🧠:[<think>Okay, let's see. I need to solve these two problems. The first one is √48 + √8 - √18 - √12. The second one is 2 times (√2 + √3) minus (√3 - √2) squared. Hmm, alright. Let me start with the first problem.For the first problem, all the terms are square roots. I remember that simplifying square roots involves factoring out perfect squares. Let me try to break down each radical step by step.Starting with √48. Let's see, 48. The factors of 48... Well, 16 times 3 is 48, right? And 16 is a perfect square. So √48 is √(16*3) which is √16 * √3. √16 is 4, so that becomes 4√3. Okay, that simplifies to 4√3.Next term is √8. Let's factor 8. 4 times 2, and 4 is a perfect square. So √8 is √(4*2) which is √4 * √2. √4 is 2, so that's 2√2. Got that.Third term is -√18. Let's do the same. 18 is 9 times 2. 9 is a perfect square. So √18 is √(9*2) = √9 * √2 = 3√2. But there's a negative sign in front, so it's -3√2.Last term is -√12. Breaking down 12: 4 times 3. 4 is a perfect square. So √12 is √(4*3) = √4 * √3 = 2√3. With the negative sign, that becomes -2√3.So now, substituting all the simplified forms back into the original expression:4√3 + 2√2 - 3√2 - 2√3.Now, let's combine like terms. The terms with √3 and the terms with √2.Starting with √3 terms: 4√3 - 2√3. That's (4 - 2)√3 = 2√3.Then the √2 terms: 2√2 - 3√2. That's (2 - 3)√2 = -1√2 or just -√2.So combining those results, the entire expression simplifies to 2√3 - √2. Hmm, let me check if I did that correctly.Original terms after simplifying: 4√3 + 2√2 - 3√2 - 2√3.Yes, combining 4√3 - 2√3 is 2√3, and 2√2 - 3√2 is -√2. So the first problem's answer is 2√3 - √2. That seems right.Now moving on to the second problem: 2(√2 + √3) - (√3 - √2)^2. Hmm, let's tackle this step by step.First, expand the terms. Let me handle each part separately. The first part is 2 times (√2 + √3). That's straightforward: 2*√2 + 2*√3, which is 2√2 + 2√3.Now the second part is (√3 - √2)^2. To expand this, I need to square the binomial. Remember that (a - b)^2 = a^2 - 2ab + b^2. Applying that here:(√3)^2 - 2*(√3)*(√2) + (√2)^2. Let's compute each term.(√3)^2 is 3. Then, -2*(√3)*(√2) is -2√6. And (√2)^2 is 2. So putting it all together: 3 - 2√6 + 2. Combining like terms here: 3 + 2 is 5, so this simplifies to 5 - 2√6.So now, the original expression becomes: 2√2 + 2√3 - (5 - 2√6). Wait, because the entire squared term is subtracted. So it's 2√2 + 2√3 minus 5 plus 2√6. Because subtracting a negative would be adding. Wait, no. Wait, when you subtract (5 - 2√6), it's the same as -5 + 2√6. Yes, that's right. So distributing the negative sign: 2√2 + 2√3 - 5 + 2√6.Now, let's see if we can combine like terms here. But the terms are all different radicals: √2, √3, √6, and a constant. So I don't think they can be combined further. So the expression would be 2√2 + 2√3 + 2√6 - 5. Let me verify the steps again.Original: 2(√2 + √3) - (√3 - √2)^2.First part: 2√2 + 2√3.Second part: (√3 - √2)^2 = (√3)^2 - 2*√3*√2 + (√2)^2 = 3 - 2√6 + 2 = 5 - 2√6.Subtracting that from the first part: 2√2 + 2√3 - (5 - 2√6) = 2√2 + 2√3 - 5 + 2√6.Yes, that's correct. So the terms are 2√2, 2√3, 2√6, and -5. Since none of the radicals are the same, that's as simplified as it gets. So the answer is 2√2 + 2√3 + 2√6 - 5.Wait, but let me check if maybe there's a mistake in the expansion of the squared term. Let's go through that again. (√3 - √2)^2.First term: (√3)^2 = 3.Second term: 2*(√3)*(√2) with a negative sign, so -2√6.Third term: (√2)^2 = 2.So 3 + 2 - 2√6 = 5 - 2√6. Yes, that's correct.Therefore, subtracting 5 - 2√6 from 2√2 + 2√3 gives 2√2 + 2√3 -5 + 2√6. So that's the answer. Hmm, maybe I can factor out a 2 from some terms? Let's see: 2√2 + 2√3 + 2√6 -5. If I factor out 2 from the first three terms, it would be 2(√2 + √3 + √6) -5, but I don't know if that's considered simpler. Probably not necessary. The question just says "Calculate," so either form might be acceptable, but usually, they prefer expanded form without factoring unless specified. So probably leave it as 2√2 + 2√3 + 2√6 -5. Let me check once more to make sure I didn't make any arithmetic errors.Original expression:2(√2 + √3) - (√3 - √2)^2First part: 2√2 + 2√3.Second part: (√3 - √2)^2. Expansion:First, square √3: 3.Then, twice the product: 2*(√3)*(√2) = 2√6. But since it's (a - b)^2, it's minus twice the product: -2√6.Then square √2: 2.So total: 3 - 2√6 + 2 = 5 - 2√6.Subtracting that from the first part:2√2 + 2√3 -5 + 2√6.Yes, that's correct. So no mistakes here.So final answers: first problem is 2√3 - √2, second problem is 2√2 + 2√3 + 2√6 -5. Let me just write them again to confirm.First problem:√48 + √8 - √18 - √12Simplifies to 4√3 + 2√2 -3√2 -2√3 = (4√3 -2√3) + (2√2 -3√2) = 2√3 -1√2 = 2√3 - √2. Correct.Second problem:2(√2 + √3) - (√3 - √2)^2First part: 2√2 + 2√3.Second part expanded: 5 - 2√6.Subtracting: 2√2 + 2√3 -5 + 2√6. Correct.So I think these are the simplified forms. The problem didn't specify to approximate numerically, so leaving in terms of radicals is appropriate.Final Answer(1) boxed{2sqrt{3} - sqrt{2}} (2) boxed{2sqrt{2} + 2sqrt{3} + 2sqrt{6} - 5}</think>To solve the given problems, we start by simplifying each term involving square roots. Problem (1): (sqrt{48} + sqrt{8} - sqrt{18} - sqrt{12})1. Simplify each term: - (sqrt{48} = sqrt{16 times 3} = 4sqrt{3}) - (sqrt{8} = sqrt{4 times 2} = 2sqrt{2}) - (sqrt{18} = sqrt{9 times 2} = 3sqrt{2}) - (sqrt{12} = sqrt{4 times 3} = 2sqrt{3})2. Substitute the simplified terms back into the expression: [ 4sqrt{3} + 2sqrt{2} - 3sqrt{2} - 2sqrt{3} ]3. Combine like terms: - Terms with (sqrt{3}): (4sqrt{3} - 2sqrt{3} = 2sqrt{3}) - Terms with (sqrt{2}): (2sqrt{2} - 3sqrt{2} = -sqrt{2})Thus, the result for problem (1) is:[boxed{2sqrt{3} - sqrt{2}}] Problem (2): (2(sqrt{2} + sqrt{3}) - (sqrt{3} - sqrt{2})^2)1. Expand and simplify each part: - First part: (2(sqrt{2} + sqrt{3}) = 2sqrt{2} + 2sqrt{3}) - Second part: ((sqrt{3} - sqrt{2})^2) - Expand using the binomial formula ((a - b)^2 = a^2 - 2ab + b^2): [ (sqrt{3})^2 - 2sqrt{3}sqrt{2} + (sqrt{2})^2 = 3 - 2sqrt{6} + 2 = 5 - 2sqrt{6} ]2. Substitute the expanded terms back into the expression: [ 2sqrt{2} + 2sqrt{3} - (5 - 2sqrt{6}) ]3. Distribute the negative sign and combine terms: [ 2sqrt{2} + 2sqrt{3} - 5 + 2sqrt{6} ]Thus, the result for problem (2) is:[boxed{2sqrt{2} + 2sqrt{3} + 2sqrt{6} - 5}],<think>Okay, so I have two problems to solve here. Let me take them one at a time. Starting with problem (1): √48 + √8 - √18 - √12. Hmm, I remember that simplifying radicals involves breaking them down into factors where one is a perfect square. That way, I can take the square root of the perfect square and bring it outside the radical. Let me try each term step by step.First term: √48. Let's factor 48. The largest perfect square that divides 48 is 16, right? Because 16*3=48. So √48 becomes √(16*3) which is √16 * √3. √16 is 4, so that simplifies to 4√3.Second term: √8. Let's see, 8 is 4*2, and 4 is a perfect square. So √8 is √(4*2) = √4 * √2 = 2√2.Third term: -√18. Breaking down 18, the largest perfect square factor is 9 (9*2=18). So √18 is √(9*2) = √9 * √2 = 3√2. But there's a negative sign in front, so it becomes -3√2.Fourth term: -√12. Let's factor 12. The largest perfect square here is 4 (4*3=12). So √12 is √(4*3) = √4 * √3 = 2√3. With the negative, it's -2√3.Now, let's rewrite the original expression with these simplified terms:4√3 + 2√2 - 3√2 - 2√3.Now, combine like terms. The terms with √3: 4√3 - 2√3. That should be (4-2)√3 = 2√3.The terms with √2: 2√2 - 3√2. That would be (2-3)√2 = -1√2 or -√2.So putting them together, the result is 2√3 - √2. Let me check if that's correct. Let me verify each step.√48: 16*3 is 48, so 4√3. Correct. √8: 4*2, so 2√2. Correct. √18: 9*2, so 3√2. Correct. √12: 4*3, so 2√3. Correct. Then combining like terms: 4√3 - 2√3 is indeed 2√3. 2√2 - 3√2 is -√2. So yes, that seems right.Okay, so problem (1) simplifies to 2√3 - √2. Got that.Now moving on to problem (2): 2(√2 + √3) - (√3 - √2)^2. Alright, this looks a bit more complex because of the squared term. Let me break it down step by step.First, expand the expression. Let me handle each part separately. The first part is 2 times (√2 + √3). The second part is subtracting the square of (√3 - √2). Let me compute each part.Starting with 2(√2 + √3). That's straightforward distribution: 2*√2 + 2*√3. Which is 2√2 + 2√3.Now, the second part is -(√3 - √2)^2. I need to expand the squared term first. Remember that (a - b)^2 = a^2 - 2ab + b^2. So applying that here:(√3)^2 - 2*(√3)*(√2) + (√2)^2. Let's compute each term.(√3)^2 is 3, since the square of a square root cancels out. Similarly, (√2)^2 is 2. Then the middle term is -2*(√3)*(√2) which is -2√6. So putting that together, (√3 - √2)^2 is 3 - 2√6 + 2. Then simplifying 3 + 2 is 5, so it becomes 5 - 2√6.But there's a negative sign in front of this whole expression, so it's -(5 - 2√6) which is -5 + 2√6 or 2√6 - 5.Now, combining both parts of the original expression: the first part was 2√2 + 2√3, and the second part after expanding is 2√6 - 5. So the entire expression becomes:2√2 + 2√3 + 2√6 - 5.Wait, hold on. Let me check that again. The first part is 2√2 + 2√3, then subtract the square which becomes adding (-5 + 2√6). So yes, 2√2 + 2√3 - 5 + 2√6. So combining like terms. But in this case, all the radical terms are different: √2, √3, √6. So they can't be combined further. So the expression simplifies to 2√2 + 2√3 + 2√6 - 5.But wait, let me double-check my expansion of (√3 - √2)^2. Let's do it step by step again:(√3 - √2)^2 = (√3)^2 - 2*√3*√2 + (√2)^2 = 3 - 2√6 + 2. 3 + 2 is 5, so 5 - 2√6. Then, the negative of that is -5 + 2√6. Correct.Then adding that to 2√2 + 2√3: so yes, 2√2 + 2√3 + (-5 + 2√6) = 2√2 + 2√3 + 2√6 -5. That seems correct.Is there a way to simplify this further? Let me see. All the radicals are different (√2, √3, √6) and the constants are just -5. So I don't think they can be combined any further. Therefore, the simplified form is 2√2 + 2√3 + 2√6 - 5.Wait, but maybe the problem expects a different form? Let me check the calculation again to ensure I didn't make a mistake.First, expanding (√3 - √2)^2:First term: (√3)^2 = 3.Second term: 2*(√3)*(√2) = 2√6, but since it's (a - b)^2, it's -2ab, so -2√6.Third term: (√2)^2 = 2.So altogether: 3 - 2√6 + 2 = 5 - 2√6. Correct.Then, the entire expression is 2(√2 + √3) - (5 - 2√6). Which is 2√2 + 2√3 -5 + 2√6. Yes, that's the same as before.So, indeed, combining like terms, but since they are all different radicals, that's as simplified as it gets. So the answer is 2√2 + 2√3 + 2√6 -5.Alternatively, we can write it as -5 + 2√2 + 2√3 + 2√6, but usually, we write the constants first, then the radicals in order. But both are correct. However, maybe there's a different approach that could combine terms?Wait, perhaps when expanding (√3 - √2)^2, maybe there's another way? Let me think. No, the expansion is straightforward. Alternatively, maybe factor something out? Let's see:Looking at the expression 2√2 + 2√3 + 2√6 -5. Is there a common factor among the radical terms? They all have a coefficient of 2. So, factor out 2: 2(√2 + √3 + √6) -5. But that doesn't really simplify it further. It's just another way of writing it. Since the problem asks to calculate, perhaps either form is acceptable, but unless there's a specific instruction, this is as simplified as it can get.Wait, but let me check once more if I did the signs correctly. The original problem is 2(√2 + √3) - (√3 - √2)^2. So expanding (√3 - √2)^2 gives 5 - 2√6, then subtracting that would be 2(√2 + √3) -5 + 2√6. So yes, the signs are correct. Because subtracting (5 - 2√6) is the same as -5 + 2√6.Therefore, the final simplified expression is 2√2 + 2√3 + 2√6 -5. Hmm. Is that the simplest form? I think so. Unless there's some other method to combine them, but I don't see how. The terms √2, √3, √6 are all irrational and distinct, so they can't be combined. Therefore, the answer should be in that form.Wait, but maybe I made a mistake in the arithmetic. Let me check each step again:Original problem: 2(√2 + √3) - (√3 - √2)^2.First part: 2(√2 + √3) = 2√2 + 2√3. Correct.Second part: (√3 - √2)^2. Let's compute this again. (√3 - √2)(√3 - √2). Multiplying term by term:First: √3*√3 = 3.Outer: √3*(-√2) = -√6.Inner: (-√2)*√3 = -√6.Last: (-√2)*(-√2) = (√2)^2 = 2.Adding those together: 3 - √6 - √6 + 2 = 3 + 2 - 2√6 = 5 - 2√6. Correct.Therefore, subtracting that gives -5 + 2√6. So total expression: 2√2 + 2√3 -5 + 2√6. Yes, that's correct.So, final answer: 2√2 + 2√3 + 2√6 -5. Alright, I think that's right. Maybe the problem expects the answer ordered differently, but the components are all there. Let me see if the question wants it in any specific order. The problem says "Calculate", so as long as it's simplified, it's okay.Alternatively, maybe there's a miscalculation here. Let me plug in approximate values to check.Compute original expression numerically:First, 2(√2 + √3) - (√3 - √2)^2.Compute each part:√2 ≈ 1.4142, √3 ≈ 1.7320, √6 ≈ 2.4495.First part: 2*(1.4142 + 1.7320) = 2*(3.1462) ≈ 6.2924.Second part: (√3 - √2)^2 ≈ (1.7320 - 1.4142)^2 ≈ (0.3178)^2 ≈ 0.1010.So the entire expression is 6.2924 - 0.1010 ≈ 6.1914.Now, compute my simplified expression: 2√2 + 2√3 + 2√6 -5.Compute each term:2√2 ≈ 2*1.4142 ≈ 2.8284,2√3 ≈ 2*1.7320 ≈ 3.4640,2√6 ≈ 2*2.4495 ≈ 4.8990,Sum these: 2.8284 + 3.4640 = 6.2924; 6.2924 + 4.8990 ≈ 11.1914; subtract 5: 11.1914 -5 = 6.1914.Which matches the original calculation. So that's a good check. Therefore, my answer is correct.Therefore, problem (2) simplifies to 2√2 + 2√3 + 2√6 -5.So summarizing:Problem (1): 2√3 - √2.Problem (2): 2√2 + 2√3 + 2√6 -5.Wait, but maybe problem (2) can be simplified further? Let me see. Let's see if combining terms is possible. For example, if there are like terms that I missed.But 2√2, 2√3, 2√6—these are all different radicals. There's no combining them. The only constant term is -5. So yes, that's the simplified form.Alternatively, maybe factor 2 out from the radicals: 2(√2 + √3 + √6) -5. But I don't know if that's considered simpler. It's a matter of form. The question just says "Calculate", so either form is acceptable, but perhaps the expanded form is preferable here.Alternatively, maybe there's a step I missed where we can combine terms through some identity. But I don't recall any identities that would allow combining √2, √3, and √6. Since √6 is the product of √2 and √3, but adding them isn't straightforward. So, I think that's as simplified as it can get.Therefore, I think both problems are solved correctly.Final Answer(1) boxed{2sqrt{3} - sqrt{2}} (2) boxed{2sqrt{2} + 2sqrt{3} + 2sqrt{6} - 5}</think>To solve problem (1): (sqrt{48} + sqrt{8} - sqrt{18} - sqrt{12}):1. Simplify each radical term: - (sqrt{48} = 4sqrt{3}) - (sqrt{8} = 2sqrt{2}) - (sqrt{18} = 3sqrt{2}) - (sqrt{12} = 2sqrt{3})2. Substitute the simplified terms back into the expression: [ 4sqrt{3} + 2sqrt{2} - 3sqrt{2} - 2sqrt{3} ]3. Combine like terms: - Terms with (sqrt{3}): (4sqrt{3} - 2sqrt{3} = 2sqrt{3}) - Terms with (sqrt{2}): (2sqrt{2} - 3sqrt{2} = -sqrt{2})Thus, the result for problem (1) is:[boxed{2sqrt{3} - sqrt{2}}]To solve problem (2): (2(sqrt{2} + sqrt{3}) - (sqrt{3} - sqrt{2})^2):1. Expand the first part: [ 2(sqrt{2} + sqrt{3}) = 2sqrt{2} + 2sqrt{3} ]2. Expand the squared term using the formula ((a - b)^2 = a^2 - 2ab + b^2): [ (sqrt{3} - sqrt{2})^2 = (sqrt{3})^2 - 2sqrt{3}sqrt{2} + (sqrt{2})^2 = 3 - 2sqrt{6} + 2 = 5 - 2sqrt{6} ]3. Substitute the expanded squared term back into the expression and simplify: [ 2sqrt{2} + 2sqrt{3} - (5 - 2sqrt{6}) = 2sqrt{2} + 2sqrt{3} - 5 + 2sqrt{6} ]Thus, the result for problem (2) is:[boxed{2sqrt{2} + 2sqrt{3} + 2sqrt{6} - 5}]]

🔑:This problem examines the mixed operations of quadratic radicals: first, convert the quadratic radicals into their simplest forms, then perform multiplication and division operations on the quadratic radicals, and finally combine them. In the mixed operations of quadratic radicals, if one can flexibly use the properties of quadratic radicals and choose the appropriate solving approach based on the characteristics of the problem, it often results in achieving more with less effort.((1)) First, convert the quadratic radicals into their simplest forms and then combine them;((2)) Use the complete square formula and the distributive law of multiplication to perform the multiplication calculation, then combine like quadratic radicals to solve. Detailed Solution((1)) [begin{align*}sqrt{48}+ sqrt{8}- sqrt{18}- sqrt{12} &= sqrt{16 cdot 3}+ sqrt{4 cdot 2}- sqrt{9 cdot 2}- sqrt{4 cdot 3} &= 4sqrt{3}+ 2sqrt{2}- 3sqrt{2}- 2sqrt{3} &= (4-2)sqrt{3}+ (2-3)sqrt{2} &= 2sqrt{3}- sqrt{2}end{align*}]So, the answer is boxed{2sqrt{3}- sqrt{2}}.((2)) [begin{align*}2( sqrt{2}+ sqrt{3})-( sqrt{3}- sqrt{2})^{2} &= 2sqrt{2}+ 2sqrt{3}- (sqrt{3}^2 - 2sqrt{3}sqrt{2}+ sqrt{2}^2) &= 2sqrt{2}+ 2sqrt{3}- (3 - 2sqrt{6}+ 2) &= 2sqrt{2}+ 2sqrt{3}- 5 + 2sqrt{6} &= 2sqrt{2}+ 2sqrt{3}+ 2sqrt{6}- 5end{align*}]Thus, the answer is boxed{2sqrt{2}+ 2sqrt{3}+ 2sqrt{6}- 5}.

🧠:[<think>Okay, let me try to solve this problem step by step. The problem says that sin α and cos α are roots of the quadratic equation 3x² - m x + 1 = 0. Then we need to find sin⁴ α + cos⁴ α. The answer choices are given as (A) 2/9, (B) 1/3, (C) 7/9, and (D) 1.First, I remember that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. So in this case, the quadratic equation is 3x² - m x + 1 = 0. Therefore, the sum of the roots (which are sin α and cos α) should be m/3, right? Because -b/a would be -(-m)/3 = m/3. And the product of the roots should be c/a, which is 1/3.So, we have sin α + cos α = m/3 and sin α * cos α = 1/3.But we need to find sin⁴ α + cos⁴ α. Hmm, how can I express sin⁴ α + cos⁴ α in terms of sin α + cos α and sin α * cos α? Let me recall some algebraic identities.I remember that sin² α + cos² α = 1. That's the Pythagorean identity. Also, sin⁴ α + cos⁴ α can be written as (sin² α)^2 + (cos² α)^2. There's an identity that relates this to (sin² α + cos² α)^2 - 2 sin² α cos² α. Let me verify that.Yes, because (a² + b²)² = a⁴ + 2a²b² + b⁴, so a⁴ + b⁴ = (a² + b²)² - 2a²b². Therefore, sin⁴ α + cos⁴ α = (sin² α + cos² α)^2 - 2 sin² α cos² α. Since sin² α + cos² α = 1, this becomes 1² - 2 sin² α cos² α = 1 - 2 sin² α cos² α.So now, we need to find sin² α cos² α. Let me see. We already know sin α + cos α and sin α * cos α. Maybe we can find sin² α cos² α from these?Wait, sin α * cos α is 1/3. Therefore, sin² α cos² α = (sin α cos α)^2 = (1/3)^2 = 1/9. Therefore, sin⁴ α + cos⁴ α = 1 - 2*(1/9) = 1 - 2/9 = 7/9. So the answer should be 7/9, which is option C.But wait, hold on. Let me make sure I didn't skip any steps or make a mistake here. Let's go through it again.Given that sin α and cos α are roots of the quadratic equation 3x² - m x + 1 = 0. Therefore, sum of roots = m/3 = sin α + cos α, and product of roots = 1/3 = sin α * cos α.We need sin⁴ α + cos⁴ α. We use the identity sin⁴ α + cos⁴ α = (sin² α + cos² α)^2 - 2 sin² α cos² α. Since sin² α + cos² α = 1, this becomes 1 - 2*(sin α cos α)^2. Since sin α cos α = 1/3, then (sin α cos α)^2 = 1/9. Therefore, 1 - 2*(1/9) = 1 - 2/9 = 7/9. So that's 7/9, which is option C. So that seems correct.But let me check if there's another way to approach this. Maybe using the original equation?Alternatively, since sin α and cos α are roots of 3x² - m x + 1 = 0, then each of them must satisfy the equation. So, 3 sin² α - m sin α + 1 = 0, and 3 cos² α - m cos α + 1 = 0.Let me write these two equations:1. 3 sin² α - m sin α + 1 = 02. 3 cos² α - m cos α + 1 = 0If I add these two equations together, I get 3(sin² α + cos² α) - m(sin α + cos α) + 2 = 0. Since sin² α + cos² α = 1, this becomes 3(1) - m(sin α + cos α) + 2 = 0. So 3 - m*(m/3) + 2 = 0. Wait, because sin α + cos α is m/3. So substituting that in:3 - m*(m/3) + 2 = 0Which simplifies to 3 + 2 - (m²)/3 = 0 => 5 - (m²)/3 = 0 => (m²)/3 = 5 => m² = 15 => m = sqrt(15) or m = -sqrt(15). But the problem states that m is a real number, so that's okay.Wait, but why am I getting m here? The problem doesn't ask for m, but maybe this could help cross-verify the previous result.Alternatively, maybe this method can help confirm if our previous answer is correct.So, from the sum and product of the roots, we found that sin⁴ α + cos⁴ α = 7/9, which is option C. Let's see if this matches with the equations above.Alternatively, perhaps we can express sin⁴ α + cos⁴ α as (sin² α + cos² α)^2 - 2 sin² α cos² α = 1 - 2*(1/3)^2 = 1 - 2/9 = 7/9, which is the same as before.Alternatively, using the two equations we had for sin α and cos α:From equation 1: 3 sin² α = m sin α - 1From equation 2: 3 cos² α = m cos α - 1So sin² α = (m sin α - 1)/3cos² α = (m cos α - 1)/3Then sin⁴ α = [ (m sin α - 1)/3 ]²Similarly, cos⁴ α = [ (m cos α - 1)/3 ]²Therefore, sin⁴ α + cos⁴ α = [ (m sin α - 1)² + (m cos α - 1)² ] / 9Let me expand this numerator:(m sin α - 1)^2 + (m cos α - 1)^2= m² sin² α - 2m sin α + 1 + m² cos² α - 2m cos α + 1= m² (sin² α + cos² α) - 2m (sin α + cos α) + 2= m² * 1 - 2m*(m/3) + 2= m² - (2m²)/3 + 2= (3m² - 2m²)/3 + 2= (m²)/3 + 2Therefore, sin⁴ α + cos⁴ α = (m²/3 + 2)/9 = (m² + 6)/27But from earlier, we found that m² = 15, so substituting that in:sin⁴ α + cos⁴ α = (15 + 6)/27 = 21/27 = 7/9. So that matches the previous result. Therefore, even when approached through this method, we get 7/9, confirming that the answer is indeed C.Alternatively, if we didn't know m², but had sin⁴ α + cos⁴ α in terms of m², we can also relate it back to the previous expressions.But since both methods give the same answer, I can be more confident that the answer is C) 7/9.Wait, but let's check once more. Suppose m² = 15, then m is either sqrt(15) or -sqrt(15). Then, sin α + cos α = m/3. Let's check if that's possible.We know that sin α + cos α = m/3. The maximum value of sin α + cos α is sqrt(2) ≈ 1.414, and the minimum is -sqrt(2). So m/3 must lie between -sqrt(2) and sqrt(2). But m is sqrt(15) ≈ 3.872, so m/3 ≈ 1.291, which is less than sqrt(2) ≈ 1.414. Similarly, if m is -sqrt(15), then m/3 ≈ -1.291, which is greater than -sqrt(2). So both values are acceptable. Therefore, there's no inconsistency here.Therefore, all steps are valid and the answer is 7/9, option C.Final Answerboxed{C}</think>Given that (sin alpha) and (cos alpha) are roots of the quadratic equation (3x^2 - mx + 1 = 0), we need to find (sin^4 alpha + cos^4 alpha).First, we use the properties of the roots of the quadratic equation:- The sum of the roots is (sin alpha + cos alpha = frac{m}{3}).- The product of the roots is (sin alpha cos alpha = frac{1}{3}).We know the identity:[sin^4 alpha + cos^4 alpha = (sin^2 alpha + cos^2 alpha)^2 - 2 sin^2 alpha cos^2 alpha]Using the Pythagorean identity (sin^2 alpha + cos^2 alpha = 1), we substitute this into the equation:[sin^4 alpha + cos^4 alpha = 1^2 - 2 (sin alpha cos alpha)^2]We already found that (sin alpha cos alpha = frac{1}{3}). Therefore,[(sin alpha cos alpha)^2 = left(frac{1}{3}right)^2 = frac{1}{9}]Substituting this back into the equation:[sin^4 alpha + cos^4 alpha = 1 - 2 left(frac{1}{9}right) = 1 - frac{2}{9} = frac{7}{9}]Thus, the answer is (boxed{C}).,<think>Okay, let's see. The problem says that sin α and cos α are roots of the quadratic equation 3x² - m x + 1 = 0. Then we need to find sin⁴α + cos⁴α. The options are given as A to D. Hmm, alright. Let me recall how to approach this.First, since sin α and cos α are roots of the quadratic equation, by Vieta's formulas, the sum of the roots is equal to m/3, and the product is 1/3. So, that would be:sin α + cos α = m/3andsin α * cos α = 1/3Wait, is that right? Vieta's formula for a quadratic equation ax² + bx + c = 0 tells us that sum of roots is -b/a and product is c/a. So in this case, the equation is 3x² - m x + 1 = 0. Therefore, sum of roots is m/3 and product is 1/3. Yeah, that's correct.So we have:sin α + cos α = m/3 ... (1)sin α * cos α = 1/3 ... (2)We need to find sin⁴α + cos⁴α. Hmm. I remember that sin⁴α + cos⁴α can be written in terms of sin²α + cos²α and sin²α cos²α. Let me recall the identity.We know that sin⁴α + cos⁴α = (sin²α + cos²α)² - 2 sin²α cos²α. Since sin²α + cos²α = 1, this simplifies to 1² - 2 sin²α cos²α = 1 - 2 (sin α cos α)^2.So, sin⁴α + cos⁴α = 1 - 2(sin α cos α)^2.From equation (2), sin α cos α = 1/3. So, substituting that in:sin⁴α + cos⁴α = 1 - 2*(1/3)^2 = 1 - 2*(1/9) = 1 - 2/9 = 7/9.Wait, so that's 7/9, which is option C. Hmm, but let me check if that's correct.Wait, but hold on. Let's verify if we can do that. Is there any condition on m here? The problem states that m is a real number, and sin α and cos α satisfy the equation. So, maybe there's a constraint on m to ensure that sin α and cos α are real numbers. But since sin α and cos α are always between -1 and 1, maybe the roots have to lie in that interval. But the problem doesn't specify any restriction on α, so maybe we just proceed with the algebra.But let me confirm the steps again. The quadratic equation has roots sin α and cos α. Therefore, sum is m/3 and product is 1/3. Then we use the identity to express sin⁴α + cos⁴α. The identity is correct: (sin²α + cos²α)^2 = sin⁴α + 2 sin²α cos²α + cos⁴α, so subtracting 2 sin²α cos²α gives sin⁴α + cos⁴α. So, sin⁴α + cos⁴α = 1 - 2*(sin α cos α)^2. Then substituting sin α cos α = 1/3, which gives 1 - 2*(1/9) = 7/9. So that would be option C. So maybe the answer is C.But wait, let me think again. Is there a possibility that m has to satisfy some condition for sin α and cos α to be valid? Because even if sin α and cos α are roots of the equation, their sum and product must satisfy certain constraints since sin α and cos α are bounded between -1 and 1.For example, the product is given as 1/3. But sin α cos α is equal to (sin 2α)/2, which has a maximum value of 1/2 and minimum of -1/2. But here, the product is 1/3, which is within that range. So that's okay.Also, the sum sin α + cos α. The maximum value of sin α + cos α is sqrt(2) ≈ 1.414, and the minimum is -sqrt(2). So, the sum m/3 must be between -sqrt(2) and sqrt(2). So m must be between -3 sqrt(2) and 3 sqrt(2). But since the problem says m is a real number, and they don't restrict us, perhaps the problem is designed such that regardless of m, sin⁴α + cos⁴α is always 7/9. But that seems odd because m is related to α. Wait, but maybe since sin α and cos α are roots, m is determined by α, but the problem is giving m as a real number and asking for sin⁴α + cos⁴α. However, through the equations, we found that sin⁴α + cos⁴α is 7/9 regardless of m, which is conflicting.Wait, but is that possible? Let me check again. Let's take an example. Suppose sin α and cos α satisfy the equation 3x² - m x + 1 = 0. Let's pick a specific α where sin α and cos α are known.For example, take α = 45°, so sin α = cos α = √2/2 ≈ 0.707. Then, the product sin α cos α would be (√2/2)^2 = 0.5. But in our problem, the product is 1/3 ≈ 0.333. So that's different. So maybe α is not 45° here.Alternatively, let's suppose that sin α and cos α are roots of 3x² - m x + 1 = 0. Let's solve for x. The roots would be [m ± sqrt(m² - 12)] / 6. Therefore, sin α and cos α must be equal to [m + sqrt(m² - 12)] / 6 and [m - sqrt(m² - 12)] / 6. However, sin α and cos α must satisfy sin²α + cos²α = 1.Let's compute sin²α + cos²α. Let s = sin α + cos α = m/3, and p = sin α cos α = 1/3. Then sin²α + cos²α = (sin α + cos α)^2 - 2 sin α cos α = s² - 2p = (m²/9) - 2*(1/3) = m²/9 - 2/3. But we know sin²α + cos²α = 1. Therefore:m²/9 - 2/3 = 1Multiply both sides by 9:m² - 6 = 9Thus m² = 15 => m = sqrt(15) or m = -sqrt(15). So even though m is given as a real number, in reality, m must be either sqrt(15) or -sqrt(15) for sin α and cos α to exist. So there's a restriction on m. However, the question is asking for sin⁴α + cos⁴α. From our previous calculation, we found that regardless of m, sin⁴α + cos⁴α = 7/9. Wait, but how is that possible? Because m is determined by α, but in our calculation, we used Vieta's formula which gives sin α + cos α = m/3 and sin α cos α = 1/3, and then the expression for sin⁴α + cos⁴α only depends on sin α cos α, which is fixed at 1/3.But in reality, if m is fixed (as sqrt(15) or -sqrt(15)), then sin α and cos α are specific numbers. However, in our calculation, sin⁴α + cos⁴α ended up being 7/9 regardless of m. Wait, but the calculation used sin α cos α = 1/3, which was given by Vieta's formula. So even if m is constrained, since the product is fixed as 1/3, then sin⁴α + cos⁴α would always be 1 - 2*(1/3)^2 = 7/9. Therefore, the answer is 7/9, which is option C. So that must be the answer.But just to double-check, let's consider if m is sqrt(15). Then, sin α + cos α = sqrt(15)/3. Let's compute sin α + cos alpha = sqrt(15)/3 ≈ 3.872/3 ≈ 1.291, which is less than sqrt(2) ≈ 1.414, so that's okay. Then, sin α and cos alpha are roots of 3x² - sqrt(15)x + 1 = 0. Let's compute the roots:x = [sqrt(15) ± sqrt(15 - 12)] / 6 = [sqrt(15) ± sqrt(3)] / 6. Let's approximate sqrt(15) ≈ 3.872, sqrt(3) ≈ 1.732. So roots are (3.872 + 1.732)/6 ≈ 5.604/6 ≈ 0.934, and (3.872 - 1.732)/6 ≈ 2.14/6 ≈ 0.357. So sin α ≈ 0.934 and cos α ≈ 0.357, or vice versa.Then sin⁴α + cos⁴α ≈ (0.934)^4 + (0.357)^4 ≈ 0.934^2 is ≈ 0.872, so 0.872^2 ≈ 0.760. Similarly, 0.357^2 ≈ 0.127, so 0.127^2 ≈ 0.016. So total ≈ 0.760 + 0.016 ≈ 0.776. Now, 7/9 ≈ 0.777..., which is very close. So that seems to check out.Similarly, if we take the other root, if sin α ≈ 0.357 and cos α ≈ 0.934, same result. So the calculation is correct. Therefore, the answer is C, 7/9.Final Answerboxed{C}</think>Given that ( sin alpha ) and ( cos alpha ) are roots of the quadratic equation ( 3x^2 - mx + 1 = 0 ), we use Vieta's formulas to find the sum and product of the roots:- The sum of the roots is ( sin alpha + cos alpha = frac{m}{3} )- The product of the roots is ( sin alpha cos alpha = frac{1}{3} )We need to find ( sin^4 alpha + cos^4 alpha ). Using the identity:[sin^4 alpha + cos^4 alpha = (sin^2 alpha + cos^2 alpha)^2 - 2 sin^2 alpha cos^2 alpha]Since ( sin^2 alpha + cos^2 alpha = 1 ), we substitute this into the identity:[sin^4 alpha + cos^4 alpha = 1^2 - 2 (sin alpha cos alpha)^2]Substituting ( sin alpha cos alpha = frac{1}{3} ):[sin^4 alpha + cos^4 alpha = 1 - 2 left( frac{1}{3} right)^2 = 1 - 2 left( frac{1}{9} right) = 1 - frac{2}{9} = frac{7}{9}]Thus, the answer is (boxed{C}).]

🔑:2.C.From the relationship between roots and coefficients, we know that sin alpha cdot cos alpha=frac{1}{3}. Therefore, we havebegin{array}{l}sin ^{4} alpha+cos ^{4} alpha =left(sin ^{2} alpha+cos ^{2} alpharight)^{2}-2(sin alpha cdot cos alpha)^{2}=frac{7}{9} .end{array}

🧠:[<think>Okay, so I need to find the greatest common divisor (gcd) of two numbers, each made up entirely of 1's. The first number has m digits, and the second has n digits. Let me think about how to approach this.First, I remember that numbers composed entirely of 1's are called repunits. Repunits have some interesting properties, especially when it comes to divisibility. Maybe the gcd of two repunits relates to the gcd of their lengths, m and n? That seems familiar, like with numbers in geometric progressions or something.Let me denote the repunit with k digits as R_k. So, R_m is the first number and R_n is the second. The problem is to find gcd(R_m, R_n).I recall that repunits can be expressed using the formula R_k = (10^k - 1)/9. Let me verify that: For example, if k=3, then R_3 = 111, and (10^3 -1)/9 = 999/9 = 111, which checks out. So, R_k = (10^k - 1)/9.So, we can rewrite the gcd as gcd((10^m - 1)/9, (10^n - 1)/9). Since both are divided by 9, maybe we can factor that out? Let's see.Let me denote A = (10^m - 1)/9 and B = (10^n - 1)/9. Then, gcd(A, B) = gcd((10^m -1)/9, (10^n -1)/9). Since 9 is a common factor in the denominators, but the numerators are 10^m -1 and 10^n -1. So, maybe the gcd of the numerators is related to 10^{gcd(m,n)} -1? Because I remember that gcd(10^m -1, 10^n -1) = 10^{gcd(m,n)} -1.Wait, is that a general formula? Let me think. For numbers of the form a^m -1 and a^n -1, their gcd is a^{gcd(m,n)} -1. Yeah, that seems right. For example, gcd(10^6 -1, 10^4 -1) = 10^{gcd(6,4)} -1 = 10^2 -1 = 99. Let me check: 10^6 -1 is 999999, 10^4 -1 is 9999. gcd(999999, 9999). Let's compute that.Divide 999999 by 9999: 9999*100 = 999900, subtract that from 999999, get 99999. Then, gcd(9999, 99999). 9999 divides into 99999 ten times, which is 99990, subtract to get 9. Then gcd(9999,9). 9999 divided by 9 is 1111, so gcd is 9. Wait, but according to the formula, it should be 99. Hmm, that contradicts. Did I do something wrong?Wait, maybe I miscalculated. Let me try again. gcd(999999, 9999). Let's use the Euclidean algorithm.First step: 999999 divided by 9999. Let's see, 9999 * 100 = 999900. 999999 - 999900 = 99. So, 999999 = 9999 * 100 + 99. Then, gcd(9999, 99).Now, 9999 divided by 99: 99 * 101 = 9999. So, remainder 0. Therefore, gcd is 99. Ah, right, I miscalculated the remainder. So, the gcd is 99, which matches 10^{gcd(6,4)} -1 = 10^2 -1 = 99. So the formula holds. So, general formula: gcd(a^m -1, a^n -1) = a^{gcd(m,n)} -1, provided that a > 1 is an integer.Therefore, applying this formula, gcd(10^m -1, 10^n -1) = 10^{gcd(m,n)} -1. Then, since A = (10^m -1)/9 and B = (10^n -1)/9, then gcd(A, B) would be [gcd(10^m -1, 10^n -1)] / 9, but only if 9 divides both 10^m -1 and 10^n -1. Wait, but 10^k -1 is always divisible by 9, since 10 ≡ 1 mod 9, so 10^k ≡ 1^k = 1 mod 9, so 10^k -1 ≡ 0 mod 9. So, yes, 10^k -1 is divisible by 9 for any k. Therefore, both A and B are integers.Therefore, gcd(A, B) = gcd((10^m -1)/9, (10^n -1)/9) = [gcd(10^m -1, 10^n -1)] / 9 = [10^{gcd(m,n)} -1]/9.But [10^{gcd(m,n)} -1]/9 is exactly the repunit with gcd(m,n) digits. For example, if gcd(m,n)=2, then R_2 = 11. If gcd(m,n)=3, R_3=111, etc. Therefore, the gcd of R_m and R_n is R_{gcd(m,n)}.Therefore, the answer should be the repunit with length equal to the gcd of m and n.Wait, let me check with an example. Let’s take m=6 and n=4. Then, R_6 = 111111, R_4=1111. The gcd should be R_{gcd(6,4)}=R_2=11. Let's compute gcd(111111, 1111). Using the Euclidean algorithm:111111 divided by 1111: 1111*100 = 111100, subtract from 111111: 11. So, gcd(1111, 11). Then, 1111 divided by 11 is 101, remainder 0. So gcd is 11. Which is correct. So, the formula works here.Another example: m=3, n=6. Then, gcd(111, 111111). Since 111 divides 111111 (111*1001=111111), so the gcd is 111. Which is R_3, and gcd(3,6)=3. So that checks out.Another test case: m=5, n=3. gcd(R_5, R_3). R_5=11111, R_3=111. Let's compute gcd(11111,111). 11111 divided by 111: 111*100=11100, subtract: 11. Then, gcd(111,11). 111 divided by 11 is 10 with remainder 1. Then gcd(11,1)=1. Therefore, the gcd is 1. Which would correspond to R_1=1, since gcd(5,3)=1. But R_1 is 1. So yes, that works.Therefore, it seems the general formula is that the gcd of R_m and R_n is R_{gcd(m,n)}.Therefore, the answer is the repunit number with gcd(m,n) digits. So, written as boxed{R_{gcd(m,n)}}, but the problem probably expects a numerical answer in terms of the number itself. Wait, but how to denote it? The problem says "Determine the gcd of...", so we need to express the gcd as a number. However, since the answer is a repunit with gcd(m,n) digits, perhaps we can express it as boxed{frac{10^{gcd(m,n)} - 1}{9}}.Alternatively, if the problem expects the answer in words as a repunit, but since mathematical notation is preferred, probably the boxed expression with the formula is acceptable.Alternatively, in the problem statement, they denote the numbers as underbraces with m digits and n digits. So maybe the answer is similar: a repunit with gcd(m,n) digits. But to write it as a boxed number, we need to express it as the number composed of gcd(m,n) ones. Since in the problem statement, they use underbrace notation, but here we need to box it.But in standard math notation, the gcd would be expressed as boxed{frac{10^{gcd(m,n)} - 1}{9}}. However, maybe the problem expects a more elementary explanation.Alternatively, if m and n are multiples of each other, say d divides both m and n, then R_d divides both R_m and R_n. So, the greatest common divisor would be R_d where d is the gcd(m,n). Therefore, the answer is R_{gcd(m,n)}. But how to write that? If the problem allows, we can write boxed{R_{gcd(m,n)}}, but since the original numbers are written as underbraces with m and n digits, perhaps the answer is also an underbrace, but I need to use LaTeX.Alternatively, the problem may expect the answer as the number consisting of gcd(m,n) ones. So, for example, if gcd(m,n) is 2, the answer is 11, if it's 3, it's 111, etc. Since the problem says "put your final answer within boxed{}", likely, they want the answer expressed as boxed{frac{10^{gcd(m,n)} - 1}{9}} or as a repunit denoted somehow. But in the problem statement, the numbers are presented as underbraces of 1's, but in the answer, we need to write the number.Alternatively, maybe just write boxed{1ldots1} with gcd(m,n) digits. But I don't know if that's standard. The most precise way is to write the formula (10^{gcd(m,n)} -1)/9. Since that's the mathematical expression for the repunit number.Alternatively, since repunits are sometimes written as R_k, but the problem didn't introduce that notation, so maybe safer to use the formula.Therefore, the gcd is (10^{gcd(m,n)} -1)/9, which is the repunit with gcd(m,n) digits. So, boxed, it should be boxed{dfrac{10^{gcd(m,n)} - 1}{9}}.Yes, that seems correct. Let me verify once more with m=6, n=4: gcd(6,4)=2, then (10^2 -1)/9=99/9=11, which is correct. For m=3, n=5: gcd(3,5)=1, so (10^1 -1)/9=9/9=1, which is correct. Another example: m=9, n=6, gcd=3, so (10^3 -1)/9=999/9=111, and gcd(999999999, 666666)... Wait, R_9 is 111111111 and R_6 is 111111. Their gcd is 111, which is R_3. Yep, that works. So the formula is solid.Final AnswerThe greatest common divisor is boxed{dfrac{10^{gcd(m,n)} - 1}{9}}.</think>To determine the gcd of two numbers composed entirely of 1's (repunits) with ( m ) and ( n ) digits respectively, we start by noting that a repunit with ( k ) digits can be expressed as ( R_k = frac{10^k - 1}{9} ).We need to find the gcd of ( R_m ) and ( R_n ), which are ( frac{10^m - 1}{9} ) and ( frac{10^n - 1}{9} ). Using the property that ( gcd(10^m - 1, 10^n - 1) = 10^{gcd(m,n)} - 1 ), we can factor out the common divisor in the numerators.Given that both ( 10^m - 1 ) and ( 10^n - 1 ) are divisible by 9, we divide the result by 9 to obtain the gcd of the repunits. Therefore, the gcd of ( frac{10^m - 1}{9} ) and ( frac{10^n - 1}{9} ) is ( frac{10^{gcd(m,n)} - 1}{9} ).This result corresponds to the repunit with ( gcd(m,n) ) digits. Thus, the greatest common divisor is:[boxed{dfrac{10^{gcd(m,n)} - 1}{9}}],<think>Okay, so I need to find the greatest common divisor (gcd) of two numbers made up entirely of ones. The first number has m digits, and the second one has n digits. Let me think about how to approach this.First, let me recall what these numbers are called. Numbers consisting of all ones are called repunits, right? Like, 11...1 with m digits is a repunit number. Repunit numbers have some interesting properties, especially when dealing with divisibility and gcd. Hmm.I remember that repunit numbers can be expressed using a formula. Specifically, the repunit with m digits is equal to (10^m - 1)/9. Let me verify that. For example, if m is 3, then (10^3 - 1)/9 = 999/9 = 111, which is correct. Similarly, for m=2, it's (100 - 1)/9 = 99/9 = 11. Yeah, that formula works. So the first number is R_m = (10^m - 1)/9, and the second is R_n = (10^n - 1)/9.So, the problem reduces to finding gcd(R_m, R_n) where R_m and R_n are repunit numbers. I need to find gcd((10^m -1)/9, (10^n -1)/9). Since both numbers are divided by 9, maybe I can first find the gcd of the numerators and then adjust for the 9?Let me write that down. Let’s denote A = 10^m - 1 and B = 10^n - 1. Then the gcd of R_m and R_n is gcd(A/9, B/9). But since both A and B are divisible by 9 (because 10 ≡ 1 mod 9, so 10^k ≡ 1^k ≡ 1 mod 9, so 10^k -1 ≡ 0 mod 9). Therefore, gcd(A/9, B/9) = gcd(A, B)/9. Wait, is that always true? Let me check.Suppose we have two numbers A and B both divisible by 9. Then gcd(A/9, B/9) = gcd(A, B)/9. Hmm, let's test with A=18, B=36. gcd(18/9, 36/9)=gcd(2,4)=2, and gcd(18,36)=18, 18/9=2. So that works. Another example: A=9, B=9. gcd(1,1)=1, and gcd(9,9)/9=9/9=1. Yes, that seems correct. So, the formula holds. Therefore, gcd(R_m, R_n) = gcd(10^m -1, 10^n -1)/9.Okay, so now I need to find gcd(10^m -1, 10^n -1). Let me recall that for numbers of the form a^k -1 and a^l -1, the gcd is a^gcd(k,l) -1. Is that a theorem? Let me confirm.Yes, I remember that in number theory, there's a formula: gcd(a^m - 1, a^n - 1) = a^gcd(m,n) - 1, provided that a > 1 is an integer. So, in this case, a is 10, so gcd(10^m -1, 10^n -1) = 10^gcd(m,n) -1.Therefore, substituting back into the previous equation, gcd(R_m, R_n) = (10^gcd(m,n) -1)/9.But wait, (10^gcd(m,n) -1)/9 is exactly the repunit number with gcd(m,n) digits. For example, if gcd(m,n) is 3, then that's (1000 -1)/9 = 111. So, indeed, the gcd of R_m and R_n is R_gcd(m,n).Therefore, the answer is that the gcd of the two repunit numbers with m and n digits is the repunit number with gcd(m,n) digits. So, written as a box, the final answer should be boxed{frac{10^{gcd(m,n)} - 1}{9}}.But let me check with some examples to make sure.Take m=2 and n=3. Then R_2=11 and R_3=111. The gcd(11,111). 111 divided by 11 is 10 with remainder 1. Then gcd(11,1)=1. So gcd is 1. According to the formula, gcd(2,3)=1, so R_1=1. Which matches. Another example: m=6, n=9. gcd(6,9)=3. So R_3=111. Let's check. R_6=111111, R_9=111111111. gcd(111111, 111111111). Let's compute.Divide 111111111 by 111111. 111111*1000=111111000. Subtract that from 111111111: 111111111 - 111111000 = 111. So 111111111 = 111111*1000 + 111. Now compute gcd(111111,111). 111111 divided by 111 is 1001. So 111*1001=111111. Therefore, gcd(111,0)=111. Therefore, gcd(R6, R9)=111=R3. Which matches the formula. So this seems correct.Another example: m=4, n=6. gcd(4,6)=2. R_2=11. Let's check. R4=1111, R6=111111. gcd(1111,111111). 111111 divided by 1111. 1111*100=111100. Subtract from 111111: 111111 - 111100 = 11. So gcd(1111,11). 1111 divided by 11 is 101, so gcd is 11. Which is R2. Correct again.Therefore, the formula seems to hold in these cases. Therefore, the answer is as I concluded: the gcd of the two repunits is the repunit with gcd(m,n) digits, which is (10^gcd(m,n) -1)/9.Final AnswerThe greatest common divisor is boxed{dfrac{10^{gcd(m,n)} - 1}{9}}.</think>To determine the greatest common divisor (gcd) of the repunit numbers with ( m ) digits and ( n ) digits, we start by noting that a repunit number with ( k ) digits can be expressed as ( R_k = frac{10^k - 1}{9} ).We need to find ( gcdleft( frac{10^m - 1}{9}, frac{10^n - 1}{9} right) ). Let ( A = 10^m - 1 ) and ( B = 10^n - 1 ). Then, the gcd of the repunit numbers is ( gcdleft( frac{A}{9}, frac{B}{9} right) ). Since both ( A ) and ( B ) are divisible by 9, we have:[gcdleft( frac{A}{9}, frac{B}{9} right) = frac{gcd(A, B)}{9}]Using the property of gcd for numbers of the form ( a^k - 1 ), we know that ( gcd(10^m - 1, 10^n - 1) = 10^{gcd(m, n)} - 1 ). Therefore, substituting back, we get:[gcdleft( frac{10^m - 1}{9}, frac{10^n - 1}{9} right) = frac{10^{gcd(m, n)} - 1}{9}]This result shows that the gcd of the two repunit numbers is another repunit number with ( gcd(m, n) ) digits. Thus, the final answer is:[boxed{dfrac{10^{gcd(m,n)} - 1}{9}}]]

🔑:Suppose m<n and write mathfrak{n}=mathrm{mq}+mathrm{r} for the Euclidean division of n by m. Then the Euclidean division of underbrace{1 ldots 1}_{n text { digits }} by underbrace{1 ldots 1}_{m text { digits }} is written asunderbrace{1 ldots 1}_{n text { digits }}=underbrace{1 ldots 1}_{m text { digits }}left(10^{m(q-1)+r}+ldots 10^{m+r}+10^{r}right)+underbrace{1 ldots 1}_{r text { digits }}By applying the Euclidean algorithm, we see that the gcd sought is underbrace{1 ldots 1}_{operatorname{gcd}(mathfrak{m}, n)}.

🧠:[<think>Okay, so I have this problem here about sets. Let me read it again. The universal set U is {1,2,3,4,5,6,7}, set M is {2,3,4,5}, and set N is {1,4,5,7}. The question is asking for M intersect with the complement of N in U, right? And then they give options A to D.Alright, first, let me recall what the complement of a set means. The complement of N with respect to U, denoted as complement_U N, should be all the elements in U that are not in N. So U has elements 1 through 7, and N has 1,4,5,7. Therefore, the complement of N would be all the elements in U that aren't in N. Let me write that out.U = {1,2,3,4,5,6,7}N = {1,4,5,7}So complement of N would be U minus N. So subtracting the elements of N from U. Let's do that:Elements in U but not in N: 2,3,6. Because 1 is in N, 4 is in N, 5 is in N, 7 is in N. So the remaining elements are 2,3,6. Therefore, complement_U N is {2,3,6}.Wait, let me confirm. Yes, 2 is not in N, 3 is not in N, 6 is not in N. So complement_U N is {2,3,6}.Now, the problem asks for M intersect with this complement. So M is {2,3,4,5}. Intersection with {2,3,6} would be the elements common to both sets.Looking at M: 2,3,4,5. The other set is 2,3,6. The common elements are 2 and 3. So the intersection should be {2,3}.Checking the answer options: Option B is {2,3}, which seems to be the answer. Let me just make sure I didn't make any mistakes.Wait, step by step again. Complement of N in U is {2,3,6}. Then M is {2,3,4,5}. Intersection is elements in both. So 2 is in both, 3 is in both. 4 and 5 are only in M, and 6 is only in the complement. So yes, only 2 and 3. So B is correct.But wait, the options given: A: {1,7}; B: {2,3}; C: {2,3,6}; D: {1,6,7}. So B is there. Therefore, the answer is B. Hmm.Alternatively, maybe I miscalculated the complement? Let's check again. N is {1,4,5,7}. So elements not in N are 2,3,6. That's correct. Then intersection with M. M has 2,3,4,5. So 2 and 3 are in both. 4 and 5 are only in M. So the intersection is indeed {2,3}. Therefore, B is the answer.Wait, but let me think again. Maybe the problem is written differently? The problem says M ∩ (complement_U N). So complement of N is {2,3,6}, then intersect with M {2,3,4,5} gives {2,3}. So yes. So answer is B. Therefore, I think the correct answer is B.Final Answerboxed{B}</think>Given the universal set ( U = {1, 2, 3, 4, 5, 6, 7} ), set ( M = {2, 3, 4, 5} ), and set ( N = {1, 4, 5, 7} ), we need to find ( M cap (complement_U N) ).First, we determine the complement of ( N ) with respect to ( U ). The complement of ( N ), denoted ( complement_U N ), consists of all elements in ( U ) that are not in ( N ). Since ( N = {1, 4, 5, 7} ), the elements not in ( N ) are ( 2, 3, 6 ). Therefore, ( complement_U N = {2, 3, 6} ).Next, we find the intersection of ( M ) with ( complement_U N ). The set ( M = {2, 3, 4, 5} ) and the set ( complement_U N = {2, 3, 6} ). The intersection of these sets includes the elements common to both, which are ( 2 ) and ( 3 ).Thus, ( M cap (complement_U N) = {2, 3} ).The correct answer is (boxed{B}).,<think>Okay, let me try to figure out this problem step by step. So, the question is asking for M intersect with the complement of N in the universal set U. The universal set U is {1,2,3,4,5,6,7}, set M is {2,3,4,5}, and set N is {1,4,5,7}. The options are A through D, and I need to figure out which one is the correct answer.First, let me recall what the complement of a set means. The complement of N with respect to U, denoted as complement_U N, should be all the elements in U that are not in N. So, U is {1,2,3,4,5,6,7}, and N is {1,4,5,7}. Therefore, the elements in U not in N would be the ones that are left when we remove 1,4,5,7 from U. Let me list them out.Original U: 1,2,3,4,5,6,7Remove the elements of N: 1,4,5,7So, remaining elements are 2,3,6. Therefore, complement_U N should be {2,3,6}. Let me check that again. If N has 1,4,5,7, then subtracting those from U gives 2,3,6. Yes, that seems right.Now, the next step is to find the intersection of set M with this complement. Set M is {2,3,4,5}, and the complement of N is {2,3,6}. The intersection of these two sets would be elements that are common to both. So let's compare each element:- 2 is in both M and complement_U N.- 3 is in both M and complement_U N.- 4 is in M but not in the complement of N.- 5 is in M but not in the complement of N.- 6 is in the complement of N but not in M.So the common elements are 2 and 3. Therefore, the intersection should be {2,3}. Let me verify once more to be sure I didn't make a mistake.First, complement of N is definitely {2,3,6} because those are the elements not in N. Then M is {2,3,4,5}. The overlap between those two sets is indeed 2 and 3. So the answer should be {2,3}, which is option B. Let me check the options again:A: {1,7} – these elements are in N, so their complement wouldn't include them. So A is not correct.B: {2,3} – yes, as I found.C: {2,3,6} – this is the complement of N itself, but we need the intersection with M, which excludes 6 because M doesn't have 6. So C is incorrect.D: {1,6,7} – these are elements not in M, so the intersection can't include them. So D is wrong.Therefore, the correct answer must be B.But wait, let me make sure I didn't mix up intersection with union or something. Intersection is elements common to both sets. So complement of N is {2,3,6}, M is {2,3,4,5}. Common elements are 2 and 3. Yes, that's right. So yes, B is correct. Hmm. I think that's solid.Final Answerboxed{B}</think>To determine ( M cap (complement_U N) ), we first need to find the complement of ( N ) with respect to the universal set ( U ). The universal set ( U ) is ( {1, 2, 3, 4, 5, 6, 7} ) and set ( N ) is ( {1, 4, 5, 7} ). The complement of ( N ), denoted ( complement_U N ), consists of all elements in ( U ) that are not in ( N ). [complement_U N = U - N = {1, 2, 3, 4, 5, 6, 7} - {1, 4, 5, 7} = {2, 3, 6}]Next, we find the intersection of set ( M ) (which is ( {2, 3, 4, 5} )) with ( complement_U N ). The intersection of two sets includes only the elements common to both sets.[M cap (complement_U N) = {2, 3, 4, 5} cap {2, 3, 6} = {2, 3}]Thus, the correct answer is (boxed{B}).]

🔑:Option BKey Point: Mixed operations of intersection, union, and complement sets. Topic: Sets. Analysis: The solution can be found by applying the basic operations of sets. Given M={2,3,4,5}, N={1,4,5,7}, Therefore, complement_U N={2,3,6}, Then Mcap (complement_U N)={2,3}, Hence, the correct option is: boxed{B} Review: This question mainly examines the basic operations of sets, requiring proficiency in the operations of intersection, union, and complement of sets, which is quite fundamental.

🧠:[<think>Alright, so I need to find the range of T = 5x + 4y + 2z given the constraints that x, y, z are non-negative real numbers and they satisfy the equations x + y + z = 30 and 3x + y - z = 50. Hmm, okay, let's break this down step by step.First, I have two equations here with three variables. Since I need to find the range of T, which is a linear combination of x, y, and z, I think I can express T in terms of variables that I can manage more easily. Maybe I can solve the system of equations to express some variables in terms of others. Let me try that.The equations are:1. x + y + z = 302. 3x + y - z = 50Let me subtract the first equation from the second to eliminate y and z. Wait, no, let's see:If I subtract equation 1 from equation 2:(3x + y - z) - (x + y + z) = 50 - 30That simplifies to:2x - 2z = 20Divide both sides by 2:x - z = 10So, x = z + 10. Okay, so x is 10 more than z. That's useful. Let me note that down: x = z + 10.Now, substitute x in terms of z into the first equation. So, original equation 1: x + y + z = 30.Substituting x:(z + 10) + y + z = 30Combine like terms:2z + y + 10 = 30Subtract 10 from both sides:2z + y = 20So, y = 20 - 2z. Alright, so y is expressed in terms of z. So now, I have x and y in terms of z. Let me write that:x = z + 10y = 20 - 2zSo now, all variables can be expressed in terms of z. Since x, y, z are all non-negative, let's find the constraints on z.Starting with x ≥ 0:x = z + 10 ≥ 0Since z is a real number and z ≥ 0 (given), then z + 10 is always at least 10, so x is automatically non-negative. So no problem there.Next, y ≥ 0:y = 20 - 2z ≥ 0So, 20 - 2z ≥ 0 → 2z ≤ 20 → z ≤ 10Also, z ≥ 0 as per the problem statement. Therefore, z is in [0, 10]So z can vary from 0 to 10. Let's note that: 0 ≤ z ≤ 10Therefore, the possible values of z are between 0 and 10 inclusive.Now, let's express T = 5x + 4y + 2z in terms of z.We have x = z + 10 and y = 20 - 2z, so substitute those into T:T = 5(z + 10) + 4(20 - 2z) + 2zLet me compute that step by step.First, 5(z + 10) = 5z + 50Then, 4(20 - 2z) = 80 - 8zThen, +2z.So, combining all terms:5z + 50 + 80 - 8z + 2z = (5z - 8z + 2z) + (50 + 80)Calculating the z terms:5z -8z is -3z, +2z is -zThen constants:50 +80 =130Therefore, T = -z +130Wait, that's interesting. So T is a linear function of z with a coefficient of -1. So as z increases, T decreases, and as z decreases, T increases.Given that z is in [0,10], then the maximum value of T occurs at the minimum z, which is z=0, and the minimum value of T occurs at the maximum z, which is z=10.So, let's compute T when z=0:T = -0 +130=130And when z=10:T = -10 +130=120Therefore, the range of T is from 120 to 130.But wait, let me verify this. Maybe I made a mistake in substitution or calculation. Let me check again.Original expression:T =5x +4y +2zExpressed in terms of z:x = z +10, so 5x =5(z +10)=5z +50y=20 -2z, so 4y=4(20 -2z)=80 -8z2z remains as 2zSo adding all together:5z +50 +80 -8z +2z = (5z -8z +2z) + (50 +80) = (-1z) +130Yes, that's correct. So T = -z +130. Therefore, since z ranges from 0 to10, T ranges from 130 -0=130 down to 130 -10=120.Therefore, T ∈ [120, 130]But wait, let me also check if there are any other constraints that might affect this. For example, when z=0, x=10, y=20, z=0. All variables are non-negative, so that's okay.When z=10, x=20, y=20 -2*10=0, z=10. So y=0 here, which is allowed since the problem states real numbers x,y,z ≥0. So y=0 is okay. So both endpoints are valid.Therefore, the maximum value of T is 130 and the minimum is 120. So the range is [120,130].But just to be thorough, let me check another point in between. Let's take z=5.Then x=15, y=20 -10=10, z=5.Compute T=5*15 +4*10 +2*5=75 +40 +10=125. Which is 130 -5=125. So that's consistent. So linear function, decreasing as z increases. So seems correct.Therefore, the answer should be that the range of T is from 120 to 130, inclusive. So written as [120, 130].But wait, let me think again. The problem says "range of T=5x +4y +2z". So we need to present the minimum and maximum values of T given the constraints.We found that T = -z +130, and z can go from 0 to10. So yes, T is between 120 and 130.Alternatively, could there be a mistake in the expression for T? Let me check the original substitution again:Original T=5x +4y +2z.With x= z +10 and y=20 -2z.So substitute:5(z +10) =5z +504(20 -2z)=80 -8z2z remains.Total: 5z +50 +80 -8z +2z = (5z -8z +2z) +130= (-1z)+130. Yep.Therefore, correct.Thus, the range of T is [120,130]. So the final answer is boxed{[120,130]}.But wait, the problem says "the range of T". In case the user expects interval notation with square brackets, which is standard for inclusive ranges. So yes, [120,130].Alternatively, if they want it written as 120 ≤ T ≤130, but in the problem statement, they use "the range of T", so interval notation is appropriate.Therefore, the answer is boxed{[120, 130]}.Final AnswerThe range of ( T ) is boxed{[120, 130]}.</think>Given the problem where real numbers ( x, y, z geq 0 ) satisfy the equations ( x + y + z = 30 ) and ( 3x + y - z = 50 ), we need to find the range of ( T = 5x + 4y + 2z ).First, we solve the system of equations:1. ( x + y + z = 30 )2. ( 3x + y - z = 50 )Subtracting the first equation from the second, we get:[(3x + y - z) - (x + y + z) = 50 - 30 implies 2x - 2z = 20 implies x - z = 10 implies x = z + 10]Substituting ( x = z + 10 ) into the first equation:[(z + 10) + y + z = 30 implies 2z + y + 10 = 30 implies 2z + y = 20 implies y = 20 - 2z]We now express ( x ), ( y ), and ( z ) in terms of ( z ):- ( x = z + 10 )- ( y = 20 - 2z )- ( z ) is a non-negative real number.Considering the non-negativity constraints:- ( x geq 0 ) is automatically satisfied since ( z geq 0 ).- ( y geq 0 implies 20 - 2z geq 0 implies z leq 10 ).Thus, ( z ) must be in the interval ([0, 10]).Next, we express ( T ) in terms of ( z ):[T = 5x + 4y + 2z = 5(z + 10) + 4(20 - 2z) + 2z = 5z + 50 + 80 - 8z + 2z = -z + 130]Since ( T = -z + 130 ) is a linear function decreasing in ( z ), the maximum value occurs at ( z = 0 ) and the minimum value occurs at ( z = 10 ):- When ( z = 0 ), ( T = 130 ).- When ( z = 10 ), ( T = 120 ).Thus, the range of ( T ) is (boxed{[120, 130]}).,<think>Okay, so I need to find the range of T = 5x + 4y + 2z given the constraints that x, y, z are non-negative real numbers and they satisfy the equations x + y + z = 30 and 3x + y - z = 50. Hmm, let me think about how to approach this.First, since we have two equations with three variables, maybe I can express two variables in terms of the third and substitute them into T to find its possible values. Let me try solving the system of equations.We have:1. x + y + z = 302. 3x + y - z = 50If I subtract the first equation from the second, that might eliminate y and z. Let's see:(3x + y - z) - (x + y + z) = 50 - 303x + y - z - x - y - z = 202x - 2z = 20Divide both sides by 2: x - z = 10So, x = z + 10. Okay, that's useful. Now I can express x in terms of z.Now substitute x = z + 10 into one of the original equations to find y. Let's take the first equation:x + y + z = 30(z + 10) + y + z = 302z + 10 + y = 30So, y = 30 - 2z - 10 = 20 - 2z. Therefore, y = 20 - 2z.Now we have expressions for x and y in terms of z:x = z + 10y = 20 - 2zBut since x, y, z are all non-negative, we need to make sure that these expressions don't result in negative values. Let's set up inequalities for each variable.Starting with x = z + 10 ≥ 0. Since z is already ≥ 0, x will automatically be ≥ 10. So that's fine.For y = 20 - 2z ≥ 0:20 - 2z ≥ 0 → 2z ≤ 20 → z ≤ 10. So z cannot exceed 10.For z itself, z ≥ 0. So combining these, z is between 0 and 10, inclusive.Therefore, z ∈ [0, 10].Now, let's express T = 5x + 4y + 2z in terms of z. Substitute the expressions for x and y:T = 5(z + 10) + 4(20 - 2z) + 2zLet me compute this step by step.First, 5(z + 10) = 5z + 50Second, 4(20 - 2z) = 80 - 8zThird, 2z remains as it is.So adding all together:5z + 50 + 80 - 8z + 2z = (5z - 8z + 2z) + (50 + 80) = (-1z) + 130 = 130 - zWait, so T simplifies to 130 - z? That's interesting. So T is a linear function of z, decreasing as z increases. Therefore, since z ranges from 0 to 10, T will range from 130 - 0 = 130 down to 130 - 10 = 120. So the range of T is [120, 130]. Is that right? Let me verify.Wait, that seems straightforward, but let me check if there's any mistake in the substitution or the inequalities.First, check the substitution again:T = 5x + 4y + 2zx = z + 10, so 5x = 5(z + 10) = 5z + 50y = 20 - 2z, so 4y = 4(20 - 2z) = 80 - 8z2z remains 2zAdding them up: 5z + 50 + 80 - 8z + 2z = (5z -8z +2z) + (50 +80) = (-1z) + 130. Yes, correct. So T = 130 - z.So since z is between 0 and 10, then when z=0, T=130; when z=10, T=120. Therefore, T can take any value from 120 to 130. But let me confirm that all intermediate values are achievable. Since z can be any real number between 0 and 10, and T is a continuous function of z (linear), then yes, T will take all values between 120 and 130. Hence, the range is [120, 130].Wait, but let me check if when z is 0, x = 10, y = 20. Are all variables non-negative? x=10, y=20, z=0. Yes, that's okay. And when z=10, x=20, y=20 -2*10=0. So x=20, y=0, z=10. All variables non-negative here as well. So both endpoints are achievable.Therefore, the range of T is from 120 to 130. So the final answer is boxed{[120,130]}.Wait, but hold on a second. Let me check if there are any other constraints that I might have missed. For example, when z is between 0 and 10, we get x = z + 10, so x ranges from 10 to 20. Similarly, y ranges from 20 down to 0. But do we have to check if there are other constraints when combining variables? For example, even though z is between 0 and 10, does that automatically ensure that x, y, z are non-negative? Let's see.If z=0: x=10, y=20, z=0. All non-negative.If z=10: x=20, y=0, z=10. All non-negative.For z between 0 and 10:x = z + 10: since z ≥0, x ≥10, which is okay.y = 20 - 2z: since z ≤10, then 2z ≤20, so y ≥0.So all variables are non-negative in this range. Therefore, all values of z in [0,10] are permissible, and corresponding x and y are non-negative. Therefore, T=130 - z varies from 130 to 120 as z goes from 0 to 10. Therefore, the range is indeed [120, 130].Therefore, the answer is boxed{[120,130]}.Final AnswerThe range of ( T ) is boxed{[120,130]}.</think>Given the problem where real numbers ( x, y, z geq 0 ) satisfy the equations ( x + y + z = 30 ) and ( 3x + y - z = 50 ), we need to find the range of ( T = 5x + 4y + 2z ).First, we solve the system of equations:1. ( x + y + z = 30 )2. ( 3x + y - z = 50 )Subtracting the first equation from the second:[(3x + y - z) - (x + y + z) = 50 - 30 2x - 2z = 20 x - z = 10 implies x = z + 10]Substituting ( x = z + 10 ) into the first equation:[(z + 10) + y + z = 30 2z + 10 + y = 30 y = 20 - 2z]We ensure non-negativity constraints:- ( x = z + 10 geq 0 ) is satisfied since ( z geq 0 ).- ( y = 20 - 2z geq 0 implies z leq 10 ).- ( z geq 0 ).Thus, ( z ) ranges from 0 to 10. Expressing ( T ) in terms of ( z ):[T = 5x + 4y + 2z = 5(z + 10) + 4(20 - 2z) + 2z = 5z + 50 + 80 - 8z + 2z = -z + 130]Since ( z ) ranges from 0 to 10, ( T ) ranges from:- When ( z = 0 ), ( T = 130 )- When ( z = 10 ), ( T = 120 )Thus, the range of ( T ) is (boxed{[120, 130]}).]

🔑:6. [120,130].Notice,begin{array}{l}T=5 x+4 y+2 z =(x+y+z)+(4 x+3 y+z) =30+(4 x+3 y+z) .end{array}From 4 x+2 y=(x+y+z)+(3 x+y-z)=80Rightarrow T=110+(y+z) text {. }Also, 20=(3 x+y-z)-(x+y+z)=2(x-z),then x-z=10.Since x, z geqslant 0, we have x geqslant 10.Thus, from x+y+z=30, we know y+z leqslant 20.Therefore, T=110+(y+z) leqslant 130 (equality holds when z=0, x=10, y=20).From 4 x+2 y=80, y geqslant 0, we know x leqslant 20.Hence, y+z=30-x geqslant 10.Therefore, T=110+(y+z) geqslant 120 (equality holds when x=20, y=0, z=10).Thus, 120 leqslant T leqslant 130.

🧠:[<think>Okay, let me try to figure out this complex number problem. The question says: The complex number z satisfies (z - 2i) = 3 + 7i. Find the complex number z. Hmm, so I need to find z here. Let me recall my knowledge about complex numbers. First, complex numbers are usually written in the form a + bi, where a is the real part and b is the imaginary part. Here, z is a complex number, and we have an equation involving z. The equation is (z - 2i) equals 3 + 7i. So, to solve for z, I need to isolate it. That should be straightforward, right? If z - 2i = 3 + 7i, then adding 2i to both sides should give z by itself.Let me write that down step by step. Starting with the equation:z - 2i = 3 + 7iTo solve for z, I can add 2i to both sides of the equation. That should cancel out the -2i on the left side. So:z - 2i + 2i = 3 + 7i + 2iSimplifying both sides, the left side becomes z, since -2i and +2i cancel each other. On the right side, we need to combine like terms. The real part is 3, and the imaginary parts are 7i and 2i. Adding those together gives 7i + 2i = 9i. So the right side becomes 3 + 9i.Therefore, z = 3 + 9i.Wait, that seems too simple. Let me check if I did that correctly. Let me substitute z back into the original equation to verify. If z is 3 + 9i, then z - 2i should be (3 + 9i) - 2i. Subtract the imaginary parts: 9i - 2i = 7i. So z - 2i = 3 + 7i, which matches the given equation. Perfect, that checks out. But let me think again. Sometimes with complex numbers, there might be a trick or something, but in this case, it seems straightforward. The equation is linear in z, so solving for z is just a matter of adding 2i to both sides. Alternatively, maybe I can think of z as a complex number with real part a and imaginary part b. Let me set z = a + bi, where a and b are real numbers. Then plug that into the equation:z - 2i = 3 + 7iSubstituting z gives:(a + bi) - 2i = 3 + 7iCombine like terms on the left side:a + (b - 2)i = 3 + 7iSince two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Therefore, setting the real parts equal:a = 3And setting the imaginary parts equal:b - 2 = 7Solving for b:b = 7 + 2 = 9Therefore, z = 3 + 9i, which matches what I found earlier. So both methods lead to the same answer, and verifying by substitution works. Hmm, so I think that's the correct answer. The problem wasn't too complicated, but it's always good to check using different methods to be sure. Sometimes, when dealing with complex numbers, especially in more complicated problems, it's easy to mix up the real and imaginary parts, but here, breaking it down into components helped confirm the result. Another way to look at this is to consider the equation as moving the point z in the complex plane. Subtracting 2i from z translates the point 2 units downward along the imaginary axis, resulting in the complex number 3 + 7i. To reverse this translation, we add 2i to 3 + 7i, moving back up 2 units along the imaginary axis, which gives 3 + 9i. This geometric interpretation also makes sense and supports the algebraic solution.Is there any other aspect I need to consider here? Maybe if there was a typo in the problem, but as stated, the equation is (z - 2i) equals 3 + 7i, so unless there's a different interpretation of the equation, this seems solid. In summary, through algebraic manipulation, substitution, component-wise analysis, and geometric interpretation, all methods lead to z being 3 + 9i. Therefore, I'm confident that the answer is correct.Final AnswerThe complex number ( z ) is boxed{3 + 9i}.</think>Given the equation ( (z - 2i) = 3 + 7i ), we need to find the complex number ( z ).Starting with the equation:[z - 2i = 3 + 7i]To isolate ( z ), we add ( 2i ) to both sides of the equation:[z - 2i + 2i = 3 + 7i + 2i]Simplifying both sides, the left side becomes ( z ), and the right side combines the imaginary parts:[z = 3 + (7i + 2i) = 3 + 9i]To verify, we substitute ( z = 3 + 9i ) back into the original equation:[(3 + 9i) - 2i = 3 + 7i]This simplifies to:[3 + 7i = 3 + 7i]Thus, the solution is confirmed. Additionally, breaking ( z ) into real and imaginary parts ( a ) and ( b ) respectively, we solve:[a + (b - 2)i = 3 + 7i]Equating real and imaginary parts gives:[a = 3 quad text{and} quad b - 2 = 7 implies b = 9]Therefore, ( z = 3 + 9i ).The complex number ( z ) is (boxed{3 + 9i}).,<think>Okay, so I have this problem here: "The complex number ( z ) satisfies ( (z - 2i) = 3 + 7i ). Find the complex number ( z )." Hmm, let's see. I need to find the complex number ( z ) such that when I subtract ( 2i ) from it, the result is ( 3 + 7i ). First, I remember that complex numbers have a real part and an imaginary part. So, if I let ( z = a + bi ), where ( a ) and ( b ) are real numbers, then maybe I can plug that into the equation and solve for ( a ) and ( b ). Let me try that.Starting with the equation:[(z - 2i) = 3 + 7i]Substituting ( z = a + bi ):[(a + bi - 2i) = 3 + 7i]Combine like terms on the left side. The imaginary parts are ( bi ) and ( -2i ), so combining them gives ( (b - 2)i ). The real part is just ( a ). So, the equation becomes:[a + (b - 2)i = 3 + 7i]Now, for two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. That means:1. Real part: ( a = 3 )2. Imaginary part: ( (b - 2) = 7 )Let me check if that makes sense. If the real part of the left side is ( a ), and the real part of the right side is 3, then ( a ) must indeed be 3. Similarly, the imaginary part on the left side is ( (b - 2) ), and on the right side, it's 7. Therefore, ( b - 2 = 7 ), so solving for ( b ), we get ( b = 7 + 2 = 9 ).Therefore, substituting back into ( z = a + bi ), we get:[z = 3 + 9i]But wait, let me verify this. If I take ( z = 3 + 9i ) and subtract ( 2i ), that would be:[(3 + 9i) - 2i = 3 + (9i - 2i) = 3 + 7i]Which matches the right side of the original equation, so that checks out. Alternatively, maybe there's a quicker way to do this without separating into real and imaginary parts. Since the equation is ( z - 2i = 3 + 7i ), I can just solve for ( z ) by adding ( 2i ) to both sides. Let's try that approach.Starting with:[z - 2i = 3 + 7i]Add ( 2i ) to both sides:[z = 3 + 7i + 2i]Combine the imaginary terms:[z = 3 + (7 + 2)i = 3 + 9i]Same result. So that works too. That seems more straightforward. Maybe the first method was breaking it down into components, but algebraically manipulating the equation directly is faster.Just to make sure I didn't make a mistake, let's go through the steps again. Original equation: ( z - 2i = 3 + 7i ). To isolate ( z ), you perform the inverse operation of subtracting ( 2i ), which is adding ( 2i ). So adding ( 2i ) to both sides cancels out the ( -2i ) on the left, leaving ( z = 3 + 7i + 2i ). Combining ( 7i + 2i ) gives ( 9i ), so ( z = 3 + 9i ). Yep, that's correct.I guess another way to think about complex numbers is that they're like vectors in a 2-dimensional space, with the real part as the x-component and the imaginary part as the y-component. So subtracting ( 2i ) from ( z ) is like subtracting the vector (0,2) from the vector (a, b), resulting in (a, b - 2), which equals (3,7). Therefore, solving for a and b gives a=3 and b=9. Same result. So all approaches lead to ( z = 3 + 9i ).I think that's solid. All methods confirm that ( z = 3 + 9i ). The answer makes sense, and the check by substituting back into the original equation works. So I don't see any issues here. Probably this problem was testing the understanding of manipulating complex numbers, which is pretty straightforward once you treat ( i ) as a variable and remember to combine like terms.Final AnswerThe complex number ( z ) is boxed{3 + 9i}.</think>Given the equation ((z - 2i) = 3 + 7i), we need to find the complex number (z).Starting with the equation:[z - 2i = 3 + 7i]To isolate (z), we add (2i) to both sides:[z = 3 + 7i + 2i]Combining the imaginary terms (7i) and (2i) gives:[z = 3 + 9i]We can verify this by substituting (z = 3 + 9i) back into the original equation:[(3 + 9i) - 2i = 3 + 7i]This confirms that the solution is correct. Therefore, the complex number (z) is (boxed{3 + 9i}).]

🔑:Let z=a+bi (where a,b in mathbb{R}). Since (z-2i)=3+7i, we have a+(b-2)i=3+7i. This implies a=3 and b-2=7. Solving these equations, we get b=9. Therefore, z=3+9i. Hence, the answer is boxed{3+9i}.