC equal to 4 cm.Wait, but BC is from (0,0) to (15,8). So, the equation of BC is needed to compute the distance from M to BC.Alternatively, maybe using coordinates complicates things. Let me see.If the distance from M to AC is 2, then in coordinates, M has y=2. So, M is somewhere along the line y=2 inside the triangle. Then, we need to compute its distance to BC, which is 4 cm. Then, once we have the coordinates, we can compute the distance to AB. Wait, but perhaps we can do this more efficiently.Wait, but maybe my initial approach was correct. Let me check.Wait, in the coordinate system, the distance from M to AC is indeed the vertical distance (since AC is horizontal). But in the problem, distances from M to the sides AC and BC are 2 cm and 4 cm. So, if we model the triangle in coordinates as above, with AC along x-axis, then:Distance from M to AC is y-coordinate = 2.Distance from M to BC is 4 cm.So, need to find point M with y=2 and distance to BC equal to 4. Then, once found, compute the distance from M to AB.Alternatively, perhaps the initial area approach is simpler. Let's confirm.In the area approach, we assumed that the sum of areas of AMC, BMC, and AMB equals the area of ABC. That gave us 21 + 34 + 5h = 84 => 5h = 29 => h = 29/5 = 5.8 cm. Which is 5.8 cm. But in the coordinate system, the distance from M to AB would be different, perhaps? Wait, maybe not. Wait, but in coordinates, AB is from (21,0) to (15,8). Let's compute the equation of AB to find the distance.Equation of AB: Points A(21,0) and B(15,8). The slope is (8 - 0)/(15 - 21) = 8 / (-6) = -4/3. So, the equation is y - 0 = (-4/3)(x - 21). So, y = (-4/3)x + 28.Expressed as 4x + 3y - 84 = 0.So, the distance from a point (x, y) to line AB is |4x + 3y - 84| / sqrt(4² + 3²) = |4x + 3y - 84| / 5.So, if point M is at (x, 2), then its distance to AB is |4x + 3*2 - 84| / 5 = |4x + 6 - 84| / 5 = |4x - 78| / 5.But also, the distance from M to BC is 4 cm. Let's compute the equation of BC. Points B(15,8) and C(0,0). The slope is (8 - 0)/(15 - 0) = 8/15. So, equation is y = (8/15)x. Or, 8x - 15y = 0.So, distance from M(x,2) to BC is |8x - 15*2| / sqrt(8² + (-15)^2) = |8x - 30| / 17. According to the problem, this distance is 4 cm. So,|8x - 30| / 17 = 4Multiply both sides by 17:|8x - 30| = 68So, 8x - 30 = 68 or 8x - 30 = -68Case 1: 8x - 30 = 68 => 8x = 98 => x = 98 / 8 = 12.25Case 2: 8x - 30 = -68 => 8x = -38 => x = -38 / 8 = -4.75But since M is inside the triangle ABC, which in coordinates spans from x=0 to x=21 (since A is at (21,0) and C is at (0,0)), so x cannot be negative. Therefore, x=12.25 cm. So, coordinates of M are (12.25, 2).Now, compute the distance from M(12.25, 2) to AB. The formula was |4x + 3y - 84| / 5.Plug in x=12.25, y=2:|4*(12.25) + 3*2 - 84| / 5Compute 4*12.25: 493*2: 6So, 49 + 6 - 84 = (55 - 84) = -29Absolute value: 29Divide by 5: 29/5 = 5.8 cm. So, 5.8 cm, which matches the earlier result.So, the distance from M to AB is 29/5 cm, which is 5.8 cm. So, the answer is 29/5 or 5.8. The problem might expect it in fractional form, so 29/5 cm. But let me check if there's another approach or something I missed.Wait, but in the initial approach, using areas, the answer came directly, while the coordinate approach confirmed it. Therefore, seems like 29/5 cm is correct.But just to make sure, let's see:In coordinate system, after finding M at (12.25, 2), then the distance to AB is 29/5=5.8 cm. So that's correct.Another way to check: Let's compute the area of AMB using coordinates.Point A(21,0), M(12.25,2), B(15,8).The area can be computed via determinant formula:Area = (1/2)| (x_A(y_M - y_B) + x_M(y_B - y_A) + x_B(y_A - y_M)) |Plug in the coordinates:x_A = 21, y_A = 0x_M = 12.25, y_M = 2x_B = 15, y_B = 8Area = (1/2)| 21*(2 - 8) + 12.25*(8 - 0) + 15*(0 - 2) |Compute each term:21*(2 - 8) = 21*(-6) = -12612.25*(8) = 9815*(-2) = -30Sum: -126 + 98 - 30 = (-126 - 30) + 98 = -156 + 98 = -58Absolute value: 58Multiply by 1/2: 29. So, area of AMB is 29 cm². Which matches (1/2)*10*h = 5h = 29 => h=29/5. So that's consistent.Therefore, all methods lead to h=29/5=5.8 cm. So, the answer is 29/5 cm, which is 5.8 cm. The problem asks to calculate the distance, so either form is acceptable, but since 29/5 is exact, probably better to put it as a fraction. Hence, the answer is boxed{dfrac{29}{5}}.Final Answerboxed{dfrac{29}{5}}</think>To solve the problem, we start by calculating the area of triangle (ABC) using Heron's formula. The sides of the triangle are (AB = 10 , text{cm}), (BC = 17 , text{cm}), and (AC = 21 , text{cm}). The semi-perimeter (s) is:[s = frac{10 + 17 + 21}{2} = 24 , text{cm}]Using Heron's formula, the area (K) of triangle (ABC) is:[K = sqrt{s(s - AB)(s - BC)(s - AC)} = sqrt{24 times 14 times 7 times 3} = sqrt{7056} = 84 , text{cm}^2]Next, we consider the distances from point (M) inside the triangle to the sides (AC) and (BC), which are 2 cm and 4 cm respectively. We denote the distance from (M) to (AB) as (h). The areas of the triangles (AMC), (BMC), and (AMB) can be calculated using these distances:- Area of (triangle AMC): (frac{1}{2} times 21 times 2 = 21 , text{cm}^2)- Area of (triangle BMC): (frac{1}{2} times 17 times 4 = 34 , text{cm}^2)- Area of (triangle AMB): (frac{1}{2} times 10 times h = 5h , text{cm}^2)Since the sum of the areas of these three triangles equals the area of triangle (ABC), we have:[21 + 34 + 5h = 84]Solving for (h):[55 + 5h = 84 implies 5h = 29 implies h = frac{29}{5} = 5.8 , text{cm}]Thus, the distance from point (M) to line (AB) is (boxed{dfrac{29}{5}}) cm.]

🔑:## Solution.Let the required distance be x (Fig. 10.63). Then S_{triangle A M B}=0.5 cdot 10 x=5 x. S_{triangle A M B}=S_{triangle A B C}-S_{triangle A M C}-S_{triangle B M C}. Let's find these areas: S_{triangle A B C}=sqrt{24 cdot 14 cdot 3 cdot 7}= =84 text{ cm}^{2}, S_{triangle A M C}=0.5 cdot 21 cdot 2=21 text{ cm}^{2}, S_{triangle B M C}=0.5 cdot 17 cdot 4=34 text{ cm}^{2}. Therefore, S_{triangle A M B}=84-55=29 text{ cm}^{2} and x=5.8 cm.Answer: 5.8 cm.

🧠:[<think>Okay, let's see. I need to solve the equation 5x² + 10y² - 12xy - 6x - 4y + 13 = 0. Hmm, solving for x and y? That seems like a quadratic equation in two variables. Maybe it's a conic section? I remember that quadratic equations in two variables can represent circles, ellipses, hyperbolas, parabolas, etc. But how do I solve this?First, maybe I can rearrange the terms to make it look more familiar. Let me group the x terms and y terms together. So, 5x² - 12xy + 10y² - 6x -4y +13 = 0. The cross term -12xy complicates things. I think if there's a cross term, the conic is rotated. To eliminate the cross term, maybe I can rotate the coordinate system. I remember that the angle of rotation θ can be found using cot(2θ) = (A - C)/B, where A, B, C are the coefficients of x², xy, and y² respectively. Let me check that formula.Yes, for the general quadratic equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, the angle θ to eliminate the xy term is given by cot(2θ) = (A - C)/B. So in this case, A = 5, B = -12, C = 10. Therefore, cot(2θ) = (5 - 10)/(-12) = (-5)/(-12) = 5/12. So cot(2θ) = 5/12. Then 2θ = arccot(5/12). Let me find θ. Alternatively, since cot(2θ) = 5/12, then tan(2θ) = 12/5. Then θ = 0.5 * arctan(12/5). Let me compute arctan(12/5). Hmm, 12/5 is 2.4. arctan(2.4) is approximately 67.38 degrees, so θ would be about 33.69 degrees. But maybe exact values would be better.Alternatively, perhaps I can use trigonometric identities. If tan(2θ) = 12/5, then sin(2θ) = 12/13 and cos(2θ) = 5/13, since (5,12,13) is a Pythagorean triple. So, cos(2θ) = 5/13. Then, using the identities for sinθ and cosθ in terms of cos(2θ):cos(2θ) = 2cos²θ - 1 => 5/13 = 2cos²θ - 1 => 2cos²θ = 18/13 => cos²θ = 9/13 => cosθ = 3/√13.Similarly, sinθ = sqrt(1 - cos²θ) = sqrt(1 - 9/13) = sqrt(4/13) = 2/√13.Alternatively, since tanθ = sinθ / cosθ. Let me see. If 2θ is such that tan(2θ) = 12/5, then tanθ can be found using the identity tan(2θ) = 2tanθ/(1 - tan²θ). Let me set t = tanθ. Then 12/5 = 2t/(1 - t²). Solving for t:12/5 = 2t/(1 - t²) => 12(1 - t²) = 10t => 12 - 12t² = 10t => 12t² +10t -12 =0.Divide by 2: 6t² +5t -6=0. Using quadratic formula: t = [-5 ± sqrt(25 + 144)]/(12) = [-5 ± sqrt(169)]/12 = [-5 ±13]/12. So two solutions: (8)/12=2/3 and (-18)/12=-3/2. Since 2θ is in the first quadrant (because tan(2θ)=12/5 is positive), θ is between 0 and 45 degrees, so tanθ should be positive and less than 1. Wait, but 2θ is arctan(12/5), which is in the first quadrant, so θ is between 0 and 45 degrees, so tanθ is positive and less than 1. However, the solutions we get are t=2/3≈0.666 and t=-3/2. The negative solution is discarded, so tanθ=2/3. Therefore, sinθ=2/√(2² +3²)=2/√13 and cosθ=3/√13. So that matches with earlier.Therefore, the rotation transformation is:x = x'cosθ - y'sinθ = (3/√13)x' - (2/√13)y'y = x'sinθ + y'cosθ = (2/√13)x' + (3/√13)y'Now, substituting these into the original equation. Let me compute each term step by step.First, compute x and y in terms of x' and y':x = (3x' - 2y')/√13y = (2x' + 3y')/√13Now, substitute into the equation:5x² +10y² -12xy -6x -4y +13=0Let me compute each term:First, x²:x² = [(3x' - 2y')/√13]^2 = (9x'² -12x'y' +4y'²)/13Similarly, y²:y² = [(2x' +3y')/√13]^2 = (4x'² +12x'y' +9y'²)/13Then xy:xy = [(3x' -2y')(2x' +3y')]/13 = [6x'² +9x'y' -4x'y' -6y'²]/13 = [6x'² +5x'y' -6y'²]/13Now, substitute into the equation:5*(9x'² -12x'y' +4y'²)/13 +10*(4x'² +12x'y' +9y'²)/13 -12*(6x'² +5x'y' -6y'²)/13 -6*(3x' -2y')/√13 -4*(2x' +3y')/√13 +13=0Let me compute each part step by step.First term: 5*(9x'² -12x'y' +4y'²)/13 = (45x'² -60x'y' +20y'²)/13Second term:10*(4x'² +12x'y' +9y'²)/13 = (40x'² +120x'y' +90y'²)/13Third term: -12*(6x'² +5x'y' -6y'²)/13 = (-72x'² -60x'y' +72y'²)/13Fourth term: -6*(3x' -2y')/√13 = (-18x' +12y')/√13Fifth term: -4*(2x' +3y')/√13 = (-8x' -12y')/√13Last term: +13Now, combine the terms over 13:First three terms:(45x'² -60x'y' +20y'² +40x'² +120x'y' +90y'² -72x'² -60x'y' +72y'²)/13Compute coefficients:x'²:45+40-72=13x'y': -60+120-60=0y'²:20+90+72=182So first three terms simplify to (13x'² + 182y'²)/13 = x'² +14y'²Fourth and fifth terms:(-18x' +12y' -8x' -12y')/√13 = (-26x')/√13So combining all terms:x'² +14y'² - (26x')/√13 +13 =0Simplify the linear term:-26x'/√13 = -26/√13 x' = - (26/√13) x' = - (2*13)/√13 x' = -2√13 x'Therefore, equation becomes:x'² +14y'² -2√13 x' +13 =0Now, complete the square for the x' terms:x'² -2√13 x' = x'² -2√13 x' + (√13)^2 - (√13)^2 = (x' - √13)^2 -13So substituting back:(x' - √13)^2 -13 +14y'² +13 =0Simplifies to:(x' - √13)^2 +14y'² =0Wait, that can't be right. The sum of squares equals zero?(x' - √13)^2 is non-negative, 14y'² is non-negative. The sum equals zero only if both terms are zero.Therefore:(x' - √13)^2 =0 => x' = √1314y'² =0 => y'=0Therefore, the only solution is x'=√13, y'=0But x' and y' are coordinates in the rotated system. So translating back to original coordinates:x = (3x' -2y')/√13 = (3*√13 -0)/√13 = 3y = (2x' +3y')/√13 = (2*√13 +0)/√13 = 2Therefore, the solution is (x, y) = (3, 2)Wait, that seems too straightforward. Let me check if that satisfies the original equation.Plug x=3, y=2 into the original equation:5*(3)^2 +10*(2)^2 -12*(3)*(2) -6*(3) -4*(2) +13=5*9 +10*4 -12*6 -18 -8 +13=45 +40 -72 -18 -8 +13Compute step by step:45 +40 =8585 -72 =1313 -18= -5-5 -8= -13-13 +13=0Yes, that works. So the equation represents a single point (3,2). But how come? If we rotated and found that in the rotated system it's a point, but in the original system, is it also a point? Because conic sections can be degenerate. So maybe this equation is a degenerate conic, just a single point.Hence, the solution is (3,2). So the equation is satisfied only by the point (3,2).But let me check if there could be more solutions. But in the rotated system, the equation reduces to (x' - √13)^2 +14y'^2=0, which can only be satisfied when both squared terms are zero, which gives only one solution. Therefore, in the original coordinates, only one point (3,2). So that's the answer.Final AnswerThe solution is boxed{(3, 2)}.</think>To solve the equation (5x^2 + 10y^2 - 12xy - 6x - 4y + 13 = 0), we recognize it as a quadratic equation in two variables, which may represent a conic section. To eliminate the cross term (-12xy), we rotate the coordinate system. The angle of rotation (theta) is determined using (cot(2theta) = frac{A - C}{B}), where (A = 5), (B = -12), and (C = 10). Calculating (cot(2theta)):[cot(2theta) = frac{5 - 10}{-12} = frac{-5}{-12} = frac{5}{12}]This gives (tan(2theta) = frac{12}{5}), leading to (theta = frac{1}{2} arctanleft(frac{12}{5}right)). Using trigonometric identities, we find (sintheta = frac{2}{sqrt{13}}) and (costheta = frac{3}{sqrt{13}}).We rotate the coordinates using:[x = frac{3x' - 2y'}{sqrt{13}}, quad y = frac{2x' + 3y'}{sqrt{13}}]Substituting these into the original equation and simplifying, we get:[x'^2 + 14y'^2 - 2sqrt{13}x' + 13 = 0]Completing the square for (x'):[(x' - sqrt{13})^2 + 14y'^2 = 0]This equation is satisfied only if both terms are zero:[x' = sqrt{13}, quad y' = 0]Converting back to original coordinates:[x = frac{3sqrt{13}}{sqrt{13}} = 3, quad y = frac{2sqrt{13}}{sqrt{13}} = 2]Verification by substituting (x = 3) and (y = 2) into the original equation confirms the solution. Thus, the solution is:[boxed{(3, 2)}],<think>Okay, so I need to solve this equation: 5x² + 10y² - 12xy - 6x - 4y + 13 = 0. Hmm, let's see. It's a quadratic equation with two variables, x and y. Since it's quadratic, maybe it represents a conic section, like an ellipse, hyperbola, parabola, or something like that. Solving it might mean finding the points (x, y) that satisfy the equation. But usually, conic sections have infinitely many solutions unless it's a degenerate case, but the problem says to solve it, so maybe there's only one solution? Or maybe they want it in some standard form?Let me think. When dealing with quadratic equations in two variables, sometimes completing the square can help. Alternatively, maybe we can treat this as a quadratic in one variable and use the discriminant condition to solve for the other variable. Let me try that approach first.Looking at the equation, maybe I can rearrange it as a quadratic in x. Let's collect terms with x², x, and constants. So:5x² - 12xy - 6x + 10y² - 4y + 13 = 0.So grouping terms:5x² + (-12y - 6)x + (10y² - 4y + 13) = 0.Yes, so this is quadratic in x: ax² + bx + c = 0, where:a = 5,b = -12y - 6,c = 10y² - 4y + 13.For this quadratic equation to have real solutions for x, the discriminant must be non-negative. The discriminant D is b² - 4ac.Let's compute that:D = (-12y - 6)² - 4 * 5 * (10y² - 4y + 13).First, expand (-12y -6)²:(-12y)^2 + 2*(-12y)*(-6) + (-6)^2 = 144y² + 144y + 36.Then compute 4ac:4 * 5 * (10y² -4y +13) = 20*(10y² -4y +13) = 200y² -80y +260.So D = (144y² + 144y +36) - (200y² -80y +260).Combine like terms:144y² - 200y² + 144y +80y +36 -260 = (-56y²) + 224y -224.So the discriminant D = -56y² + 224y -224.For the equation to have real solutions, D ≥ 0. So:-56y² + 224y -224 ≥ 0.Let me factor out -56:-56(y² - 4y + 4) ≥ 0.Note that y² -4y +4 = (y - 2)^2, so:-56(y - 2)^2 ≥ 0.Since (y - 2)^2 is always non-negative, multiplying by -56 reverses the inequality:-56(something non-negative) is non-positive. So the left side is ≤ 0. But we need D ≥ 0. Therefore, the only way this can be true is if (y - 2)^2 = 0, so y = 2.Therefore, the discriminant is zero only when y = 2. So there's exactly one real solution for x when y = 2, and no real solutions otherwise. So that suggests that the original equation has exactly one solution (x, y) where y must be 2, and x is determined from the quadratic equation when y=2.So let's substitute y = 2 into the original equation to find x:5x² +10*(2)^2 -12x*2 -6x -4*2 +13=0.Compute term by term:5x² +10*4 -24x -6x -8 +13=0.Simplify:5x² +40 -24x -6x -8 +13=0.Combine like terms:5x² + (40 -8 +13) + (-24x -6x) = 0.So:5x² +45 -30x =0.Simplify:5x² -30x +45 =0.Divide both sides by 5:x² -6x +9 =0.Factor:(x -3)^2 =0.Therefore, x=3.So the only solution is x=3, y=2. Let me check if this satisfies the original equation:5*(3)^2 +10*(2)^2 -12*(3)*(2) -6*(3) -4*(2) +13.Calculate each term:5*9 =45; 10*4=40; -12*6= -72; -6*3= -18; -4*2= -8; +13.So sum: 45 +40 -72 -18 -8 +13.Compute step by step:45 +40 =85; 85 -72=13; 13 -18= -5; -5 -8= -13; -13 +13=0. Yes, it checks out.So the only solution is (3, 2). Therefore, the equation represents a single point, which makes sense because the discriminant led us to only one possible y value and then x is uniquely determined. So the solution is x=3, y=2.Wait, but let me confirm if there's another approach. Maybe completing the square in two variables? Let's try that.Starting with the original equation:5x² +10y² -12xy -6x -4y +13=0.Hmm, quadratic terms: 5x² -12xy +10y². Maybe we can write this as a quadratic form and diagonalize it? But that might be more complicated. Alternatively, group terms with x and y.Alternatively, think of this as 5x² -12xy +10y² -6x -4y +13=0. Maybe complete the square for x and y.First, group the x terms:5x² -12xy -6x. Let's factor out the 5:5(x² - (12/5)xy - (6/5)x). Hmm, not sure. Alternatively, write the quadratic part as a matrix and diagonalize. But that might be too involved.Alternatively, think of the equation as:5x² - (12y +6)x +10y² -4y +13 =0. Wait, that's similar to the earlier approach. Since we already solved it by discriminant condition, and got a unique solution, maybe that's sufficient. But just to make sure, let's see if we can write the equation in terms of squares.Alternatively, maybe use substitution. Suppose we let u = x - a and v = y - b, trying to eliminate the linear terms. Let me try translating the variables.Let’s assume we can write the equation in terms of (x - h) and (y - k). Let me set x = u + h, y = v + k, and substitute into the equation to eliminate the linear terms. Let's see.Original equation:5x² +10y² -12xy -6x -4y +13=0.Substituting x = u + h, y = v + k:5(u + h)^2 +10(v + k)^2 -12(u + h)(v + k) -6(u + h) -4(v + k) +13 =0.Expand the terms:5(u² + 2hu + h²) +10(v² + 2kv +k²) -12(uv +uk + hv + hk) -6u -6h -4v -4k +13=0.Now collect like terms:Quadratic terms in u and v:5u² +10v² -12uv.Linear terms in u:5*2h u -12v u -6u.Wait, wait, hold on. Let me reorganize:Quadratic terms:5u² +10v² -12uv.Linear terms in u:(10h -12k -6)u.Linear terms in v:(20k -12h -4)v.Constant terms:5h² +10k² -12hk -6h -4k +13.Now, to eliminate the linear terms in u and v, we set the coefficients to zero:10h -12k -6 =0,20k -12h -4 =0.So we have a system:10h -12k =6,-12h +20k =4.Let me write these equations:Equation 1: 10h -12k =6,Equation 2: -12h +20k =4.Let me solve this system. Let's multiply Equation 1 by 12 and Equation 2 by 10 to eliminate h:Equation1 *12: 120h -144k =72,Equation2 *10: -120h +200k =40.Now add the two equations:(120h -144k) + (-120h +200k) =72 +40,56k =112,Thus, k =112 /56 =2.Then substitute k=2 into Equation1:10h -12*2 =6,10h -24 =6,10h =30,h=3.So h=3, k=2. Therefore, substituting x =u +3, y =v +2.So substituting back into the original equation, the linear terms in u and v disappear, and we have:5u² +10v² -12uv + [constant terms] =0.Compute the constant terms:5h² +10k² -12hk -6h -4k +13.Plugging h=3, k=2:5*(9) +10*(4) -12*(3*2) -6*3 -4*2 +13=45 +40 -72 -18 -8 +13.Same as before: 45+40=85; 85-72=13; 13-18=-5; -5-8=-13; -13+13=0.So the transformed equation is:5u² +10v² -12uv =0.So now we have 5u² -12uv +10v² =0.This is a homogeneous quadratic equation in u and v. Let's see if this can be factored or represents a single point.Let me check if this equation has only the trivial solution u=0, v=0.Suppose that there exists some (u, v) ≠ (0,0) satisfying 5u² -12uv +10v²=0.Treat this as a quadratic in u:5u² -12v u +10v²=0.Discriminant D = ( -12v )² -4*5*10v² =144v² -200v²= -56v².Since discriminant is negative unless v=0.If v=0, then equation becomes 5u²=0 ⇒ u=0. So the only solution is u=0, v=0.Therefore, the transformed equation reduces to u=0, v=0, which gives x =3 +0=3, y=2 +0=2. Hence, the only solution is (3, 2), which matches our previous result.Therefore, by translating the coordinates, we have shown that the equation represents a single point (3, 2). So this confirms the solution we found earlier using the discriminant method.Alternatively, maybe there's another way to see this. The original equation is a quadratic in two variables. If it represents a single point, then it's a degenerate conic. The fact that both methods lead to the same single solution makes sense.So yes, the solution is x=3 and y=2.Final AnswerThe solution is boxed{(3, 2)}.</think>To solve the equation (5x^2 + 10y^2 - 12xy - 6x - 4y + 13 = 0), we start by treating it as a quadratic equation in (x):[5x^2 + (-12y - 6)x + (10y^2 - 4y + 13) = 0.]For this quadratic equation to have real solutions for (x), the discriminant must be non-negative. The discriminant (D) is given by:[D = (-12y - 6)^2 - 4 cdot 5 cdot (10y^2 - 4y + 13).]Expanding and simplifying the discriminant:[D = 144y^2 + 144y + 36 - 20(10y^2 - 4y + 13),][D = 144y^2 + 144y + 36 - 200y^2 + 80y - 260,][D = -56y^2 + 224y - 224.]Factoring out (-56):[D = -56(y^2 - 4y + 4) = -56(y - 2)^2.]For (D geq 0), we need (-56(y - 2)^2 geq 0), which implies (y = 2). Substituting (y = 2) into the original equation:[5x^2 + 10(2)^2 - 12x(2) - 6x - 4(2) + 13 = 0,][5x^2 + 40 - 24x - 6x - 8 + 13 = 0,][5x^2 - 30x + 45 = 0,][x^2 - 6x + 9 = 0,][(x - 3)^2 = 0.]Thus, (x = 3). Therefore, the only solution is ((x, y) = (3, 2)).To confirm, we substitute (x = 3) and (y = 2) back into the original equation and verify that all terms cancel out to zero. Additionally, by translating the coordinates and completing the square, we confirm that the equation reduces to a single point ((3, 2)).The solution is (boxed{(3, 2)}).]

🔑:I. The original equation is transformed into (2 x-3 y)^{2}+(x-3)^{2}+(y- 2)^{2}=0, solving it yields x=3, y=2.

🧠:[<think>Okay, so I need to solve the inequality (1 - x)(3 + x) > 0 and figure out which of the given options is correct. Let me start by understanding what this inequality is asking. It's looking for all real numbers x where the product of (1 - x) and (3 + x) is positive. That means both factors have to be either both positive or both negative. First, maybe I should find the roots of the expression (1 - x)(3 + x). The roots are the values of x that make each factor zero. So, setting each factor equal to zero:1 - x = 0 => x = 13 + x = 0 => x = -3So the critical points are at x = -3 and x = 1. These points divide the real number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞). To determine where the product is positive, I need to test each interval.Let me start with the interval (-∞, -3). Let's pick a test point, say x = -4.Plugging into (1 - x): 1 - (-4) = 1 + 4 = 5, which is positive.Plugging into (3 + x): 3 + (-4) = -1, which is negative.So the product of positive and negative is negative. Therefore, this interval does not satisfy the inequality.Next, the interval (-3, 1). Let's choose x = 0.(1 - 0) = 1 (positive)(3 + 0) = 3 (positive)Product is positive, so this interval satisfies the inequality.Now the interval (1, ∞). Let's take x = 2.(1 - 2) = -1 (negative)(3 + 2) = 5 (positive)Product is negative, so this interval doesn't work.Wait, but the original inequality is greater than zero, so only the interval (-3, 1) works. But let me check the critical points. At x = -3 and x = 1, the product is zero, which is not greater than zero, so they are not included.Looking at the options, A is (-3, 1), which matches my result. But let me double-check because sometimes when dealing with quadratic inequalities, the sign can alternate depending on the leading coefficient. Wait, the original expression is (1 - x)(3 + x). Let's expand it to make sure.Multiplying out:(1)(3) + (1)(x) - x(3) - x(x) = 3 + x - 3x - x² = 3 - 2x - x². So in standard form, it's -x² - 2x + 3. The quadratic coefficient is -1, which is negative, so the parabola opens downward. That means the graph is a downward opening parabola with roots at x = -3 and x = 1. Since it opens downward, the parabola is above the x-axis (positive) between its roots. So the solution set is indeed between -3 and 1. So A is correct.But wait, hold on, the options given include B: (-∞, -3) ∪ (1, ∞). That would be the case if the quadratic was positive outside the roots, but since the coefficient is negative, the quadratic is positive between the roots. So the answer should be A. Hmm, but maybe I made a mistake in the test points?Wait, let me verify again. If I take x = -4: (1 - (-4))(3 + (-4)) = (5)(-1) = -5 < 0. Correct. So that interval is negative. Then between -3 and 1, x = 0 gives (1)(3) = 3 > 0. Correct. Then x = 2 gives (-1)(5) = -5 < 0. Correct. So the product is positive only in (-3, 1). So A is the answer.But looking back at the options, B is the union of the two intervals outside. Maybe I confused the direction? Wait, let's check the original inequality again: (1 - x)(3 + x) > 0. If I were to write this as (-x +1)(x + 3) > 0, which is the same as (-1)(x -1)(x + 3) > 0. So factoring out the negative sign, we can write it as - (x -1)(x +3) > 0. Then multiplying both sides by -1 reverses the inequality: (x -1)(x +3) < 0. Now, solving (x -1)(x +3) < 0. The roots are still at x = 1 and x = -3. Since the quadratic here is (x -1)(x +3), which is a parabola opening upward. So it is below zero between the roots. Wait, but hold on, this seems conflicting.Wait, perhaps there's a mistake here. Let me re-express the original inequality step by step. The original inequality is (1 - x)(3 + x) > 0. Let's consider two factors: A = (1 - x) and B = (3 + x). The product A*B > 0.Alternatively, maybe I can factor out a negative sign from (1 - x). So (1 - x) = -(x - 1). So the original inequality becomes -(x - 1)(x + 3) > 0. Then, multiplying both sides by -1 (and reversing the inequality) gives (x - 1)(x + 3) < 0. Now, solving (x -1)(x +3) < 0. The roots are x = 1 and x = -3. The parabola opens upward (since the coefficient is positive), so it's negative between the roots. Therefore, the solution is (-3, 1). So again, the same result. So answer A.But why is option B there? Maybe a common mistake if someone doesn't account for the direction of the parabola or the sign flipping. Let me check if when expanding, perhaps I made a mistake.Original expression: (1 - x)(3 + x) = 1*3 + 1*x - x*3 - x*x = 3 + x - 3x -x² = 3 -2x -x². That's correct. So in standard form, it's -x² -2x +3. So leading coefficient is -1. So the parabola opens downward, which means the graph is positive between the roots. Therefore, solution is between -3 and 1, which is A. So answer is A. But wait, the options given are A: (-3,1), B: (-∞, -3) ∪ (1, ∞), etc.But maybe there's a miscalculation. Let me try another test point in (1, ∞). Let's take x = 10. Then (1 -10)(3 +10) = (-9)(13) = -117 <0. So negative. So not part of solution. Similarly, left interval x = -10: (1 - (-10))(3 + (-10)) = (11)(-7) = -77 <0. So negative. So only the middle interval is positive. So A is correct. Therefore, the answer is A. Hmm. But the initial options had B as another possibility. But according to all steps, it should be A. Wait, but maybe the problem is written differently?Wait, the problem is written as (1 - x)(3 + x) > 0, which is the same as -(x -1)(x +3) >0. So (x -1)(x +3) <0. Then, since (x -1)(x +3) is a quadratic opening upwards, it is negative between its roots. So x between -3 and1. Therefore, solution is (-3,1), which is A. Therefore, the answer is A. So why is B an option? Maybe a trick to test if someone inverts the roots? Hmm. Alternatively, perhaps I made a mistake in the direction when factoring out the negative sign.Wait, let's check again. (1 - x)(3 + x) = [ - (x -1) ] (x +3) = - (x -1)(x +3). So the inequality is - (x -1)(x +3) >0. Then, multiplying both sides by -1 reverses the inequality: (x -1)(x +3) <0. So solving that. Since (x -1)(x +3) is a quadratic with roots at 1 and -3, opening upwards. So it's negative between -3 and 1. So solution is (-3,1). Therefore, answer A. So the correct answer should be A. But the options are given as A, B, C, D.But let me cross-verify with another method. Let's make a sign chart.The critical points are x = -3 and x =1. So we can divide the number line into intervals:1. x < -32. -3 < x <13. x >1Testing each interval:1. x < -3: pick x = -4. Then (1 - (-4)) =5, (3 + (-4))=-1. Product is 5*(-1)=-5 <02. -3 <x <1: pick x=0. (1-0)=1, (3+0)=3. Product 1*3=3 >03. x>1: pick x=2. (1-2)=-1, (3+2)=5. Product -1*5=-5 <0Thus, the inequality (1 -x)(3 +x) >0 holds when -3 <x <1. Therefore, the answer is A.But the system initially showed the options, and B is (-infty, -3) union (1, infty). That's where the product is negative. So if someone by mistake thought that the product is positive outside the roots, they would choose B. But according to all calculations, the correct answer is A.Therefore, the answer should be A.Final Answerboxed{A}</think>To solve the inequality ((1 - x)(3 + x) > 0), we first find the critical points by setting each factor to zero:- (1 - x = 0 implies x = 1)- (3 + x = 0 implies x = -3)These critical points divide the real number line into three intervals: ((-infty, -3)), ((-3, 1)), and ((1, infty)). We test each interval to determine where the product is positive.1. Interval ((-infty, -3)): Choose (x = -4). - ((1 - (-4)) = 5) (positive) - ((3 + (-4)) = -1) (negative) - Product: (5 times -1 = -5) (negative)2. Interval ((-3, 1)): Choose (x = 0). - ((1 - 0) = 1) (positive) - ((3 + 0) = 3) (positive) - Product: (1 times 3 = 3) (positive)3. Interval ((1, infty)): Choose (x = 2). - ((1 - 2) = -1) (negative) - ((3 + 2) = 5) (positive) - Product: (-1 times 5 = -5) (negative)The product is positive only in the interval ((-3, 1)). The critical points (x = -3) and (x = 1) are not included because the inequality is strict ((>)).Expanding the original expression confirms it is a quadratic with a negative leading coefficient, indicating the parabola opens downward. This means the quadratic is positive between its roots, confirming the solution is ((-3, 1)).Thus, the correct answer is (boxed{A}).,<think>Okay, so I need to solve the inequality (1 - x)(3 + x) > 0 and choose the correct answer from the options given. Let me think step by step how to approach this.First, I remember that when dealing with inequalities involving products of two terms, like (a)(b) > 0, the product will be positive if both factors are positive or both are negative. So, I need to find the intervals where both (1 - x) and (3 + x) are either both positive or both negative.Let me write down the two factors:1. (1 - x)2. (3 + x)I should find the critical points where each factor equals zero because those points divide the number line into intervals where the sign of each factor doesn't change. For (1 - x) = 0:1 - x = 0 => x = 1For (3 + x) = 0:3 + x = 0 => x = -3So, the critical points are at x = -3 and x = 1. These points divide the real number line into three intervals:1. (-∞, -3)2. (-3, 1)3. (1, ∞)Now, I need to test the sign of each factor in each interval to determine the sign of the product.Let's start with the first interval: (-∞, -3)Pick a test point, say x = -4.Compute (1 - x) when x = -4: 1 - (-4) = 1 + 4 = 5 (positive)Compute (3 + x) when x = -4: 3 + (-4) = -1 (negative)Product: positive * negative = negativeSo, in (-∞, -3), the product is negative.Second interval: (-3, 1)Test point x = 0.(1 - 0) = 1 (positive)(3 + 0) = 3 (positive)Product: positive * positive = positiveSo, in (-3, 1), the product is positive.Third interval: (1, ∞)Test point x = 2.(1 - 2) = -1 (negative)(3 + 2) = 5 (positive)Product: negative * positive = negativeSo, in (1, ∞), the product is negative.Therefore, the product (1 - x)(3 + x) is positive only in the interval (-3, 1). However, the original inequality is strict (> 0), so we don't include the endpoints where the product is zero.But wait, let me check the answer options again. The options are:A: (-3, 1)B: (-∞, -3) ∪ (1, +∞)C: (-1, 3)D: (-∞, -1) ∪ (3, +∞)From my analysis, the solution is (-3, 1), which is option A. However, I recall sometimes when multiplying inequalities, if the leading coefficients are negative, the direction can change. Wait, but here it's factored as (1 - x)(3 + x). Let me double-check my test points to make sure I didn't make a mistake.First interval (-∞, -3): x = -4, got positive * negative = negative. Correct.Second interval (-3, 1): x = 0, positive * positive = positive. Correct.Third interval (1, ∞): x = 2, negative * positive = negative. Correct.So, the product is positive only in (-3, 1), so option A. But wait, the answer given in the options is A and B. Wait, let me check again.Wait, maybe there's a miscalculation here. Let me check another test point in the second interval. Let's pick x = -2, which is in (-3,1).(1 - (-2)) = 1 + 2 = 3 (positive)(3 + (-2)) = 1 (positive)Product: 3*1 = 3 (positive). Correct.Another test point in (-∞, -3): x = -5.(1 - (-5)) = 6 (positive)(3 + (-5)) = -2 (negative)Product: 6*(-2) = -12 (negative). Correct.Third interval, x=3 (though it's in (1, ∞)):(1 - 3) = -2 (negative)(3 + 3) = 6 (positive)Product: -12 (negative). Correct.So the product is positive only in (-3,1). Therefore, the solution set is (-3,1), which is option A. However, I need to check again because sometimes when you factor, if the leading term is negative, the inequality flips. Wait, the original inequality is (1 - x)(3 + x) > 0. Let me expand this to standard form to check.Expanding (1 - x)(3 + x):First multiply: 1*(3 + x) = 3 + xThen multiply -x*(3 + x) = -3x - x²Combine terms: 3 + x -3x -x² = 3 - 2x - x²So, the quadratic is -x² -2x +3. Let me write it as -x² -2x +3 >0.Alternatively, factor out the negative sign: - (x² + 2x -3) >0Which is equivalent to x² + 2x -3 <0 (since multiplying both sides by -1 reverses the inequality)Wait, so maybe that's another approach. Let me solve x² + 2x -3 <0.First, find the roots of x² +2x -3 =0.Using quadratic formula: x = [-2 ± sqrt(4 +12)] / 2 = [-2 ± sqrt(16)] /2 = [-2 ±4]/2.So, x = (-2 +4)/2= 2/2=1 and x=(-2 -4)/2=-6/2=-3.So, the quadratic x² +2x -3 factors as (x +3)(x -1). Therefore, x² +2x -3 <0 is (x +3)(x -1) <0.So, solving (x +3)(x -1) <0.The critical points are x = -3 and x =1. The parabola opens upwards (since coefficient of x² is positive), so it is below zero between the roots. Therefore, the solution is (-3,1). Which matches the previous result.But in the original inequality, we had (1 -x)(3 +x) >0. Expanding gives -x² -2x +3 >0, which is the same as x² +2x -3 <0. So, same solution (-3,1). Therefore, answer is A.But the options given are A: (-3,1), B: (-∞,-3) ∪ (1, ∞), C: (-1,3), D: (-∞,-1) ∪ (3, ∞). So according to this, the answer should be A. But wait, in some textbooks, sometimes similar problems might have different answers. Let me check once again to ensure.Alternatively, maybe I made a mistake in the sign analysis.Original inequality: (1 - x)(3 + x) >0.Another approach: Find when the product is positive.The critical points are at x =1 and x = -3. The number line is divided into three intervals as before.In (-∞, -3), both factors: (1 - x) is positive because x <1, so 1 - x is positive. (3 +x) is negative because x < -3. So positive * negative = negative. Not greater than zero.In (-3,1): (1 - x) is still positive (since x <1), and (3 +x) is now positive (since x >-3). So positive * positive = positive. Satisfies the inequality.In (1, ∞): (1 -x) becomes negative (x >1), and (3 +x) is still positive (x >-3). So negative * positive = negative. Doesn't satisfy the inequality.Thus, the solution is indeed (-3,1), option A. But why is option B presented as a possible answer? Let me check if I have the factors correct.Original inequality: (1 -x)(3 +x) >0. Wait, if I rewrite the factors, maybe it's easier to see.Alternatively, factor as (-1)(x -1)(x +3). But no, that's complicating. Alternatively, let me consider that (1 -x) is equal to -(x -1), so the inequality can be written as -(x -1)(x +3) >0. Multiply both sides by -1 (remembering to reverse the inequality): (x -1)(x +3) <0.So, solving (x -1)(x +3) <0. The critical points are x=1 and x=-3. The product is negative when one factor is positive and the other is negative.For (x -1)(x +3) <0, the solution is the interval between -3 and 1 where the product is negative? Wait, no. Wait, the product (x -1)(x +3) is a quadratic opening upwards. It is negative between its roots. So, solving (x -1)(x +3) <0 gives (-3,1). Therefore, the original inequality (1 -x)(3 +x) >0 is equivalent to (x -1)(x +3) <0, which is (-3,1). Therefore, answer A.So, all methods point to A being the correct answer. Therefore, I think the answer is A. But the system might have different options. Wait, let me check the problem again.The options are:A: (-3, 1)B: (-∞, -3) ∪ (1, +∞)C: (-1, 3)D: (-∞, -1) ∪ (3, +∞)Yes, so according to all the steps, the correct answer is A. However, sometimes when I see similar problems, sometimes the answer is B, but in this case, according to the math, it's A. Maybe there was a mistake in the critical points?Wait, let me confirm once again with another test point in each interval.First interval: x = -4 (in (-∞, -3)):(1 - (-4)) = 5 (positive)(3 + (-4)) = -1 (negative)Product: 5*(-1) = -5 <0. So, not satisfying the inequality.Second interval: x = 0 (in (-3,1)):(1 -0) =1 (positive)(3 +0) =3 (positive)Product:1*3=3 >0. Satisfies.Third interval: x=2 (in (1, ∞)):(1 -2)=-1 (negative)(3 +2)=5 (positive)Product: -1*5=-5 <0. Doesn't satisfy.Thus, only (-3,1) satisfies the inequality. So, answer is A.But maybe the user made a typo? Wait, let's check the original inequality again: (1 -x)(3 +x) >0. Yes, that's correct.Alternatively, if the inequality were (x -1)(x +3) >0, then the solution would be B. But here, it's (1 -x)(3 +x) >0, which is equivalent to -(x -1)(x +3) >0, which is (x -1)(x +3) <0, solution (-3,1). Therefore, answer A.But perhaps the options are swapped? Wait, no. The options are as given. So, unless I made a mistake in the sign analysis, which I don't see, the answer should be A. However, let me check again.Alternatively, maybe I messed up the direction of the inequality when I multiplied by -1.Original inequality: (1 -x)(3 +x) >0Let me expand this again:(1)(3) + (1)(x) -x(3) -x(x) = 3 + x -3x -x² = 3 -2x -x².So, the quadratic is -x² -2x +3 >0.Multiply both sides by -1 (remembering to reverse the inequality):x² +2x -3 <0.Which factors to (x +3)(x -1) <0.The solution to this inequality is where the product is negative, which is between -3 and 1. Therefore, (-3,1). So, answer A.Alternatively, if we graph the quadratic y = (1 -x)(3 +x), which is a downward opening parabola (since the coefficient of x² is -1). The parabola opens downward, with roots at x =1 and x=-3. The parabola is above the x-axis (y>0) between its roots. Wait, but since it opens downward, it's above the x-axis between x = -3 and x=1. So, y>0 in (-3,1). Therefore, the solution is indeed (-3,1). Therefore, answer A.So, after multiple checks, I confirm that the solution set is (-3,1), which is option A. Therefore, the correct answer is A.Wait, but I just thought of something. When I expanded (1 -x)(3 +x), I got -x² -2x +3. If I consider the quadratic function f(x) = -x² -2x +3, then it's a downward opening parabola, so the graph is a "frown" shape. The vertex is at x = -b/(2a) = -(-2)/(2*(-1)) = 2/(-2) = -1. So, the vertex is at x = -1. Plugging back, f(-1) = -(-1)^2 -2*(-1) +3 = -1 +2 +3 =4. So, the vertex is at (-1,4). Since it opens downward, the parabola is positive between its roots x=-3 and x=1. Therefore, the inequality f(x) >0 is satisfied in (-3,1), which again confirms the answer is A.Therefore, all methods confirm that the solution is A. However, the option B is presented as (-∞, -3) ∪ (1, ∞), which would be the solution if the inequality were (x -1)(x +3) >0. But in this case, since the original inequality is (1 -x)(x +3) >0, which flips the sign compared to (x -1)(x +3). Therefore, the solution is inverted. So, in this case, answer A is correct.Thus, after thorough analysis, the answer must be A.Final Answerboxed{A}</think>To solve the inequality ((1 - x)(3 + x) > 0), we first identify the critical points where each factor is zero: (x = 1) and (x = -3). These points divide the real number line into three intervals: ((-∞, -3)), ((-3, 1)), and ((1, ∞)).We test the sign of the product ((1 - x)(3 + x)) in each interval:1. Interval ((-∞, -3)): - Choose (x = -4): - (1 - (-4) = 5) (positive) - (3 + (-4) = -1) (negative) - Product: Positive (times) Negative = Negative2. Interval ((-3, 1)): - Choose (x = 0): - (1 - 0 = 1) (positive) - (3 + 0 = 3) (positive) - Product: Positive (times) Positive = Positive3. Interval ((1, ∞)): - Choose (x = 2): - (1 - 2 = -1) (negative) - (3 + 2 = 5) (positive) - Product: Negative (times) Positive = NegativeThe product ((1 - x)(3 + x)) is positive only in the interval ((-3, 1)). Expanding the product confirms this result, showing the quadratic (-x^2 - 2x + 3), which is positive between its roots (x = -3) and (x = 1).Thus, the solution set is ((-3, 1)), corresponding to option A.[boxed{A}]]

🔑:The inequality (1-x)(3+x) > 0 can be transformed into (x-1)(x+3) < 0,The two roots of the equation (x-1)(x+3) = 0 are -3 and 1,Therefore, the solution set of the original inequality is {x | -3 < x < 1}.Hence, the correct choice is boxed{text{A}}.

🧠:[<think>Okay, so I have this problem here in the Cartesian coordinate system. Let me read it again carefully.First part (1): The sum of distances from point P to the two points (0, -√3) and (0, √3) equals 4. I need to find the equation of the trajectory of point P, which is called C.Second part (2): The line y = kx + 1 intersects C at points A and B. If vectors OA and OB are perpendicular, find the value of k.Starting with part (1). Hmm, the sum of distances from a point to two fixed points is constant. That sounds like the definition of an ellipse. Yeah, in an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant. So, the trajectory C should be an ellipse with foci at (0, -√3) and (0, √3), and the major axis length is 4.Wait, let me recall. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1, where a and b are the semi-major and semi-minor axes. But first, I need to figure out the major axis direction. Since the foci are on the y-axis (both have x-coordinate 0), the major axis is along the y-axis.In an ellipse, the distance between the foci is 2c, where c is the focal distance. Here, the two foci are (0, -√3) and (0, √3), so the distance between them is 2√3. Therefore, 2c = 2√3, so c = √3.The sum of distances from any point on the ellipse to the foci is 2a, which is given as 4. Therefore, 2a = 4 => a = 2.For an ellipse, the relationship between a, b, and c is a² = b² + c². Wait, no, actually, if the major axis is along the y-axis, then the standard form is (x²/b²) + (y²/a²) = 1, where a > b. Then, in that case, the relationship is a² = b² + c². So, since we have a = 2, c = √3, then we can solve for b.So, a² = 4 = b² + (√3)² => 4 = b² + 3 => b² = 1 => b = 1.Therefore, the equation of the ellipse is (x²/1²) + (y²/2²) = 1, which simplifies to x² + (y²)/4 = 1.Wait, let me verify. Let's check with a point on the ellipse. For example, when x = 0, y must be ±2. Let's compute the sum of distances from (0,2) to the foci (0, -√3) and (0, √3). The distance from (0,2) to (0, √3) is |2 - √3|, and to (0, -√3) is |2 + √3|. The sum is (2 - √3) + (2 + √3) = 4, which matches. Similarly, at y=0, x must be ±1. The distance from (1,0) to each focus is sqrt(1² + (√3)^2) = sqrt(1 + 3) = 2. So sum is 2 + 2 = 4. Correct. So equation is x² + y²/4 = 1. So part (1) answer is x² + y²/4 = 1.Now part (2): The line y = kx + 1 intersects C (which is the ellipse) at points A and B. If OA and OB are perpendicular, find k.So OA and OB are vectors from the origin to points A and B. For them to be perpendicular, their dot product must be zero. Let's denote A as (x1, y1) and B as (x2, y2). Then OA · OB = x1x2 + y1y2 = 0.Given that A and B lie on both the ellipse and the line y = kx + 1. So we can set up the system of equations:1. x² + y²/4 = 12. y = kx + 1Substitute equation 2 into equation 1:x² + (kx + 1)² / 4 = 1Let me expand that:x² + (k²x² + 2kx + 1)/4 = 1Multiply through by 4 to eliminate the denominator:4x² + k²x² + 2kx + 1 = 4Combine like terms:(4 + k²)x² + 2kx + (1 - 4) = 0Simplify:(4 + k²)x² + 2kx - 3 = 0So this is a quadratic equation in x. Let's denote this as:A x² + B x + C = 0, where A = 4 + k², B = 2k, C = -3.The solutions to this equation are the x-coordinates of points A and B. Let’s denote them as x1 and x2.From quadratic equation theory, we know that:x1 + x2 = -B/A = -(2k)/(4 + k²)x1 x2 = C/A = (-3)/(4 + k²)Similarly, the y-coordinates of A and B are y1 = kx1 + 1 and y2 = kx2 + 1.So, OA · OB = x1x2 + y1y2 = x1x2 + (kx1 + 1)(kx2 + 1) = 0.Let me compute that:First, compute (kx1 + 1)(kx2 + 1):= k²x1x2 + k(x1 + x2) + 1Therefore, OA · OB = x1x2 + k²x1x2 + k(x1 + x2) + 1= (1 + k²)x1x2 + k(x1 + x2) + 1Now, substitute x1 + x2 = -2k/(4 + k²) and x1x2 = -3/(4 + k²):So plug those in:= (1 + k²)(-3/(4 + k²)) + k*(-2k/(4 + k²)) + 1Let me compute each term:First term: (1 + k²)(-3)/(4 + k²) = -3(1 + k²)/(4 + k²)Second term: k*(-2k)/(4 + k²) = -2k²/(4 + k²)Third term: +1Combine all terms:= [ -3(1 + k²) - 2k² ] / (4 + k²) + 1Simplify numerator of first fraction:-3 - 3k² - 2k² = -3 -5k²So:= (-3 -5k²)/(4 + k²) + 1Convert 1 to (4 + k²)/(4 + k²):= [ -3 -5k² + 4 + k² ] / (4 + k²)Combine like terms in numerator:(-3 + 4) + (-5k² + k²) = 1 -4k²Thus:= (1 -4k²)/(4 + k²)Set this equal to zero (since OA · OB = 0):(1 -4k²)/(4 + k²) = 0A fraction is zero when numerator is zero (and denominator is not zero). So:1 -4k² = 0Solve for k:4k² = 1 => k² = 1/4 => k = ±1/2So k can be 1/2 or -1/2.But we need to check if these values are valid. The denominator in the original equation is 4 + k², which is always positive, so no problem there.Additionally, we need to ensure that the line y = kx + 1 actually intersects the ellipse at two distinct points A and B. The quadratic equation in x should have two real roots. The discriminant D = B² -4AC.For the quadratic equation (4 + k²)x² + 2kx -3 =0, discriminant D is:(2k)^2 -4*(4 + k²)*(-3) = 4k² +12*(4 + k²) = 4k² +48 +12k² = 16k² +48.Since 16k² +48 is always positive (as k² ≥0), so the quadratic equation always has two distinct real roots. Therefore, k = ±1/2 are valid solutions.Hence, the values of k are 1/2 and -1/2.Wait, but the problem says "find the value of k". So it could be both. Let me check the problem statement again.Yes, it says "find the value of k". Since the problem didn't specify any constraints on k being positive or negative, both solutions are acceptable. So the answer is k = ±1/2.But let me check my calculations again to make sure I didn't make a mistake.Starting from OA · OB = x1x2 + y1y2 = 0.Computed y1y2 = (kx1 +1)(kx2 +1) = k²x1x2 +k(x1 +x2) +1.Then OA · OB = x1x2 +k²x1x2 +k(x1 +x2) +1 = (1 +k²)x1x2 +k(x1 +x2) +1.Substituted x1 +x2 = -B/A = -2k/(4 +k²) and x1x2 = C/A = -3/(4 +k²).Plug in, get:(1 +k²)(-3/(4 +k²)) +k*(-2k/(4 +k²)) +1Compute term by term:First term: -3(1 +k²)/(4 +k²)Second term: -2k²/(4 +k²)Third term: +1Combine:[-3 -3k² -2k²]/(4 +k²) +1 = (-3 -5k²)/(4 +k²) +1Convert 1 to denominator (4 +k²):(-3 -5k² +4 +k²)/(4 +k²) = (1 -4k²)/(4 +k²) =0Hence, 1 -4k²=0 => k²=1/4 =>k=±1/2.All steps seem correct. So answer for part (2) is k=±1/2.Final Answer(1) The equation of ( C ) is (boxed{x^2 + frac{y^2}{4} = 1}).(2) The value of ( k ) is (boxed{pm dfrac{1}{2}}).</think>(1) The sum of distances from point ( P ) to the points ( (0, -sqrt{3}) ) and ( (0, sqrt{3}) ) is constant and equals 4. This defines an ellipse with foci at ( (0, -sqrt{3}) ) and ( (0, sqrt{3}) ). The major axis length is 4, so ( 2a = 4 ) gives ( a = 2 ). The distance between the foci is ( 2c = 2sqrt{3} ), so ( c = sqrt{3} ). Using the relationship ( a^2 = b^2 + c^2 ), we find ( b ):[2^2 = b^2 + (sqrt{3})^2 implies 4 = b^2 + 3 implies b^2 = 1 implies b = 1]The equation of the ellipse is:[boxed{x^2 + frac{y^2}{4} = 1}](2) The line ( y = kx + 1 ) intersects the ellipse at points ( A ) and ( B ). Substituting ( y = kx + 1 ) into the ellipse equation:[x^2 + frac{(kx + 1)^2}{4} = 1]Expanding and simplifying:[4x^2 + k^2x^2 + 2kx + 1 = 4 implies (4 + k^2)x^2 + 2kx - 3 = 0]Let ( x_1 ) and ( x_2 ) be the roots. Then:[x_1 + x_2 = -frac{2k}{4 + k^2}, quad x_1 x_2 = -frac{3}{4 + k^2}]The vectors ( overrightarrow{OA} ) and ( overrightarrow{OB} ) are perpendicular, so their dot product is zero:[x_1 x_2 + (kx_1 + 1)(kx_2 + 1) = 0]Expanding and substituting the roots:[x_1 x_2 + k^2 x_1 x_2 + k(x_1 + x_2) + 1 = 0]Substituting ( x_1 + x_2 ) and ( x_1 x_2 ):[left(1 + k^2right)left(-frac{3}{4 + k^2}right) + kleft(-frac{2k}{4 + k^2}right) + 1 = 0]Simplifying:[frac{-3(1 + k^2) - 2k^2 + 4 + k^2}{4 + k^2} = 0 implies frac{1 - 4k^2}{4 + k^2} = 0]Solving ( 1 - 4k^2 = 0 ):[k^2 = frac{1}{4} implies k = pm frac{1}{2}]The value of ( k ) is:[boxed{pm dfrac{1}{2}}],<think>Okay, so I have this problem here. Let me try to understand it step by step. First, part (1) says: In the Cartesian coordinate system xOy, the sum of distances from point P to the two points (0, -√3) and (0, √3) equals 4. Let the trajectory of point P be C. I need to write the equation of C. Hmm, this sounds familiar. The sum of distances from a point to two fixed points is constant. That's the definition of an ellipse, right? The two fixed points are the foci, and the constant sum is the major axis length. So, maybe C is an ellipse. Let me recall the standard form of an ellipse. The standard equation of an ellipse centered at (h, k) is ((x-h)^2)/a^2 + ((y-k)^2)/b^2 = 1, where a and b are the semi-major and semi-minor axes. The distance between the foci is 2c, and the relationship between a, b, and c is c^2 = a^2 - b^2 for an ellipse where a > b. In this case, the two foci are at (0, -√3) and (0, √3). So, they are both on the y-axis, symmetric about the origin. That suggests the center of the ellipse is at the origin (0,0). The distance between the two foci is 2c. Let me compute c. The distance between (0, √3) and (0, -√3) is 2√3. Therefore, 2c = 2√3 => c = √3. The sum of distances from any point on the ellipse to the two foci is 2a. The problem states this sum is 4, so 2a = 4 => a = 2. Now, we need to find b. Using the relationship c^2 = a^2 - b^2. We have c = √3, so c^2 = 3. a = 2, so a^2 = 4. Therefore:3 = 4 - b^2 => b^2 = 4 - 3 = 1 => b = 1. Since the foci are on the y-axis, the major axis is vertical. Therefore, the standard equation of the ellipse is (x^2)/b^2 + (y^2)/a^2 = 1. Plugging in a and b:x^2/1 + y^2/4 = 1 => x^2 + y^2/4 = 1. Wait, let me confirm. If major axis is along the y-axis, then the denominator under y^2 should be a^2, which is 4. Yes, so the equation is x^2 + (y^2)/4 = 1. So, that's the equation for trajectory C. I think that's part (1) done. Let me check once again. If a point P is on this ellipse, the sum of distances to (0, √3) and (0, -√3) should be 2a = 4. Let's take a point on the ellipse, say (0, 2). The distance to both foci is |2 - √3| and |2 + √3|. Wait, adding those: (2 + √3) + (2 - √3) = 4. Correct. Similarly, take (1, 0). Distance to both foci is √(1^2 + (√3)^2) = √(1 + 3) = 2. So sum is 2 + 2 = 4. Correct. So the equation x^2 + y^2/4 = 1 is correct. Okay, part (1) seems solved. Now, part (2): The line y = kx + 1 intersects C at points A and B. If OA is perpendicular to OB, find the value of k. So, OA and OB are vectors from the origin to points A and B. Their dot product should be zero since they are perpendicular. First, let me note that points A and B are the intersection points of the line y = kx + 1 and the ellipse x^2 + y^2/4 = 1. So, substitute y = kx + 1 into the ellipse equation. So substituting:x^2 + ( (kx + 1)^2 ) / 4 = 1Let me expand that:x^2 + (k^2x^2 + 2kx + 1)/4 = 1Multiply through by 4 to eliminate denominator:4x^2 + k^2x^2 + 2kx + 1 = 4Combine like terms:(4 + k^2)x^2 + 2kx + (1 - 4) = 0Which simplifies to:(4 + k^2)x^2 + 2kx - 3 = 0So, that's a quadratic equation in x. Let's denote this as:A x^2 + B x + C = 0, whereA = 4 + k^2B = 2kC = -3The solutions x1 and x2 correspond to the x-coordinates of points A and B. Let me denote the coordinates of A as (x1, y1) and B as (x2, y2). Then y1 = kx1 + 1 and y2 = kx2 + 1.We need OA perpendicular to OB. The vectors OA and OB are (x1, y1) and (x2, y2). Their dot product is x1x2 + y1y2 = 0.Therefore, x1x2 + y1y2 = 0.But since y1 = kx1 + 1 and y2 = kx2 + 1, substituting:x1x2 + (k x1 + 1)(k x2 + 1) = 0Let me expand this:x1x2 + k^2 x1x2 + k x1 + k x2 + 1 = 0Combine like terms:(1 + k^2) x1x2 + k (x1 + x2) + 1 = 0So, the equation we need is (1 + k^2) x1x2 + k (x1 + x2) + 1 = 0.But from the quadratic equation, we know that:x1 + x2 = -B/A = - (2k) / (4 + k^2)x1x2 = C/A = (-3)/(4 + k^2)Therefore, substituting these into the equation:(1 + k^2) * (-3)/(4 + k^2) + k * (-2k)/(4 + k^2) + 1 = 0Let me compute each term:First term: (1 + k^2)*(-3)/(4 + k^2) = -3(1 + k^2)/(4 + k^2)Second term: k*(-2k)/(4 + k^2) = -2k^2/(4 + k^2)Third term: 1Combine all terms:[ -3(1 + k^2) - 2k^2 ] / (4 + k^2) + 1 = 0Let me compute the numerator:-3(1 + k^2) - 2k^2 = -3 - 3k^2 - 2k^2 = -3 -5k^2So, the expression becomes:(-3 -5k^2)/(4 + k^2) + 1 = 0Combine the terms:[ (-3 -5k^2) + (4 + k^2) ] / (4 + k^2) = 0Simplify numerator:(-3 -5k^2 + 4 + k^2) = (1 -4k^2)Therefore:(1 -4k^2)/(4 + k^2) = 0For this fraction to be zero, the numerator must be zero:1 -4k^2 = 0 => 4k^2 = 1 => k^2 = 1/4 => k = ±1/2So, the possible values for k are 1/2 and -1/2.But we need to check if these values of k actually result in the line intersecting the ellipse at two real points A and B. Since the quadratic equation in x had real solutions, the discriminant must be non-negative.Let me compute the discriminant D for the quadratic equation (4 + k^2)x^2 + 2kx -3 =0.D = B^2 -4AC = (2k)^2 -4*(4 + k^2)*(-3)= 4k^2 +12*(4 + k^2)= 4k^2 +48 +12k^2= 16k^2 +48Since 16k^2 is always non-negative, and 48 is positive, D =16k^2 +48 >0 for all real k. Therefore, regardless of k, the line y =kx +1 intersects the ellipse at two real points. Hence, both k=1/2 and k=-1/2 are valid.Therefore, the values of k are 1/2 and -1/2.Wait, but let me check for k=1/2 and k=-1/2 if OA is indeed perpendicular to OB. Maybe there's some mistake here.Let me take k=1/2 first. Then the line is y=(1/2)x +1. Let's find points A and B.Substitute into the ellipse equation:x^2 + [(1/2 x +1)^2]/4 =1x^2 + ( (1/4 x^2 + x +1 ) ) /4 =1x^2 + (1/16 x^2 + 1/4 x + 1/4 ) =1Multiply all terms by 16 to eliminate denominators:16x^2 + x^2 +4x +4 =1617x^2 +4x +4 -16=017x^2 +4x -12 =0Compute discriminant D=16 +4*17*12=16 +816=832>0. So two real roots.Then x1 +x2= -4/17, x1x2= -12/17Then vectors OA and OB: (x1, (1/2)x1 +1) and (x2, (1/2)x2 +1)Dot product:x1x2 + [(1/2 x1 +1)(1/2 x2 +1)]= x1x2 + (1/4 x1x2 +1/2 x1 +1/2 x2 +1 )= (1 +1/4)x1x2 +1/2(x1 +x2) +1= (5/4)(-12/17) +1/2*(-4/17) +1= (-15/17) + (-2/17) +1= (-17/17) +1 = -1 +1=0Perfect, the dot product is zero. Similarly for k=-1/2, the same steps would follow. So the answer is k=±1/2.Therefore, part (2) answer is k=±1/2.I think that's all. Let me recap.For part (1), since the sum of distances is constant, it's an ellipse with foci on y-axis, center at origin, a=2, c=√3, so b=1. Equation x² + y²/4 =1.For part (2), substituting the line into ellipse, getting quadratic in x, using Vieta's formulas for sum and product of roots, and setting the dot product OA.OB=0 leads to equation which solves to k=±1/2. Checked with actual substitution, works. So values are correct.Final Answer(1) The equation of ( C ) is boxed{dfrac{x^2}{1} + dfrac{y^2}{4} = 1}.(2) The value of ( k ) is boxed{pm dfrac{1}{2}}.</think>To solve the given problem, we start by analyzing the sum of distances from point ( P ) to the two points ((0, -sqrt{3})) and ((0, sqrt{3})). This sum is constant and equals 4, which indicates that the trajectory of point ( P ) is an ellipse.1. Equation of the Ellipse: - The foci of the ellipse are at ((0, sqrt{3})) and ((0, -sqrt{3})), so the center of the ellipse is at the origin ((0, 0)). - The distance between the foci is (2sqrt{3}), giving ( c = sqrt{3} ). - The sum of the distances from any point on the ellipse to the foci is 4, which means ( 2a = 4 ) hence ( a = 2 ). - Using the relationship ( c^2 = a^2 - b^2 ), we find ( b ): [ c^2 = 3, quad a^2 = 4 implies 3 = 4 - b^2 implies b^2 = 1 implies b = 1 ] - Since the major axis is vertical, the equation of the ellipse is: [ frac{x^2}{1} + frac{y^2}{4} = 1 ]2. Finding the value of ( k ): - The line ( y = kx + 1 ) intersects the ellipse. Substituting ( y = kx + 1 ) into the ellipse equation: [ x^2 + frac{(kx + 1)^2}{4} = 1 ] - Simplifying this equation: [ 4x^2 + (k^2x^2 + 2kx + 1) = 4 implies (4 + k^2)x^2 + 2kx - 3 = 0 ] - Let ( x_1 ) and ( x_2 ) be the roots of this quadratic equation. Using Vieta's formulas: [ x_1 + x_2 = -frac{2k}{4 + k^2}, quad x_1x_2 = -frac{3}{4 + k^2} ] - For vectors (overrightarrow{OA}) and (overrightarrow{OB}) to be perpendicular, their dot product must be zero: [ x_1x_2 + (kx_1 + 1)(kx_2 + 1) = 0 ] - Expanding and substituting the values from Vieta's formulas: [ (1 + k^2)x_1x_2 + k(x_1 + x_2) + 1 = 0 ] [ (1 + k^2)left(-frac{3}{4 + k^2}right) + kleft(-frac{2k}{4 + k^2}right) + 1 = 0 ] - Simplifying this equation leads to: [ frac{1 - 4k^2}{4 + k^2} = 0 implies 1 - 4k^2 = 0 implies k^2 = frac{1}{4} implies k = pm frac{1}{2} ] Final Answers(1) The equation of ( C ) is boxed{dfrac{x^2}{1} + dfrac{y^2}{4} = 1}.(2) The value of ( k ) is boxed{pm dfrac{1}{2}}.]

🔑:Solution: (1) Let P(x,y), according to the definition of an ellipse, the trajectory of point P, C, is an ellipse with foci at (0, -sqrt{3}) and (0, sqrt{3}), and its major axis is 2. Its minor axis b= sqrt{2^2-(sqrt{3})^2}=1, thus the equation of curve C is x^2+ frac{y^2}{4}=1. (2) Let A(x_{1},y_{1}), B(x_{2},y_{2}), their coordinates satisfy begin{cases} x^2+ frac{y^2}{4}=1 y=kx+1 end{cases}. Eliminating y and rearranging gives (k^2+4)x^2+2kx-3=0, thus x_{1}+x_{2}=-frac{2k}{k^2+4}, x_{1}x_{2}=-frac{3}{k^2+4}. Since overrightarrow{OA} perp overrightarrow{OB}, then x_{1}x_{2}+y_{1}y_{2}=0. Since y_{1}y_{2}=k^2x_{1}x_{2}+k(x_{1}+x_{2})+1, then x_{1}x_{2}+y_{1}y_{2}=-frac{3}{k^2+4}-frac{3k^2}{k^2+4}-frac{2k^2}{k^2+4}+1=0, simplifying gives -4k^2+1=0, therefore k=pm frac{1}{2}. Thus, the final answers are: (1) The equation of C is boxed{x^2+ frac{y^2}{4}=1}. (2) The value of k is boxed{pm frac{1}{2}}.

🧠:[<think>Okay, let me try to figure out these equations step by step. So, the first one is 13 + a = 15. Hmm, I need to find the value of 'a' here. Since 13 plus something equals 15, I can subtract 13 from both sides to solve for 'a'. Let me do that: 15 minus 13 is 2. So, a should be 2. Let me check: 13 + 2 is 15. Yep, that works.Moving on to the second equation: 13 - e = 10. Alright, here I need to find 'e'. If I subtract e from 13 and get 10, then e must be the number that when subtracted from 13 gives 10. So, maybe I can rearrange the equation. If I add e to both sides and subtract 10 from both sides, I get 13 - 10 = e. That means e is 3. Let me verify: 13 - 3 is 10. That's correct. So, e is 3 here.Now the third part is an inequality: 13 + e < 15. We need to find the value of 'e' that makes this true. Wait, but didn't we just find e as 3 in the previous equation? Hmm, but maybe this is a different 'e' or maybe it's the same? The problem doesn't specify, but usually variables in the same problem are the same unless stated otherwise. But let's see. If we use the same 'e' as before which is 3, then 13 + 3 is 16, which is not less than 15. So that doesn't work. That means there might be a mistake here, or maybe it's a separate question.Wait, looking back at the problem, the three equations are listed under Task 1. So maybe each is separate, and they want the value of 'a' and 'e' in each case. So the first two are equations, and the third is an inequality. So in the third one, we have to solve for 'e' such that 13 + e is less than 15. So let's solve that. Subtract 13 from both sides: e < 15 - 13. Which is e < 2. So e has to be less than 2. But in the second equation, e was 3. That seems contradictory. So perhaps each of these is separate, and the variables are local to each equation? Like, maybe each line is a separate problem? That must be it. So in the first line, a is 2. In the second line, e is 3. In the third line, e is any number less than 2. But the problem says "e = ..." which usually implies a specific answer. Maybe they want the maximum integer value e can take? If e is an integer, then the largest integer less than 2 is 1. But the problem doesn't specify if e has to be an integer or not. So strictly speaking, e can be any real number less than 2. But maybe in the context of the problem, since the first two equations had integer solutions, they expect e to be an integer here as well. So, if that's the case, e would be 1. But I need to check if there's more context. Since the problem is presented as three separate equations, perhaps each is independent. Therefore, the first two have specific solutions, and the third is an inequality where e can be any number less than 2. But the way it's written with "e = ..." might suggest they want the maximum integer solution. Hmm. Alternatively, maybe there's a typo and the third inequality was meant to use a different variable. But as per the given problem, I need to answer based on what's provided. So, for the third part, solving 13 + e < 15 gives e < 2. So, depending on how the answer is expected, it's either e < 2 or if integers, e ≤1. But since the problem didn't specify, the answer is e < 2. But let me check again.First equation: straightforward, a=2. Second equation: e=3. Third inequality: e needs to be less than 2. So even though in the second equation e was 3, in the third one it's a different context. So the answers are a=2, e=3, and e<2. So when writing the answers, for the third one, we just state e is less than 2. But maybe they want it in the same format as the equations. Let me look at the problem again:They have:13 + a = 15 a = …13 - e = 10 e = …13 + e < 15 e = …So for the inequality, it's written as "e = ...", but inequalities don't have a single solution. So perhaps it's a mistake in the problem, or they expect the range. But how to write that in the answer? If it's supposed to be similar to the equations, maybe they just want the expression, like e < 2. But if they expect a specific number, then perhaps the maximum integer, which is 1. But without more context, it's safer to answer e < 2.Alternatively, maybe there's an error in the problem and the third inequality was meant to use a different variable, but as given, we have to work with what's there. So I'll go with e < 2 for the third part.So summarizing:First equation: a = 2Second equation: e = 3Third inequality: e < 2Therefore, the answers are 2, 3, and e must be less than 2. But since the third is written as "e = ...", maybe they want to express it as e < 2 in the blank. So writing e = values less than 2. But in standard math notation, we just write e < 2. So that's probably how they want it.</think>For the given equations and inequality:1. Solving (13 + a = 15): [ a = 15 - 13 = boxed{2} ]2. Solving (13 - e = 10): [ e = 13 - 10 = boxed{3} ]3. Solving the inequality (13 + e < 15): [ e < 15 - 13 implies e < boxed{2} ]Final Answers- (a = boxed{2})- (e = boxed{3})- (e < boxed{2}),<think>Okay, let's tackle these equations one by one. Hmm, first up is 13 + a = 15. Alright, so I need to find the value of 'a' here. Let me think. If I subtract 13 from both sides of the equation, that should isolate 'a'. So, 15 minus 13 is 2. Therefore, a must be 2. Let me check: 13 + 2 is indeed 15. Yep, that works.Next, the second equation is 13 - e = 10. Alright, solving for 'e' this time. To get 'e' by itself, maybe I can subtract 13 from both sides? Wait, that would give me -e = 10 - 13, which is -3. Then, to solve for 'e', I should multiply both sides by -1. So, e = 3. Let me verify: 13 - 3 equals 10. Perfect, that's correct.Now, the third one is an inequality: 13 + e < 15. Hmm, similar to the first equation but with a less than sign instead of equals. Let's solve for 'e' here. Subtract 13 from both sides, so e < 15 - 13. That simplifies to e < 2. So, any value of 'e' that's less than 2 will satisfy this inequality. But wait, in the second equation, we found that e is 3. Is there a conflict here? Wait, maybe these are separate problems? The variables might not be connected across different equations. Let me re-read the problem to check.Looking back, the three are listed as separate equations. So each one is independent unless specified otherwise. So, in the third inequality, e just needs to be less than 2. It doesn't relate to the previous e=3 because each equation is separate. So, the answer here is e < 2. But maybe the question expects a specific number? Wait, no, it's an inequality, so the solution is a range. But the problem says "e = ..." for the first two, but the third one is "e = ..." as well? Wait, let me check the original problem again.Wait, the original task shows:13+a=15 a=...13-e=10 e=...13+e<15 e=...So the first two are equations, and the third is an inequality, but all three lines are presented with "e=..." at the end. That's a bit confusing. Maybe in the third case, since it's an inequality, they expect the possible values for e? But the way it's written is "e=...", which usually denotes equality. Maybe there's a typo, or perhaps they want the maximum integer value e can take? Let me think.If the third problem is 13 + e < 15, then solving for e gives e < 2. If e has to be an integer, the largest integer less than 2 is 1. But the problem doesn't specify that e has to be an integer. However, in the first two problems, a and e were integers. Maybe this is a continuation where e is the same variable? Wait, in the second equation, e was found to be 3, but the third equation requires e < 2. That would be a contradiction. But if each equation is separate, then e can be different. Hmmm.Wait, perhaps in the context of the problem, the variables are independent unless stated otherwise. So, in each line, a and e are separate variables. So, in the first equation, a is 2. In the second equation, e is 3. In the third equation, e must be less than 2. But since the third line uses e again, maybe they want the possible values for e here, given the previous equations? Wait, but in the second equation, e was 3, which conflicts with e needing to be less than 2 in the third. But if they are separate, then maybe each line is its own problem.But the way it's presented is three separate tasks. So perhaps each line is a different problem. So:First problem: Solve 13 + a = 15 for a. Answer: 2.Second problem: Solve 13 - e = 10 for e. Answer: 3.Third problem: Solve 13 + e < 15 for e. Answer: e < 2.But in the original problem, all three are listed under "Task 1". So maybe they are connected? Wait, variables usually persist unless specified. If all three are part of the same task, then variables a and e are the same across all equations. That complicates things because in the second equation e is 3, but the third equation requires e < 2, which would be impossible. Therefore, there must be a mistake.Alternatively, maybe each equation is separate. So in the first equation, a is 2. In the second equation, e is 3. In the third equation, e is a variable independent from the second equation, so e can be any number less than 2. But the problem might expect an equality answer even for the inequality, which is confusing. Alternatively, maybe they made a typo and meant to use a different variable in the third equation? But the problem as stated uses 'e' again.Alternatively, perhaps the third inequality is part of a system with the previous equations. Let's see. If we have e from the second equation as 3, then substituting into the third inequality gives 13 + 3 = 16, which is not less than 15. So that's a contradiction. Therefore, if all three are part of a system, there's no solution. But that seems unlikely. The problem is presented as three separate equations, so probably each is independent.So, probably, the first two are equations to solve for a and e, and the third is an inequality to solve for e, each separate. Therefore, the answers are:a = 2, e = 3, and e < 2.But since the third is an inequality, maybe they expect a range. However, the problem uses "e=..." which might be a formatting issue. But given the problem's structure, I think each is separate, so e in the third problem is different from the second. So the answers are:First line: a=2Second line: e=3Third line: e<2So I should present them as such. Let me check again:13 + a =15: subtract 13, a=2. Correct.13 - e =10: subtract 13, -e= -3, so e=3. Correct.13 + e <15: subtract 13, e<2. Correct.Yes, that seems right. So the final answers are 2, 3, and e being less than 2. But how to present the last one? The problem says "e=...", but since it's an inequality, maybe they expect writing "e < 2". However, the original problem uses "e=..." for all three, which might be a mistake. Alternatively, if it's expecting possible integer solutions, but that's unclear. Given the problem's format, perhaps it's expecting expressions in the same style. Maybe in the third case, they just write "e < 2" in the blank. Since the user hasn't specified any particular format for inequalities, but in the problem statement it's written as "e=...", but the third one is an inequality, so probably the answer is e < 2. Therefore, the answers are 2, 3, and e < 2. I think that's it.</think>For the given equations and inequality:1. Solving (13 + a = 15): [ a = 15 - 13 = boxed{2} ]2. Solving (13 - e = 10): [ -e = 10 - 13 implies -e = -3 implies e = boxed{3} ]3. Solving the inequality (13 + e < 15): [ e < 15 - 13 implies e < boxed{2} ]Answers: (a = boxed{2}) (e = boxed{3}) (e < boxed{2})]

🔑:13+a=15, a=2 ; 13-e=10, e=3 ; 13+e<15, e=0,1