🧠:[<think>Okay, so I need to solve this fractional equation: 3/2 minus (2x)/(3x - 1) equals 7/(6x - 2). Hmm, let me start by writing it down again to make sure I have it right. The equation is:(3/2) - (2x)/(3x - 1) = 7/(6x - 2)First, I should probably figure out the denominators here so I can find any restrictions on the variable x. The denominators are 2, (3x - 1), and (6x - 2). Let's see when each denominator equals zero because those values would make the equation undefined.Starting with 3x - 1 = 0:3x - 1 = 03x = 1x = 1/3So x can't be 1/3. Next, 6x - 2 = 0:6x - 2 = 06x = 2x = 2/6 = 1/3Oh, both denominators 3x - 1 and 6x - 2 are zero when x is 1/3. So the restriction is x ≠ 1/3. Alright, I need to keep that in mind for the end. If my solution is x = 1/3, I have to reject it.Now, to solve the equation. Since there are denominators, maybe I should eliminate them by finding a common denominator and multiplying both sides by it. Let me note the denominators again: 2, (3x - 1), and (6x - 2). Wait, 6x - 2 is actually 2*(3x - 1). Let me verify that:6x - 2 = 2*(3x - 1) = 6x - 2. Yep, that's correct. So the denominators are 2, (3x - 1), and 2*(3x - 1). Therefore, the least common denominator (LCD) would be 2*(3x - 1). That's probably the best one to use.So, if I multiply every term in the equation by the LCD, which is 2*(3x - 1), that should eliminate all the denominators. Let's try that.Multiplying each term by 2*(3x - 1):First term: (3/2) * 2*(3x - 1) = 3*(3x - 1)Second term: - (2x)/(3x - 1) * 2*(3x - 1) = -2x * 2 = -4xThird term: 7/(6x - 2) * 2*(3x - 1). Wait, 6x - 2 is 2*(3x - 1), so substituting that in, the denominator becomes 2*(3x - 1), so 7 divided by that times 2*(3x - 1) would be 7/(2*(3x -1)) * 2*(3x -1) = 7. So that simplifies to 7.Putting it all together:3*(3x - 1) - 4x = 7Now, simplify the left side:First, distribute the 3 in 3*(3x -1):3*3x = 9x3*(-1) = -3So that's 9x - 3Then subtract 4x: 9x - 3 - 4x = (9x - 4x) - 3 = 5x - 3So the equation becomes:5x - 3 = 7Now, solve for x:Add 3 to both sides:5x - 3 + 3 = 7 + 35x = 10Divide both sides by 5:x = 10/5 = 2So x equals 2. Now, check if this violates any restrictions. Earlier, we saw that x cannot be 1/3, which 2 isn't, so that's okay.But just to be thorough, let me plug x = 2 back into the original equation to verify it works.Original equation:(3/2) - (2x)/(3x -1) = 7/(6x -2)Plugging x = 2:Left side: 3/2 - (2*2)/(3*2 -1) = 3/2 - 4/(6 -1) = 3/2 - 4/5Right side: 7/(6*2 -2) = 7/(12 -2) = 7/10Calculate left side:3/2 - 4/5. To subtract these, common denominator is 10.3/2 = 15/10, 4/5 = 8/10. So 15/10 - 8/10 = 7/10.Right side is 7/10. So both sides equal 7/10. That checks out. Therefore, x = 2 is the solution.Wait, but let me double-check my algebra when I multiplied through by the LCD. Let me go back through the steps again to be sure.Original equation:3/2 - (2x)/(3x -1) = 7/(6x -2)Multiply each term by 2*(3x -1):First term: (3/2)*2*(3x -1) = 3*(3x -1) [since 3/2 * 2 = 3]Second term: - (2x)/(3x -1) * 2*(3x -1) = -2x *2 = -4x [since (3x -1) cancels out]Third term: 7/(6x -2) * 2*(3x -1). Since 6x -2 = 2*(3x -1), substitute:7/(2*(3x -1)) * 2*(3x -1) = 7. Correct, because the denominators cancel.So equation becomes 3*(3x -1) -4x =7. Then expanding:9x -3 -4x =7 → 5x -3=7 → 5x=10 → x=2. All steps correct.Check in original equation as above: correct. So seems like x=2 is the right answer.But just to be absolutely thorough, maybe there's an alternative approach? Let me try cross-multiplying after getting a common denominator or something else.Alternatively, perhaps I can manipulate the equation first.Original equation:3/2 - (2x)/(3x -1) = 7/(6x -2)Notice that 6x -2 is 2*(3x -1). Let me write the equation as:3/2 - (2x)/(3x -1) = 7/(2*(3x -1))So perhaps combine the left side first. To do that, find a common denominator for 3/2 and (2x)/(3x -1). The denominators are 2 and (3x -1), so LCD is 2*(3x -1).Express both terms with that denominator:3/2 = 3*(3x -1)/[2*(3x -1)](2x)/(3x -1) = 2x*2/[2*(3x -1)] = 4x/[2*(3x -1)]So the left side becomes:[3*(3x -1) - 4x]/[2*(3x -1)] = 7/[2*(3x -1)]Since the denominators are the same, set the numerators equal:3*(3x -1) -4x =7Which is exactly the same equation as before. So same result: 9x -3 -4x =7 →5x=10→x=2.So regardless of the method, same answer. So x=2 is correct.Alternatively, maybe cross-multiplying step by step.But seems like all approaches lead to x=2. So confident that's the solution. But since fractional equations can sometimes have extraneous solutions, even though we checked, let's check once again.Original equation with x=2:Left side: 3/2 - (2*2)/(3*2 -1) = 3/2 - 4/5.Convert to decimals to check:3/2 = 1.5, 4/5=0.8. So 1.5 - 0.8 = 0.7.Right side: 7/(6*2 -2) =7/(12-2)=7/10=0.7.Yep, 0.7=0.7. So it's correct.Therefore, the solution is x=2.Final AnswerThe solution is boxed{2}.</think>To solve the fractional equation (dfrac{3}{2} - dfrac{2x}{3x-1} = dfrac{7}{6x-2}):1. Identify restrictions on (x): - Denominators (3x - 1) and (6x - 2) must not be zero. - Solving (3x - 1 = 0) gives (x = dfrac{1}{3}). Thus, (x neq dfrac{1}{3}).2. Find the least common denominator (LCD): - The denominators are (2), (3x - 1), and (6x - 2) (which is (2(3x - 1))). - The LCD is (2(3x - 1)).3. Multiply each term by the LCD: [ 2(3x - 1) left( dfrac{3}{2} right) - 2(3x - 1) left( dfrac{2x}{3x - 1} right) = 2(3x - 1) left( dfrac{7}{6x - 2} right) ] - Simplifying each term: - (2(3x - 1) cdot dfrac{3}{2} = 3(3x - 1)) - (2(3x - 1) cdot dfrac{2x}{3x - 1} = 4x) - (2(3x - 1) cdot dfrac{7}{2(3x - 1)} = 7)4. Form the simplified equation: [ 3(3x - 1) - 4x = 7 ] - Expand and simplify: [ 9x - 3 - 4x = 7 implies 5x - 3 = 7 implies 5x = 10 implies x = 2 ]5. Verify the solution: - Substitute (x = 2) back into the original equation: - Left side: (dfrac{3}{2} - dfrac{4}{5} = dfrac{15}{10} - dfrac{8}{10} = dfrac{7}{10}) - Right side: (dfrac{7}{10}) - Both sides are equal, confirming (x = 2) is correct.The solution is (boxed{2}).,<think>Alright, so I need to solve this fractional equation: 3/2 minus (2x)/(3x - 1) equals 7/(6x - 2). Hmm, okay. Let me try to figure this out step by step. First, fractional equations can sometimes be tricky because of the variables in the denominators. I remember that a good strategy is to eliminate the denominators by finding the least common denominator (LCD) and multiplying each term by the LCD. That should help me get rid of the fractions and make the equation easier to handle.Let me look at the denominators here. The denominators are 2, (3x - 1), and (6x - 2). Wait, 6x - 2 can be factored, maybe? Let me check. 6x - 2 is equal to 2*(3x - 1). Yes, that's right. So, 6x - 2 is 2 times (3x - 1). So, the denominators are 2, (3x - 1), and 2*(3x - 1). Therefore, the LCD would be 2*(3x - 1), since that's the product of the highest powers of all the factors present in the denominators. So, if I multiply each term in the equation by 2*(3x - 1), that should eliminate all the denominators. Let me write that down.Starting with the original equation:3/2 - (2x)/(3x - 1) = 7/(6x - 2)First, note that 6x - 2 is 2*(3x - 1). So, replacing that:3/2 - (2x)/(3x - 1) = 7/[2*(3x - 1)]Now, multiply each term by the LCD, which is 2*(3x - 1):2*(3x - 1)*(3/2) - 2*(3x - 1)*(2x)/(3x - 1) = 2*(3x - 1)*7/[2*(3x - 1)]Simplify each term:First term: 2*(3x - 1)*(3/2). The 2 and the denominator 2 cancel each other, so it becomes (3x - 1)*3 = 3*(3x - 1) = 9x - 3.Second term: -2*(3x - 1)*(2x)/(3x - 1). The (3x - 1) cancels out in numerator and denominator, so we have -2*2x = -4x.Third term: On the right side, 2*(3x - 1)*7/[2*(3x - 1)]. Here, the 2 and (3x - 1) cancel out top and bottom, leaving just 7.So, after simplifying all terms, the equation becomes:9x - 3 - 4x = 7Combine like terms on the left side:(9x - 4x) - 3 = 75x - 3 = 7Now, solve for x. Add 3 to both sides:5x = 10Divide both sides by 5:x = 2Wait, but before I conclude, I need to check if this solution is valid. Since there were denominators in the original equation, I need to make sure that x = 2 doesn't make any denominator zero. Let's check each denominator:For 3x - 1: When x = 2, 3*2 - 1 = 6 - 1 = 5 ≠ 0.For 6x - 2: When x = 2, 6*2 - 2 = 12 - 2 = 10 ≠ 0.The denominators are 2, 5, and 10, none of which are zero. So, x = 2 is a valid solution.Let me double-check by substituting x = 2 into the original equation to ensure both sides are equal.Left side: 3/2 - (2x)/(3x - 1) = 3/2 - (2*2)/(3*2 - 1) = 3/2 - 4/(6 - 1) = 3/2 - 4/5.Convert to common denominator, which is 10:3/2 = 15/10, 4/5 = 8/10. So, 15/10 - 8/10 = 7/10.Right side: 7/(6x - 2) = 7/(6*2 - 2) = 7/(12 - 2) = 7/10.Yes, both sides equal 7/10 when x = 2. So, the solution checks out.Therefore, x = 2 is the correct solution.Final AnswerThe solution is boxed{2}.</think>To solve the fractional equation (dfrac{3}{2} - dfrac{2x}{3x-1} = dfrac{7}{6x-2}), we start by identifying the denominators: (2), (3x - 1), and (6x - 2). Noting that (6x - 2 = 2(3x - 1)), the least common denominator (LCD) is (2(3x - 1)).We multiply each term by the LCD (2(3x - 1)):[2(3x - 1) cdot dfrac{3}{2} - 2(3x - 1) cdot dfrac{2x}{3x - 1} = 2(3x - 1) cdot dfrac{7}{2(3x - 1)}]Simplifying each term:1. The first term: (2(3x - 1) cdot dfrac{3}{2} = 3(3x - 1) = 9x - 3)2. The second term: (-2(3x - 1) cdot dfrac{2x}{3x - 1} = -4x)3. The third term: (2(3x - 1) cdot dfrac{7}{2(3x - 1)} = 7)This results in the equation:[9x - 3 - 4x = 7]Combining like terms:[5x - 3 = 7]Solving for (x):[5x = 10 implies x = 2]We verify that (x = 2) does not make any denominator zero. Substituting (x = 2) into the original equation confirms both sides are equal. Thus, the solution is (boxed{2}).]

🔑:To solve the original equation, we can transform it into 3(3x-1)-4x=7, After simplifying, we get: 5x=10, Solving this, we find: x=2, Upon verification, x=2 is a solution to the original equation, Therefore, the solution to the original equation is boxed{x=2}.

🧠:[<think>Okay, so I need to solve this expression: the square root of 2 and 1/4 minus (-9.6) to the power of 0, minus (3 and 3/8) to the power of -2/3, plus log base 4 of 3 multiplied by log base 9 of 2. Hmm, that's a bit of a mouthful. Let me break it down step by step.First, let's handle each term individually. Maybe if I tackle them one by one, it'll make more sense. Let's start with the first term: (2 1/4)^(1/2). That's the square root of 2 and 1/4. Wait, 2 and 1/4 is a mixed number. I should convert that to an improper fraction or a decimal to make it easier to work with. Let me convert it to a decimal. 1/4 is 0.25, so 2.25. So the square root of 2.25. Hmm, what's the square root of 2.25? I remember that 1.5 squared is 2.25 because 1.5 times 1.5 is 2.25. So that part is 1.5. Got that.Next term is (-9.6)^0. Any number to the power of 0 is 1, right? But wait, the base here is negative: -9.6. But even if it's negative, as long as it's not zero, anything to the power of 0 is 1. So this term is 1. So that's straightforward.Third term is (3 3/8)^(-2/3). Hmm, negative exponent, so that's the reciprocal. So first, let's convert 3 3/8 to an improper fraction. 3 is 24/8, so 24/8 + 3/8 is 27/8. So it's (27/8)^(-2/3). Which is the same as (8/27)^(2/3). Let me remember how exponents work here. When you have a fraction exponent, the denominator is the root, and the numerator is the power. So 2/3 exponent means cube root first, then square. So let's do that.First, cube root of 8/27. The cube root of 8 is 2, and the cube root of 27 is 3. So that's 2/3. Then squaring that gives (2/3)^2 = 4/9. So this term is 4/9. Wait, but let me check: since it's (27/8)^(-2/3), we flip the base to 8/27 and make the exponent positive, then take the 2/3 power. Yeah, so cube root of 8/27 is 2/3, then squaring gives 4/9. Correct. So the third term is 4/9. Got it.Fourth term is log base 4 of 3 multiplied by log base 9 of 2. Hmm, logarithms with different bases. Maybe I can use the change of base formula or some properties of logarithms to simplify this product. Let me recall that log_b a * log_c b can sometimes be simplified, but here it's log_4 3 * log_9 2. Not sure if there's a direct relation. Let me think.Alternatively, I can express both logs in terms of natural logs or common logs. Let's try that. The change of base formula says log_b a = ln a / ln b. So log_4 3 is ln 3 / ln 4, and log_9 2 is ln 2 / ln 9. So multiplying them together gives (ln 3 / ln 4) * (ln 2 / ln 9). Let me see if that simplifies.Alternatively, maybe express both logs with base 2 or base 3? Since 4 is 2^2, 9 is 3^2. Let's try that. So log base 4 of 3 can be written as log_{2^2} 3, which is (1/2) log_2 3. Similarly, log base 9 of 2 is log_{3^2} 2, which is (1/2) log_3 2. So then the product becomes (1/2 log_2 3) * (1/2 log_3 2) = (1/4) (log_2 3 * log_3 2). Now, log_b a * log_a b = 1, right? Because log_b a = 1 / log_a b. So log_2 3 * log_3 2 = 1. Therefore, this product is (1/4) * 1 = 1/4. Oh! That's a nice simplification. So the fourth term is 1/4. Cool, so that part is 1/4.Okay, let's recap all the terms:1. sqrt(2 1/4) = 1.52. (-9.6)^0 = 13. (3 3/8)^(-2/3) = 4/9 ≈ 0.444...4. log_4 3 * log_9 2 = 1/4 = 0.25Now, putting it all together:First term: 1.5Second term: -1 (because it's minus (-9.6)^0, which is minus 1)Third term: -4/9 (because it's minus (3 3/8)^(-2/3))Fourth term: +1/4So the entire expression is 1.5 - 1 - 4/9 + 1/4.Let me compute that step by step.First, 1.5 - 1 = 0.5.Then, 0.5 - 4/9. Let's convert 0.5 to 1/2. So 1/2 - 4/9. To subtract these, find a common denominator. The common denominator of 2 and 9 is 18.1/2 = 9/18, 4/9 = 8/18. So 9/18 - 8/18 = 1/18 ≈ 0.0555...Then, add 1/4. So 1/18 + 1/4. Convert to common denominator, which is 36.1/18 = 2/36, 1/4 = 9/36. So 2/36 + 9/36 = 11/36 ≈ 0.3055...Wait, so 1.5 -1 is 0.5. Then 0.5 is 1/2. Then 1/2 - 4/9 is 1/18. Then 1/18 + 1/4 is 11/36. So the final result is 11/36. Is that correct?Wait, let me verify each step again.First, 1.5 (which is 3/2) minus 1 is 1/2.Then, 1/2 minus 4/9. Let me compute that:1/2 - 4/9 = (9/18 - 8/18) = 1/18. Correct.Then, 1/18 plus 1/4:1/18 + 1/4 = (2/36 + 9/36) = 11/36. Correct.So the total is 11/36, which is approximately 0.30555...But maybe the answer is supposed to be in fractions? Let me check if 11/36 can be simplified. 11 is a prime number, so unless 36 is a multiple of 11, which it's not. So 11/36 is the simplified fraction.Alternatively, maybe convert all terms to fractions with denominators of 36 to see:1.5 is 3/2 = 54/36-1 is -36/36-4/9 is -16/36+1/4 is +9/36So adding them up: 54 -36 -16 +9 = (54 -36) + (-16 +9) = 18 -7 = 11. So 11/36. Yep, that checks out.So the final answer is 11/36. Let me see if there's another way to present that, but 11/36 is already in simplest terms. So I think that's the answer. Let me just confirm each step once more to be safe.First term: sqrt(2.25) is indeed 1.5. Check.Second term: (-9.6)^0 is 1. Check.Third term: (27/8)^(-2/3) becomes (8/27)^(2/3). Cube root of 8 is 2, cube root of 27 is 3. Squaring gives 4/9. Check.Fourth term: log_4 3 * log_9 2. Converting to base 2 and 3 gives us (1/2 log_2 3)(1/2 log_3 2) = 1/4 (log_2 3 * log_3 2) = 1/4 *1 = 1/4. Check.Then, combining all terms:1.5 -1 -4/9 +1/4 = (1.5 -1) + (-4/9 +1/4) = 0.5 + (-16/36 +9/36) = 0.5 -7/36.Wait, 0.5 is 18/36. So 18/36 -7/36 =11/36. Yep, same result.Okay, so all steps check out. So the answer is 11/36. I think that's it. I don't see any mistakes in the reasoning.Final Answerboxed{dfrac{11}{36}}</think>First, we convert (2 dfrac{1}{4}) to a decimal, which is (2.25). The square root of (2.25) is (1.5).Next, ((-9.6)^0) is (1) because any non-zero number raised to the power of 0 is 1.For the term ((3 dfrac{3}{8})^{-frac{2}{3}}), we convert (3 dfrac{3}{8}) to the improper fraction (dfrac{27}{8}). Taking the reciprocal and applying the exponent, we get (left(dfrac{8}{27}right)^{frac{2}{3}}). The cube root of (dfrac{8}{27}) is (dfrac{2}{3}), and squaring this gives (left(dfrac{2}{3}right)^2 = dfrac{4}{9}).Next, we simplify (log_4 3 cdot log_9 2). Using the change of base formula, we convert these logarithms to bases 2 and 3: (log_{2^2} 3 = dfrac{1}{2} log_2 3) and (log_{3^2} 2 = dfrac{1}{2} log_3 2). The product becomes (dfrac{1}{4} (log_2 3 cdot log_3 2)). Since (log_2 3 cdot log_3 2 = 1), this term simplifies to (dfrac{1}{4}).Combining all terms:[1.5 - 1 - dfrac{4}{9} + dfrac{1}{4}]Converting each term to fractions with a common denominator of 36:[1.5 = dfrac{54}{36}, quad -1 = -dfrac{36}{36}, quad -dfrac{4}{9} = -dfrac{16}{36}, quad dfrac{1}{4} = dfrac{9}{36}]Adding these together:[dfrac{54}{36} - dfrac{36}{36} - dfrac{16}{36} + dfrac{9}{36} = dfrac{11}{36}]Thus, the final answer is (boxed{dfrac{11}{36}}).,<think>Okay, let's try to solve this problem step by step. The expression given is:({{(2dfrac{1}{4})}^{frac{1}{2}}} - {{(-9.6)}^{0}} - {{(3dfrac{3}{8})}^{-frac{2}{3}}} + {{log }_{4}}3 cdot {{log }_{9}}2).Hmm, that's a bit of a mouthful. Let's break it down term by term. I need to handle each part separately and then combine them. Let me start with the first term: ({{(2dfrac{1}{4})}^{frac{1}{2}}}).First, converting the mixed number (2dfrac{1}{4}) into an improper fraction. 2 is equal to (dfrac{8}{4}), so adding (dfrac{1}{4}) gives (dfrac{9}{4}). So, the first term is (left(dfrac{9}{4}right)^{frac{1}{2}}).The square root of (dfrac{9}{4}) is straightforward. The square root of 9 is 3, and the square root of 4 is 2, so this simplifies to (dfrac{3}{2}) or 1.5. Got that.Next term is (-{{(-9.6)}^{0}}). Any non-zero number raised to the power of 0 is 1. Wait, but the base here is (-9.6), which is negative. However, even if the base is negative, as long as it's not zero, the result of raising it to the power of 0 is still 1. So, ((-9.6)^0 = 1). Therefore, this term becomes (-1). So far, so good.Moving on to the third term: (-{{(3dfrac{3}{8})}^{-frac{2}{3}}}). Let me parse this. First, convert (3dfrac{3}{8}) to an improper fraction. 3 is (dfrac{24}{8}), so adding (dfrac{3}{8}) gives (dfrac{27}{8}). So the term is (-left(dfrac{27}{8}right)^{-frac{2}{3}}).Negative exponents mean reciprocals. So, (left(dfrac{27}{8}right)^{-frac{2}{3}} = left(dfrac{8}{27}right)^{frac{2}{3}}). Let's compute that. The exponent (frac{2}{3}) can be broken down into taking the cube root first and then squaring the result. So, cube root of (dfrac{8}{27}) is (dfrac{2}{3}) because (2^3 = 8) and (3^3 = 27). Then squaring (dfrac{2}{3}) gives (dfrac{4}{9}). Therefore, this term becomes (-dfrac{4}{9}). But wait, there's already a negative sign in front, so the entire term is (- times dfrac{4}{9} = -dfrac{4}{9}). Wait, let me double-check that. The original term is (-{{(3dfrac{3}{8})}^{-frac{2}{3}}}). So, ({{(3dfrac{3}{8})}^{-frac{2}{3}}}) is (dfrac{4}{9}), so the term is (-dfrac{4}{9}). Yes, that's correct.Alright, now the last term: ({{log }_{4}}3 cdot {{log }_{9}}2). This looks a bit tricky with logarithms. Let me recall logarithmic identities. Maybe using the change of base formula? Let's see.The change of base formula says that ({{log }_{a}b} = dfrac{ln b}{ln a}) or (dfrac{{{log }_{c}b}}{{{log }_{c}a}}) for any positive c ≠ 1. Alternatively, maybe we can express both logarithms with the same base or find a relationship between them.Alternatively, notice that 4 is (2^2) and 9 is (3^2). Let me express the logs in terms of base 2 and base 3. Let's try that.First, ({{log }_{4}3}). Since 4 is (2^2), we can write this as (dfrac{{{log}_{2}3}}{{{log}_{2}4}} = dfrac{{{log}_{2}3}}{2}).Similarly, ({{log }_{9}2}). Since 9 is (3^2), we can write this as (dfrac{{{log}_{3}2}}{{{log}_{3}9}} = dfrac{{{log}_{3}2}}{2}).Therefore, the product ({{log }_{4}3} cdot {{log }_{9}2}) becomes (left(dfrac{{{log}_{2}3}}{2}right) cdot left(dfrac{{{log}_{3}2}}{2}right)) = (dfrac{{{log}_{2}3} cdot {{log}_{3}2}}{4}).Hmm, now ({{log}_{2}3} cdot {{log}_{3}2}). There's a logarithm identity that states ({{log}_{a}b} cdot {{log}_{b}a} = 1). Let me verify that. Let ({{log}_{a}b} = x), then (a^x = b). Similarly, ({{log}_{b}a} = 1/x) because (b^{1/x} = a). So, their product is (x cdot (1/x) = 1). Yes, that's correct. Therefore, ({{log}_{2}3} cdot {{log}_{3}2} = 1). Therefore, the entire product becomes (dfrac{1}{4}).So, the last term is (dfrac{1}{4}).Now, putting all the terms together:First term: (dfrac{3}{2})Second term: (-1)Third term: (-dfrac{4}{9})Fourth term: (+dfrac{1}{4})So, the expression simplifies to:(dfrac{3}{2} - 1 - dfrac{4}{9} + dfrac{1}{4})Let me convert all fractions to have a common denominator to add them up. The denominators are 2, 1, 9, 4. The least common multiple (LCM) of 2, 9, and 4. Let's see. Prime factors:2: 29: 3²4: 2²So LCM is 2² * 3² = 4 * 9 = 36. So, converting all fractions to 36ths.First term: (dfrac{3}{2} = dfrac{3 times 18}{2 times 18} = dfrac{54}{36})Second term: (-1 = -dfrac{36}{36})Third term: (-dfrac{4}{9} = -dfrac{4 times 4}{9 times 4} = -dfrac{16}{36})Fourth term: (+dfrac{1}{4} = dfrac{9}{36})Adding them all together:(dfrac{54}{36} - dfrac{36}{36} - dfrac{16}{36} + dfrac{9}{36})Calculating step by step:Start with 54 - 36 = 1818 - 16 = 22 + 9 = 11So total is (dfrac{11}{36})Wait, let me check that again. Wait, 54 - 36 is 18, then 18 -16 is 2, then 2 +9 is 11. So, 11/36.But let me verify once again:First term: 54/36Second term: -36/36Third term: -16/36Fourth term: +9/36So adding all numerators:54 - 36 -16 +9 = (54 -36) + (-16 +9) = 18 + (-7) = 11. Yes, 11/36.So, the final result is 11/36.Wait, but let me check if I converted all terms correctly. Let me verify each term:First term: 3/2 = 1.5, which is 54/36. Correct.Second term: -1 is -36/36. Correct.Third term: -4/9 is approximately -0.444..., which is -16/36. Correct.Fourth term: 1/4 is 0.25, which is 9/36. Correct.So adding them all: 54 -36 = 18; 18 -16=2; 2+9=11. So 11/36 is correct. So the answer is 11/36.But let me check once again the logarithmic term. Let me recap:({{log }_{4}3} cdot {{log }_{9}2}). So, log base 4 of 3 is log3 / log4, and log base 9 of 2 is log2 / log9. Then, their product is (log3 / log4)(log2 / log9). Let's compute this with natural logs for verification.Let me compute each logarithm:log4(3) = ln3 / ln4log9(2) = ln2 / ln9So, product = (ln3 / ln4)(ln2 / ln9) = (ln3 * ln2) / (ln4 * ln9)But ln4 = 2 ln2, ln9 = 2 ln3.Therefore, denominator is (2 ln2)(2 ln3) = 4 ln2 ln3Numerator is ln3 ln2Thus, the product is (ln3 ln2) / (4 ln2 ln3) = 1/4. Correct. So the logarithmic term is indeed 1/4.Therefore, the calculation is correct.Hence, the answer is 11/36.Final Answerboxed{dfrac{11}{36}}</think>First, convert the mixed numbers to improper fractions:- (2dfrac{1}{4} = dfrac{9}{4})- (3dfrac{3}{8} = dfrac{27}{8})Calculate each term separately:1. ({{(2dfrac{1}{4})}^{frac{1}{2}}} = left(dfrac{9}{4}right)^{frac{1}{2}} = dfrac{3}{2})2. ({{(-9.6)}^{0}} = 1), so the term is (-1)3. ({{(3dfrac{3}{8})}^{-frac{2}{3}}} = left(dfrac{27}{8}right)^{-frac{2}{3}} = left(dfrac{8}{27}right)^{frac{2}{3}} = left(dfrac{2}{3}right)^2 = dfrac{4}{9}), so the term is (-dfrac{4}{9})4. ({{log }_{4}}3 cdot {{log }_{9}}2): - ({{log }_{4}3} = dfrac{{{log}_{2}3}}{2}) - ({{log }_{9}2} = dfrac{{{log}_{3}2}}{2}) - Product: (left(dfrac{{{log}_{2}3}}{2}right) cdot left(dfrac{{{log}_{3}2}}{2}right) = dfrac{1}{4})Combine all terms:- (dfrac{3}{2} - 1 - dfrac{4}{9} + dfrac{1}{4})Convert to a common denominator (36):- (dfrac{3}{2} = dfrac{54}{36})- (-1 = -dfrac{36}{36})- (-dfrac{4}{9} = -dfrac{16}{36})- (dfrac{1}{4} = dfrac{9}{36})Add the fractions:- (dfrac{54}{36} - dfrac{36}{36} - dfrac{16}{36} + dfrac{9}{36} = dfrac{11}{36})Final answer:(boxed{dfrac{11}{36}})]

🔑:AnalysisThis question tests the operations of powers and logarithms.According to the rules of operations for powers and logarithms, simplification can be done.SolutionSolution: ({(2 dfrac{1}{4})}^{frac{1}{2}}−{(−9.6)}^{0}−{(3 dfrac{3}{8})}^{-frac{2}{3}}+{log }_{4}3cdot{log }_{9}2 ={left( dfrac{9}{4}right)}^{frac{1}{2}}-1-{left( dfrac{27}{8}right)}^{-frac{2}{3}}+ dfrac{1}{4}{log }_{2}3{log }_{3}2= dfrac{3}{2}-1- dfrac{4}{9}+ dfrac{1}{4}= dfrac{11}{36}),Therefore, the answer is boxed{dfrac{11}{36}}.

🧠:[<think>Okay, so I need to figure out how many different ways I can arrange 3 red beads and 2 blue beads on a ring. The problem mentions it's a ring, which makes me think that rotations and reflections might count as the same arrangement. Hmm, right, because if you can rotate the ring or flip it over, some arrangements that look different when laid out straight might actually be the same on a ring.First, let me recall if there's a formula for counting distinct arrangements on a ring. I think this is related to circular permutations and maybe something called Burnside's lemma? Burnside's lemma helps count distinct arrangements accounting for symmetry operations like rotations and reflections. Let me try to remember how that works.For circular arrangements, the number of distinct necklaces (or in this case, bead arrangements) can be calculated by considering the group of symmetries, which includes rotations and reflections. Burnside's lemma states that the number of distinct arrangements is equal to the average number of fixed points of these group actions. So, I need to calculate the number of colorings fixed by each symmetry operation and then take the average.First, let me note that there are 5 beads total (3 red and 2 blue). Since it's a ring, the group of symmetries is the dihedral group D5, which has 10 elements: 5 rotations and 5 reflections.The rotations are by 0°, 72°, 144°, 216°, and 288° (i.e., 0, 1, 2, 3, 4 steps in terms of positions). The reflections are over 5 axes of symmetry. For a ring with an odd number of beads, each reflection axis goes through one bead and the midpoint between two beads. Wait, but 5 is a prime number? Wait, 5 is prime, so maybe that's useful.But let me get back to Burnside. The formula is:Number of distinct arrangements = (1/|G|) * Σ (number of colorings fixed by each group element)Where |G| is the order of the group, which here is 10.So, I need to calculate for each rotation and reflection, how many colorings remain fixed under that symmetry.Starting with the rotations:1. Identity rotation (0°): Every coloring is fixed by the identity. So the number of fixed colorings here is the total number of linear arrangements divided by... Wait, no. Wait, actually, for the identity, all possible colorings are fixed. But since the beads are arranged on a ring, does that affect the count? Wait, no, the total number of colorings without considering symmetry is C(5,3) = 10, since we choose 3 positions out of 5 to be red, and the rest blue. So, there are 10 possible colorings.Therefore, the identity element fixes all 10 colorings.2. Rotation by 72° (1 step): For a coloring to be fixed under this rotation, rotating the ring by one bead must result in the same coloring. So, all beads must be the same color, but we have 3 red and 2 blue beads. Since the counts are different, there's no way a rotation by 72° can fix a coloring unless all beads are the same color. But since we have different numbers of red and blue beads, there are no colorings fixed by a rotation of 72°. So, fixed colorings here: 0.3. Rotation by 144° (2 steps): Similarly, rotating by two positions. For a coloring to be fixed under this rotation, the color of each bead must be the same as the bead two positions away. So, this would imply that the beads are partitioned into cycles. Since 5 is prime, rotating by 2 steps is a generator of the cyclic group, so the cycle structure is a single cycle of length 5. Wait, but 5 is prime, so any rotation except the identity will cycle all beads. Therefore, similar to rotation by 1 step, unless all beads are the same color, which they aren't, there are no fixed colorings. So, fixed colorings: 0.Same logic applies to rotations by 216° (3 steps) and 288° (4 steps). Since 3 and 4 are co-prime with 5, those rotations also generate the entire cyclic group. Therefore, no fixed colorings except if all beads are the same color, which isn't the case. So, fixed colorings for each of these rotations: 0.So, total fixed colorings from rotations: 10 (identity) + 0 + 0 + 0 + 0 = 10.Now, moving on to reflections. There are 5 reflections, each corresponding to a line of symmetry. For a ring with 5 beads, each reflection will fix one bead and swap the others in pairs. Wait, but 5 is odd, so each reflection axis passes through one bead and the midpoint between two beads. Therefore, each reflection will fix one bead and swap the other two pairs.So, for each reflection, a fixed coloring must have the fixed bead's color unchanged, and the other beads must be symmetric with respect to the axis. Since the other beads are swapped in pairs, their colors must be the same as their pair.So, let's model this. For a given reflection, bead 1 is fixed, beads 2 and 5 are swapped, beads 3 and 4 are swapped. So, to have a coloring fixed by this reflection, bead 1 can be either red or blue, beads 2 and 5 must be the same color, beads 3 and 4 must be the same color.So, the coloring is determined by the color of bead 1, the color of beads 2 and 5, and the color of beads 3 and 4. However, the total number of red beads must be 3 and blue beads 2.So, let's compute the number of colorings fixed by each reflection. Let's suppose bead 1 is fixed. Then:Case 1: Bead 1 is red. Then, beads 2 and 5 must be either both red or both blue, and beads 3 and 4 must be either both red or both blue. The total number of red beads would be 1 (bead1) + 2*(beads2&5) + 2*(beads3&4). Wait, no, beads 2 and 5 are two beads, so if they are both red, that's 2 reds. Similarly for beads 3 and 4.So, bead1 is red. Then, we have 2 more red beads needed (since total is 3). Let me denote:Let a = color of beads 2 and 5 (each can be R or B)Let b = color of beads 3 and 4 (each can be R or B)Total red beads: 1 (bead1) + (if a=R then 2 else 0) + (if b=R then 2 else 0). This must equal 3.So, 1 + 2*(number of R in a and b) = 3.Thus, 2*(number of R in a and b) = 2 => number of R in a and b is 1.So, either a=R and b=B, or a=B and b=R.So, there are 2 possibilities when bead1 is red.Case 2: Bead1 is blue. Then, we need 3 red beads in the remaining beads. Since bead1 is blue, beads 2,5 and 3,4 must contribute 3 reds. Each pair (2,5 and 3,4) can be either both R or both B.Total red beads: 0 (bead1) + 2*(number of R in a and b) = 3.But 2*(number of R in a and b) = 3. But 3 is not divisible by 2, so this is impossible. Therefore, if bead1 is blue, there are no colorings fixed by the reflection.Therefore, for each reflection, the number of fixed colorings is 2 (when bead1 is red and exactly one pair is R) plus 0 (when bead1 is blue) = 2.But wait, each reflection might fix a different bead, right? The axes of reflection in a 5-bead ring go through each bead and the midpoint between two others. So, each reflection fixes a different bead. Therefore, for each reflection, depending on which bead is fixed, the count might be the same. But since the beads are indistinct except for their positions, and we are considering colorings up to rotation and reflection, perhaps each reflection will have the same number of fixed colorings. Let me check.Suppose another reflection that fixes bead2 instead of bead1. Then, similar logic applies: bead2 is fixed, beads1 and3 are swapped, beads4 and5 are swapped. Then, bead2's color is fixed, beads1 and3 must be the same, beads4 and5 must be the same. Then, total red beads would be:If bead2 is red: 1 + 2*(number of R in a and b) = 3, so same as before, leading to 2 possibilities.If bead2 is blue: 0 + 2*(number of R in a and b) = 3, which is impossible.Therefore, regardless of which bead is fixed by the reflection, the number of fixed colorings is 2. Since there are 5 reflections, each contributing 2 fixed colorings, the total fixed colorings from reflections is 5*2=10.Therefore, total fixed colorings from all group elements: rotations (10) + reflections (10) = 20.Then, by Burnside's lemma, the number of distinct colorings is 20 / 10 = 2.Wait, that seems low. Wait, but let me check again. So according to this calculation, there are only 2 distinct arrangements. But let's see. Let's try to list them.If all red beads are together: RRRBB. Since it's a ring, arranging three red beads together and the two blue beads next to each other. But on a ring, rotating or reflecting this would give the same arrangement. Alternatively, if the red beads are not all together. Wait, but with three red beads and two blue beads on a ring, can they be arranged in a non-symmetric way?Wait, let's consider possible configurations:1. All three red beads are consecutive: R R R B B. On a ring, any rotation is equivalent, so this is one arrangement.2. Two red beads are consecutive, and the third is separated by one blue bead: R R B R B. But on a ring, this might be equivalent to another arrangement. Wait, let's see. Let's fix a bead: Starting with a red bead, then another red, then blue, then red, then blue. If we rotate this, starting at the next bead: R B R B R. But this is the same as the original if we rotate it by two positions. Wait, actually, maybe this is equivalent to the first case?Wait, no. Wait, R R B R B. If you rotate by two positions, you get B R B R R. Which is a reflection of the original. So, depending on reflection, maybe these are equivalent. Wait, this is confusing.Alternatively, maybe there are two distinct arrangements: one where the red beads are all together, and one where they are separated. But in the case of 3 red and 2 blue beads, is there a way to arrange them so that the red beads are not all consecutive?Let me try to imagine. If we spread out the red beads, since there are two blue beads, the maximum spacing between red beads would be limited. For example: R B R B R. That uses three red beads and two blue beads. But wait, in a ring, this would place the red beads each separated by one blue bead. However, with three red beads and two blue beads, this would actually require three blue beads (since each red is separated by a blue), but we only have two blue beads. Wait, no, let's see:Wait, if we have R B R B R, that is a sequence of beads: red, blue, red, blue, red. Then, on a ring, this uses three reds and two blues. Each red is separated by a blue. But since it's a ring, the last red connects back to the first red, which is separated by a blue. Wait, but between the last red and the first red, there's a blue? Wait, no. Wait, let's count:Positions 1: R2: B3: R4: B5: RThen, position 5 is adjacent to position 1. So between position 5 and 1, there's a connection. Since position 5 is R and position 1 is R, so actually, there is no blue bead between the last and first red. Therefore, actually, this arrangement has two blue beads separating some red beads but not all. Wait, the red beads at positions 1,3,5 are each separated by a blue bead between 1 and 2, 3 and 4, but between 5 and 1, they are adjacent. So actually, this arrangement has two blue beads and three red beads with two instances of red beads separated by a blue and one instance of two red beads adjacent (positions 5 and 1). Wait, but in reality, on the ring, positions 5 and 1 are adjacent, so in this arrangement, there are two single blue beads between some reds and two reds adjacent. Wait, perhaps this is actually equivalent to the arrangement where two reds are adjacent and the third is separated by a blue. Wait, but how?Alternatively, perhaps this is a different arrangement. Wait, if you have three red beads arranged such that two are adjacent and the third is separated by one blue bead from the pair. For example: R R B R B. On the ring, this would have two reds together, then a blue, then a red, then a blue. But since it's a ring, the last blue is adjacent to the first R. So, between the last blue and the first R, there's no bead. So, actually, this is equivalent to R R B R B, which has a blue between the first two reds and the third red, and another blue next to the third red. Wait, but in this case, the two blue beads are adjacent? No, in R R B R B, the blues are separated by a red and a blue. Wait, maybe my visualization is off.Let me list the beads in order:1: R2: R3: B4: R5: BThen, position 5 is adjacent to position 1. So, between 5 (B) and 1 (R), there's an adjacency. So, the blue at 5 is next to red at 1 and red at 4. Wait, no, position 5 is adjacent to 4 and 1. So, position 5 (B) is between 4 (R) and 1 (R). So, in this arrangement, the blues are at positions 3 and 5. Position 3 is between R2 and R4, and position 5 is between R4 and R1.So, in this arrangement, the two blue beads are each between a pair of red beads. But how is that different from the case where all three red beads are together?If all three red beads are together: positions 1,2,3 are R, R, R, then positions 4 and 5 are B, B. Then, the blues are adjacent. So, in this case, the blues are next to each other, whereas in the other arrangement, the blues are separated by red beads.Therefore, these are two different arrangements: one where the blue beads are adjacent, and one where they are separated. Therefore, there should be at least two distinct colorings, which matches the Burnside's lemma result of 2. So, that seems correct.Therefore, according to Burnside's lemma, the answer is 2. But let me confirm once again.Total fixed colorings:- Identity: 10- Rotations (non-identity): 0 each, total 0- Reflections: 2 each, total 10Total: 10 + 0 + 10 = 20Average: 20 / 10 = 2So, the answer is 2. Therefore, there are 2 distinct schemes.Wait, but the example in the problem statement is Example 4-9, and the figure is 4-11. The user might be referring to a specific figure, but since I don't have access to the figure, I have to assume the problem is general. But according to the calculation, it's 2.Alternatively, maybe I made a mistake in the reflection part. Let me check again.For each reflection, fixed colorings: If the fixed bead is red, then the remaining beads must be arranged such that two pairs are colored, with exactly one pair red and the other pair blue. So, two possibilities. If the fixed bead is blue, it's impossible. So, 2 per reflection.But in our case, the colorings are with 3 red and 2 blue beads, so when the fixed bead is red, the remaining four beads (two pairs) need to contribute 2 reds. Since each pair is either both red or both blue, we can have one pair red and one pair blue, which gives exactly 2 reds. Hence, two colorings: one where the first pair is red and the second is blue, and vice versa. But since the pairs are determined by the reflection axis, swapping the pairs would give a different coloring, but maybe under rotation they are equivalent? Wait, no. Because the reflection axis fixes a bead, so different reflection axes fix different beads. But since we are considering all reflections, each reflection's fixed colorings are counted separately.But in reality, when considering all symmetries, some colorings might be counted multiple times across different reflections. However, Burnside's lemma accounts for all fixed colorings across all symmetries, regardless of overlap.Wait, but if each reflection contributes 2 fixed colorings, and there are 5 reflections, that's 10. Adding the 10 from identity gives 20. Divided by 10, gives 2. So the conclusion holds.Alternatively, perhaps the two distinct colorings are the ones where the two blue beads are adjacent or not. Which would correspond to the two cases we thought of earlier: all reds together (blues adjacent) and reds separated with blues apart.Therefore, the answer is 2. So, I think the calculation is correct.Final AnswerThe number of different schemes is boxed{2}.</think>To determine the number of distinct schemes for arranging 3 red beads and 2 blue beads on a ring, we use Burnside's lemma to account for rotational and reflectional symmetries. The dihedral group (D_5) has 10 elements (5 rotations and 5 reflections).1. Rotations: - Identity rotation (0°): All 10 possible colorings (choosing 3 positions out of 5 for red beads) are fixed. - Rotations by 72°, 144°, 216°, and 288°: No coloring can be fixed because the number of red and blue beads is not equal, so there are 0 fixed colorings for each rotation. Total fixed colorings from rotations: (10 + 0 + 0 + 0 + 0 = 10).2. Reflections: - Each reflection fixes one bead and swaps the others in pairs. For a fixed bead (red or blue), the remaining pairs must be colored such that the total number of red beads is 3. - If the fixed bead is red, the remaining two pairs must contribute 2 red beads (one pair red and one pair blue), giving 2 possibilities per reflection. - If the fixed bead is blue, it is impossible to achieve 3 red beads, so there are 0 possibilities. Each reflection contributes 2 fixed colorings. With 5 reflections, the total fixed colorings from reflections: (5 times 2 = 10).Using Burnside's lemma, the number of distinct colorings is the average number of fixed colorings:[frac{10 text{ (rotations)} + 10 text{ (reflections)}}{10} = frac{20}{10} = 2]Thus, the number of different schemes is (boxed{2}).,<think>Alright, let's try to figure out how many different ways we can arrange 3 red beads and 2 blue beads on a ring. So, the problem is about counting distinct color arrangements considering rotational symmetry. Hmm, since it's a ring, rotations that make the same color sequence should be considered identical. Also, maybe reflections? The problem doesn't specify, but usually, unless stated otherwise, I think we consider both rotations and reflections as equivalent. But wait, the example in the book (since it mentions Figure 4-11) might have a specific method. Let me recall Burnside's lemma because that's often used in counting distinct arrangements under group actions.Burnside's lemma says the number of distinct colorings is equal to the average number of colorings fixed by each element of the group. The group here is the dihedral group D5, since there are 5 beads. The dihedral group has 10 elements: 5 rotations and 5 reflections. Wait, no, actually for n beads, the dihedral group Dn has 2n elements. So here, n=5, so 10 elements. The group includes rotations by 0°, 72°, 144°, 216°, 288°, and reflections over 5 axes. So, to apply Burnside's lemma, I need to compute for each symmetry operation (each element of the group) the number of colorings fixed by that operation, then take the average.First, let's list all the group elements:1. The identity rotation (0°).2. Rotation by 72°.3. Rotation by 144°.4. Rotation by 216°.5. Rotation by 288°.6. Reflection over 5 different axes.For each of these, compute the number of colorings fixed by that symmetry.Starting with the identity rotation. Every coloring is fixed by the identity, so the number here is the total number of colorings without considering symmetry. Since there are 5 beads, 3 red and 2 blue, the number is C(5,3) = 10.Next, rotations by 72°, 144°, 216°, 288°. These are the non-identity rotations. For a rotation to fix a coloring, the coloring must be invariant under that rotation. For a rotation by k*72°, the cycle structure is a single cycle of length 5 if 5 is coprime to k. Wait, 5 is prime, so except for the identity rotation, all other rotations have a single cycle of length 5. Therefore, for a coloring to be fixed under such a rotation, all beads must be the same color. But we have 3 red and 2 blue beads, which are not all the same. Therefore, the number of colorings fixed by these rotations is 0.So, all non-identity rotations fix 0 colorings.Now, reflections. There are 5 reflections, each corresponding to a reflection over an axis that passes through one bead and the midpoint between two beads. For a reflection to fix a coloring, the coloring must be symmetric with respect to that axis. Each reflection in D5 is a flip that fixes one bead and swaps the other two pairs. So, the cycle structure for each reflection is a fixed bead and two 2-cycles. Therefore, the beads swapped by the reflection must be the same color.To count the number of colorings fixed by a reflection, we need to consider the beads in each cycle. The fixed bead can be either red or blue. Then, each pair of beads swapped by the reflection must be the same color. But since each pair is swapped, the two beads in each pair must be the same color. However, we have an odd number of beads here? Wait, the ring has 5 beads. So, when you reflect, one bead is fixed, and the other four are swapped in pairs. So, for example, bead 1 is fixed, beads 2 and 5 are swapped, beads 3 and 4 are swapped. So, bead 1 can be colored either red or blue, beads 2 and 5 must be the same color, beads 3 and 4 must be the same color. So, the coloring is determined by the color of bead 1, and the colors of the pairs (2,5) and (3,4). Each of these can be red or blue. However, we need the total number of red beads to be 3 and blue to be 2.So, let's model this. Let's say the fixed bead is color A (either red or blue), pair (2,5) is color B, and pair (3,4) is color C. Then total beads: 1*A + 2*B + 2*C. We need this to equal 3 red and 2 blue.So, possible cases:Case 1: Fixed bead is red (A=red). Then total red beads: 1 + 2*B_red + 2*C_red. Total blue beads: 2*B_blue + 2*C_blue. But since we have 3 red and 2 blue, so 1 + 2*(B_red + C_red) = 3 and 2*(B_blue + C_blue) = 2. From the first equation: 2*(B_red + C_red) = 2 => B_red + C_red = 1. From the second equation: B_blue + C_blue = 1. Since each pair (B and C) must be either red or blue, and B and C can be each red or blue. So, B_red + C_red =1, which means exactly one of B or C is red, the other is blue. Similarly, since B_blue = 1 - B_red, and same for C. So, how many possibilities? We have two pairs (B and C), each pair is a single color. We need exactly one of them to be red. So, there are 2 choices: either B is red and C is blue, or B is blue and C is red. Therefore, if the fixed bead is red, there are 2 colorings.Case 2: Fixed bead is blue (A=blue). Then total red beads: 0 + 2*B_red + 2*C_red = 3. But 2*(B_red + C_red) = 3. However, 3 is odd, and 2*(something) must be even. Therefore, impossible. So, no solutions here.Therefore, for each reflection, the number of colorings fixed is 2 (when the fixed bead is red, and one of the two pairs is red).Since there are 5 reflections, each contributing 2 fixed colorings, total fixed colorings for reflections is 5*2=10.Wait, but hold on. Is that correct? Wait, each reflection is a different axis. However, depending on the axis, the fixed bead is different. But when we count fixed colorings for each reflection, we have to see if the axis passes through a bead or between beads. Wait, in D5, the reflections can be either through a bead and the midpoint between two beads, or through midpoints between two beads. Wait, no, for odd n, all reflections pass through one bead and the midpoint between two others. So, in our case, each reflection passes through one bead and the midpoint between two others. Therefore, in each reflection, there is one fixed bead, and the other four beads are swapped in two pairs. Therefore, for each reflection, the analysis is as above: fixed bead can be red or blue, but only when fixed bead is red, you can have 2 colorings. If fixed bead is blue, you can't get 3 red beads. Therefore, each reflection fixes 2 colorings. So, 5 reflections, 2 each, total 10.So, summarizing:- Identity: 10 fixed colorings- Rotations (non-identity): 0 each, total 0- Reflections: 2 each, total 10Total fixed colorings: 10 + 0 + 10 = 20Then, by Burnside's lemma, the number of distinct colorings is total fixed colorings divided by the group size. The group size is 10. So, 20 /10=2. Wait, that can't be. The answer is 2? But that seems low. Wait, but let's check.Wait, if you have 3 red and 2 blue beads on a ring, how many distinct colorings are there? Let's think manually.Suppose we fix one bead as red (since rotations are allowed). Then, the arrangements depend on the positions of the other two red beads and two blue beads. But since it's a ring, we can fix one bead and arrange the rest. Wait, but since we can also reflect, maybe the number is even smaller.Alternatively, using Burnside's count, if the average is 2, then answer is 2. But let's verify.Alternatively, maybe I made a mistake in counting fixed colorings under reflections.Wait, for each reflection, the fixed bead is red, and the two pairs. Then, the total red beads are 1 + number of red pairs *2. Since we need 3 red beads, 1 + 2*(number of red pairs)=3. Therefore, 2*(number of red pairs)=2, so number of red pairs=1. Therefore, exactly one of the two pairs is red, the other is blue. So, two possibilities: first pair red, second pair blue; or first pair blue, second pair red. So, two possibilities for each reflection. However, but in the reflection, the pairs are specific. For example, in a reflection that fixes bead 1, pairs are (2,5) and (3,4). If we fix bead 1 as red, then (2,5) can be red and (3,4) blue, or (2,5) blue and (3,4) red. Each of these is a distinct coloring. However, but when considering the reflection symmetry, these colorings are fixed by that reflection. However, when considering different reflections, different beads are fixed. So, for example, a coloring fixed by reflection over bead 1 is different from a coloring fixed by reflection over bead 2. But in Burnside's lemma, we count all colorings fixed by any reflection. So, if each reflection fixes two colorings, and there are five reflections, but some colorings might be fixed by multiple reflections. Wait, but actually, in this case, is there any coloring that is fixed by more than one reflection? For example, a coloring with a bead fixed as red and two adjacent red beads might be fixed by one reflection, but not others. Wait, let's take an example. Suppose bead 1 is red, beads 2 and 5 are red, beads 3 and 4 are blue. This coloring is fixed by the reflection that fixes bead 1. But if we try to apply another reflection, say fixing bead 2, then bead 2 is fixed, and pairs (1,3) and (4,5). But in our coloring, bead 1 is red, bead 3 is blue, so they are different; bead 4 is blue, bead5 is red, which are different. Therefore, this coloring is not fixed by the reflection fixing bead2. Similarly, other reflections would not fix this coloring. Therefore, each of the colorings fixed by a reflection is unique to that reflection. Therefore, total fixed colorings across all reflections are 5*2=10, and none overlapping.Therefore, total fixed colorings are 10 (identity) +0 (rotations)+10 (reflections)=20. Divide by group order 10, gives 2. So, answer is 2.But let me check with another approach. Suppose we use the formula for necklaces with n beads and k colors, but here we have two colors, and the number is C(n, k)/n when n is prime? Wait, no, that's for the case where we have exactly one color, but maybe not. Alternatively, for circular permutations with indistinct beads, the formula is (n-1)! for distinct beads, but here beads are colored with repetition. Wait, perhaps it's better to use the formula from combinatorics.The number of distinct necklaces with n beads, k colors, accounting for rotations and reflections is given by (1/(2n))*(sum_{d|n} phi(d)*C(n/d, k/d)) + ...) Hmm, maybe not. Alternatively, when n is prime, the formula simplifies. Since 5 is prime, the number of distinct necklaces with 5 beads, 2 colors, is (2^5 + 5*2^((5+1)/2))/10. Wait, that's the formula for the number of binary necklaces considering rotations and reflections. Wait, the formula is (1/(2n))*(sum_{d|n} phi(d)*2^(n/d)) + n*2^{1})). Wait, no, maybe I need to recall the formula correctly.Wait, the number of binary necklaces (considering rotations) is (1/n)*sum_{d|n} phi(d)*2^{n/d}). For n=5, that's (1/5)(phi(1)*2^5 + phi(5)*2^1)= (1/5)(32 +4*2)= (32+8)/5=40/5=8. Then, considering reflections, the formula becomes (number of necklaces considering rotations + number of palindromic necklaces)/2. Palindromic necklaces are those invariant under reversal. For n odd, the number of palindromic necklaces is 2^{(n+1)/2}. So, for n=5, it's 2^{3}=8. Therefore, total necklaces considering rotations and reflections would be (8 +8)/2=8. Wait, but that's for all binary necklaces with 5 beads, any number of colors. But our problem is with exactly 3 red and 2 blue beads, so fixed content. So, maybe the formula isn't directly applicable here.Alternatively, another approach. Since the beads are arranged on a ring, we can fix one bead to eliminate rotational symmetry. Let's fix bead1 as red. Then, we have 4 remaining beads: 2 red and 2 blue. Arrangements are considered distinct if they cannot be rotated into each other. But since we fixed bead1, rotations are accounted for. However, reflections are still possible. So, now, the problem reduces to arranging 2 red and 2 blue beads in a line, considering reflection symmetry. Wait, but it's a ring, but since we fixed bead1, the remaining beads are in a line from bead2 to bead5. Wait, no, actually, fixing bead1, the rest are in a circle, but we can consider linear arrangements with equivalence under rotation and reflection. Hmm, maybe not straightforward.Alternatively, with bead1 fixed as red, the remaining beads are bead2, bead3, bead4, bead5, which need to have 2 red and 2 blue. The number of linear arrangements is C(4,2)=6. But considering rotational symmetry around the fixed bead, but bead1 is fixed, so rotations would correspond to shifting the remaining beads. Wait, but since bead1 is fixed, the remaining beads form a line? Wait, no. On a ring, fixing bead1 would make the rest of the beads in a circle, but since we already fixed bead1, rotations would not affect bead1. Wait, perhaps this approach is not correct.Alternatively, since we have fixed bead1 as red, the remaining beads form a linear sequence from bead2 to bead5, but it's actually still a circle. However, since bead1 is fixed, rotations are eliminated, but reflections would reverse the order. So, the problem reduces to counting the number of necklaces with bead1 fixed as red, and the rest 4 beads (2 red, 2 blue), considering reflection. Wait, but bead1 is fixed, so reflections would flip the order of beads2-5. Therefore, the number of distinct colorings is equal to the number of distinct arrangements of beads2-5 with 2 red and 2 blue, considering reflection. So, the number is equal to (number of linear arrangements + number of symmetric arrangements)/2. The number of linear arrangements is C(4,2)=6. The number of symmetric arrangements (palindromic) is C(2,1)=2, since beads2-5 must be a palindrome. For a palindrome of length 4, beads2 must equal beads5, and beads3 must equal beads4. So, beads2 and beads5 are the same color, beads3 and beads4 are the same color. We need 2 red and 2 blue beads in total (bead1 is already red). So, bead2 and beads5: if they are red, then beads3 and beads4 must be blue (total red beads: 1+2=3, blue:2). If bead2 and beads5 are blue, then beads3 and beads4 must be red (total red beads:1+2=3, blue:2+2=4, which is invalid). Wait, because we need exactly 2 blue beads. Wait, bead1 is red, so total red beads:1 + beads2-5. We need total red beads:3, so beads2-5 must have 2 red and 2 blue. So, if beads2 and beads5 are red, then beads3 and beads4 must be blue. That gives 2 red in beads2-5. If beads2 and beads5 are blue, then beads3 and beads4 must be red, but that gives 2 blue and 2 red in beads2-5, which is acceptable. Wait, no. Wait, if beads2 and beads5 are blue, then beads3 and beads4 are red, so total blue beads: 2 (from beads2 and beads5) and total red beads: 2 (from beads3 and beads4), plus bead1 red. So total red beads:3, blue beads:2. That works. Wait, but if beads2 and beads5 are blue, beads3 and beads4 are red, then total beads: bead1=red, beads2=blue, beads3=red, beads4=red, beads5=blue. Wait, that's 3 red and 2 blue. So, actually, two possibilities for symmetric arrangements: either beads2 and beads5 are red, beads3 and beads4 are blue; or beads2 and beads5 are blue, beads3 and beads4 are red. Therefore, 2 symmetric arrangements. So, the number of distinct colorings after fixing bead1 is (6 +2)/2=4. But wait, when we fix bead1, are we considering reflection? Wait, if we fix bead1, the reflection would map to reversing beads2-5. So, the distinct colorings under reflection would be the orbits under reflection of the remaining beads. So, the formula is indeed (number of linear arrangements + number of palindromic arrangements)/2= (6+2)/2=4. Therefore, total distinct colorings are 4. But according to Burnside's lemma, we had 2. There's a discrepancy here. Which is correct?Wait, perhaps I confused something in the Burnside approach. Let me check again. According to Burnside, the answer was 2. According to fixing bead1 and considering reflections, we get 4. But these can't both be right. Where is the mistake?Wait, maybe in the Burnside calculation, when we fixed bead1, but Burnside's lemma accounts for all symmetries, including those that move bead1. So, perhaps the correct answer is 2, but when we fix bead1, we're only considering rotational symmetry, but Burnside's lemma considers both rotations and reflections. Wait, no, when we fix bead1, we're eliminating rotational symmetry but still need to account for reflections. However, in the Burnside calculation, the group includes both rotations and reflections, so the orbit under the entire dihedral group is considered. If fixing bead1 and considering reflections gives 4, but Burnside's says 2, there must be a miscalculation.Wait, let's go back. When I fixed bead1 as red, and computed 4 colorings, but Burnside's lemma gives 2. However, maybe fixing bead1 as red is not the same as the orbit count. Because in the Burnside count, colorings can have bead1 in any position, but when we fix bead1, we're essentially choosing a representative from each orbit where bead1 is red. However, some colorings might not have any red bead, but in our case, all colorings have 3 red beads, so every coloring has at least one red bead. Therefore, the number of orbits should be equal to the number of colorings with bead1 fixed as red divided by the number of symmetries that fix bead1. Wait, this is getting complicated. Alternatively, let's list all possible colorings.Suppose we have beads labeled 1 to 5 on a ring. We need to color 3 red and 2 blue. Let's try to find distinct colorings under rotation and reflection.First, consider the spacing between the blue beads. In a ring with 5 beads, placing 2 blue beads. The possible distinct arrangements are determined by the distance between them. Since it's a ring, the distance can be 1, 2, or same as 3 (since 5-2=3). Wait, in a 5-bead ring, the distance between two beads can be measured as the minimal number of steps between them. So, for two blue beads, the possible distances are 1 or 2. Because distance 3 is equivalent to distance 2 in the other direction. So, there are two distinct arrangements: adjacent blue beads (distance 1) or non-adjacent (distance 2). However, when considering reflection, does this matter?Wait, if the two blue beads are adjacent, then the coloring is symmetric with respect to the reflection that fixes one of the blue beads. If they are separated by one bead (distance 2), then they form a sort of "opposite" position. Wait, but with 5 beads, there is no true opposite. Let me visualize.Case 1: Blue beads are adjacent. Let's say beads 1 and 2 are blue. Then, the red beads are 3,4,5. Under rotation and reflection, this arrangement can be rotated so that any two adjacent beads are blue. Similarly, a reflection would fix a blue bead and swap the other beads. But would this be equivalent to another arrangement?Case 2: Blue beads are separated by one bead. For example, beads 1 and 3 are blue. Then, red beads are 2,4,5. This configuration can be rotated to have any two beads with one bead in between. However, in this case, the two blue beads are separated by one red bead. But in a ring of 5, this is different from the adjacent case.Wait, but are these two cases equivalent under reflection? Let's see. If we have blue beads at 1 and 3, and reflect over the axis through bead1 and the midpoint between beads3 and 4, would that make blue beads at 1 and3 swap with others? Wait, maybe not. Alternatively, reflecting might map this to another arrangement. Wait, but in any case, perhaps these two cases are distinct. So, there are two distinct colorings: one with adjacent blue beads and one with blue beads separated by one bead. Therefore, answer is 2. So, this matches the Burnside's lemma result.But when I fixed bead1 as red and considered the remaining beads, I found 4 colorings. But these must be grouped into orbits under the dihedral group. Each orbit corresponds to a distinct coloring considering rotations and reflections. If the answer is 2, then those 4 colorings when bead1 is fixed must be grouped into 2 pairs, each pair being equivalent under reflections.Wait, let's list them. If bead1 is fixed as red, then the remaining beads are 2,3,4,5 with 2 red and 2 blue. The possible arrangements are:1. beads2 and3 red, beads4 and5 blue2. beads2 and4 red, beads3 and5 blue3. beads2 and5 red, beads3 and4 blue4. beads3 and4 red, beads2 and5 blue5. beads3 and5 red, beads2 and4 blue6. beads4 and5 red, beads2 and3 blueBut we fixed bead1 as red, and the remaining 4 beads have 2 red and 2 blue. So, 6 possibilities. However, considering reflections (which would reverse the order of beads2-5), some of these are equivalent.For example, arrangement1: beads2,3 red; beads4,5 blue. Its reflection would be beads2,5 blue; beads3,4 red. Which is arrangement4.Similarly, arrangement2: beads2,4 red; beads3,5 blue. Its reflection would be beads2,5 blue; beads3,4 red. Wait, no. Wait, reflecting would fix bead1 and reverse beads2-5. So, original beads2,3,4,5: arrangement1 is R,R,B,B. Reflection would reverse to B,B,R,R, which is beads5,4,3,2: B,B,R,R. But since it's a ring, after reflection, beads2 becomes beads5, beads3 becomes beads4, etc. So, arrangement1 reflected is beads5,4 blue and beads3,2 red. But bead1 is still red. So, in terms of the fixed bead1, the reflected arrangement would have beads2 and5 blue, beads3 and4 red. Which is arrangement3: beads2 and5 blue, beads3 and4 red. Wait, arrangement3 is beads2 and5 blue. Wait, no, in arrangement3, beads2 and5 are red? Wait, no, when we fixed bead1 as red, the remaining beads are 2,3,4,5 with 2 red and 2 blue. So arrangement1: beads2 and3 red (R,R,B,B)Arrangement2: beads2 and4 red (R,B,R,B)Arrangement3: beads2 and5 red (R,B,B,R)Arrangement4: beads3 and4 red (B,R,R,B)Arrangement5: beads3 and5 red (B,R,B,R)Arrangement6: beads4 and5 red (B,B,R,R)Now, applying reflection (reversing beads2-5), we get:Reflection of arrangement1 (R,R,B,B) becomes (B,B,R,R), which is arrangement6.Similarly, reflection of arrangement6 is arrangement1.Reflection of arrangement2 (R,B,R,B) becomes (B,R,B,R), which is arrangement5.Reflection of arrangement5 is arrangement2.Reflection of arrangement3 (R,B,B,R) becomes (R,B,B,R), which is itself? Wait, no. Wait, original arrangement3 is beads2 and5 red. Wait, beads2 is red, beads3 is blue, beads4 is blue, beads5 is red. Reflecting would reverse the order of beads2-5, so beads5 becomes beads2, beads4 becomes beads3, beads3 becomes beads4, beads2 becomes beads5. So, beads2 (originally red) becomes beads5 (red), beads3 (blue) becomes beads4 (blue), beads4 (blue) becomes beads3 (blue), beads5 (red) becomes beads2 (red). So, after reflection, the coloring is beads2=red, beads3=blue, beads4=blue, beads5=red, which is the same as arrangement3. Therefore, arrangement3 is symmetric under reflection.Similarly, arrangement4 (B,R,R,B): beads2=blue, beads3=red, beads4=red, beads5=blue. Reflecting would give beads2=blue, beads5=blue, beads4=red, beads3=red, which is the same as arrangement4. So, arrangements3 and4 are symmetric under reflection.Wait, let's check arrangement3: beads2=red, beads3=blue, beads4=blue, beads5=red. This is symmetric because beads2 and5 are red, beads3 and4 are blue. Similarly, arrangement4: beads2=blue, beads3=red, beads4=red, beads5=blue. Beads2 and5 are blue, beads3 and4 are red. Therefore, both arrangements3 and4 are symmetric under reflection.Therefore, the orbits under the dihedral group (rotations and reflections) when bead1 is fixed as red are:- Orbit1: arrangement1 and arrangement6 (related by reflection)- Orbit2: arrangement2 and arrangement5 (related by reflection)- Orbit3: arrangement3 (symmetric)- Orbit4: arrangement4 (symmetric)But wait, that's four orbits. However, according to Burnside's lemma, there should be 2 orbits. Wait, this is confusing.But maybe when bead1 is fixed, we are not considering all symmetries. Because the dihedral group allows rotations and reflections that move bead1. So, fixing bead1 reduces the symmetry group, but the original problem allows any symmetries. Therefore, when we fix bead1, we are actually counting distinct colorings under the subgroup that fixes bead1, which is only the identity and the reflection that fixes bead1. But the full dihedral group includes rotations and reflections that move bead1. Therefore, the orbits under the full group may merge some of these orbits.For example, arrangement1 (beads2,3 red) can be rotated to place red beads at beads3,4, etc. However, since we fixed bead1, rotation would move bead1, which is not allowed in this fixed scenario. Therefore, fixing bead1 restricts the symmetries considered.This suggests that fixing bead1 is not the right way to count orbits under the full dihedral group. Instead, we need to consider all possible symmetries.Given the confusion, I think the Burnside's lemma result of 2 is correct, given that when considering all symmetries, the two distinct arrangements are: blue beads adjacent or non-adjacent.Another way to see: in a ring with 5 beads, the number of distinct ways to place 2 blue beads is 2: either they are adjacent or not. Because with 5 beads, the cyclic symmetry makes all adjacent pairs equivalent, and all non-adjacent pairs also equivalent (since 5 is prime, the step between the beads can be 1 or 2, which are distinct). Therefore, answer is 2.Yes, this makes sense. Therefore, the final answer is 2.Final AnswerThe number of different schemes is boxed{2}.</think>To determine the number of distinct schemes for embedding 3 red beads and 2 blue beads on a ring, we use Burnside's lemma to account for rotational and reflectional symmetries. 1. Group Elements: The dihedral group (D_5) has 10 elements (5 rotations and 5 reflections).2. Fixed Colorings: - Identity Rotation: All permutations are fixed. There are (binom{5}{3} = 10) fixed colorings. - Non-Identity Rotations (72°, 144°, 216°, 288°): No coloring with 3 red and 2 blue beads can be fixed by these rotations, so 0 fixed colorings. - Reflections: Each reflection fixes one bead and swaps the others. For each reflection, the fixed bead must be red (to maintain 3 red beads), and the remaining pairs must be one red and one blue. This gives 2 fixed colorings per reflection. With 5 reflections, we get (5 times 2 = 10) fixed colorings.3. Total Fixed Colorings: Summing up, we have (10 + 0 + 10 = 20) fixed colorings.4. Applying Burnside's Lemma: The number of distinct colorings is the average number of fixed colorings: [ frac{20}{10} = 2 ]Thus, the number of different schemes is (boxed{2}).]

🔑:begin{array}{l}G:(1)(2)(3)(4)(5),(12345),(13524) quad(14253),(15432) quad(1)(25)(34),(2)(13)(45) quad(3)(15)(34),(4)(12)(35)end{array}text { (5) }(14)(23) text {. }Corresponding to g_{0}=(1)(2)(3)(4)(5), there are C(5,3)=10 schemes with 3 red and 2 blue tiles, which remain unchanged under g_{0}. For (12345) and the other 4 schemes of (5) format, the number of schemes is 0. For the (1)^{1}(2)^{2} format, the number of schemes is 2, henceN=frac{1}{10} times[10+5 times 2]=2

🧠:[<think>Okay, let's tackle this problem step by step. So, we have two sequences here: an arithmetic sequence {bₙ} starting with 1, and another sequence {aₙ} that satisfies a recurrence relation aₙ₊₁ - 3aₙ - 1 = 0. We also know that b₃ + 1 = a₂ and a₁ = 1. The problem has two parts: first, we need to show that {aₙ + 1/2} is a geometric sequence, and second, find the sum Tₙ of the first n terms of cₙ = aₙ·bₙ.Starting with part (1). We need to prove that {aₙ + 1/2} is a geometric sequence. A geometric sequence has a common ratio between consecutive terms. So, if I can show that (aₙ₊₁ + 1/2)/(aₙ + 1/2) is a constant, then it's a geometric sequence.Given the recurrence relation: aₙ₊₁ - 3aₙ - 1 = 0. Let me rearrange this to express aₙ₊₁ in terms of aₙ: aₙ₊₁ = 3aₙ + 1. Hmm. So each term is three times the previous term plus 1. That seems like a linear recurrence relation. Maybe I can solve it to find an explicit formula for aₙ, which might help in showing that {aₙ + 1/2} is geometric.Alternatively, maybe manipulate the given recurrence to see if adding 1/2 to both sides helps. Let's try that. Let's consider aₙ₊₁ + 1/2. From the recurrence, aₙ₊₁ = 3aₙ + 1. So substituting that in, we have:aₙ₊₁ + 1/2 = 3aₙ + 1 + 1/2 = 3aₙ + 3/2.Now, factor out 3 on the right side: 3(aₙ + 1/2). So:aₙ₊₁ + 1/2 = 3(aₙ + 1/2).Ah! That's exactly the form we need. This shows that each term aₙ₊₁ + 1/2 is 3 times the previous term aₙ + 1/2. Therefore, the sequence {aₙ + 1/2} is a geometric sequence with common ratio 3. That should suffice for part (1). Let me double-check.Given a₁ = 1. So a₁ + 1/2 = 1 + 1/2 = 3/2. Then a₂ + 1/2 = 3(a₁ + 1/2) = 3*(3/2) = 9/2. Let's verify with the given information. Wait, we also know that b₃ + 1 = a₂. Let's compute b₃ first. Since {bₙ} is arithmetic with first term 1. The general term of an arithmetic sequence is bₙ = b₁ + (n - 1)d, where d is the common difference. Here, b₁ = 1, so b₃ = 1 + 2d. Then, a₂ = b₃ + 1 = 1 + 2d + 1 = 2 + 2d. But according to the recurrence relation, a₂ = 3a₁ + 1 = 3*1 + 1 = 4. So 2 + 2d = 4 ⇒ 2d = 2 ⇒ d = 1. Therefore, the arithmetic sequence {bₙ} has common difference 1, so bₙ = 1 + (n - 1)*1 = n. So bₙ = n. That will be helpful for part (2).But back to part (1). Since we derived that aₙ₊₁ + 1/2 = 3(aₙ + 1/2), and the first term when n=1 is a₁ + 1/2 = 3/2, then yes, it's a geometric sequence with first term 3/2 and ratio 3. Therefore, part (1) is proven.Moving on to part (2). We need to find the sum Tₙ = c₁ + c₂ + ... + cₙ where cₙ = aₙ·bₙ. Since we already found that bₙ = n, and we need aₙ. From part (1), we know {aₙ + 1/2} is a geometric sequence with first term 3/2 and ratio 3. Therefore, the general term is (aₙ + 1/2) = (3/2)*3^{n-1} = (3/2)*3^{n-1} = 3^{n}/2. Therefore, aₙ = 3^{n}/2 - 1/2 = (3^{n} - 1)/2.So, cₙ = aₙ·bₙ = [(3^{n} - 1)/2]·n = (n·3^{n} - n)/2.Therefore, Tₙ is the sum from k=1 to n of (k·3^{k} - k)/2. Let's split the sum into two parts:Tₙ = (1/2)[Σ_{k=1}^n k·3^{k} - Σ_{k=1}^n k].We can compute each sum separately. The second sum Σk from 1 to n is n(n + 1)/2. The first sum Σk·3^{k} from 1 to n is a bit trickier. Let me recall the formula for the sum of k·r^{k}. There is a standard formula for this. Let me derive it.Let S = Σ_{k=1}^n k·r^{k}.We can write S = r + 2r² + 3r³ + ... + n·r^n.Multiply both sides by r:rS = r² + 2r³ + 3r^4 + ... + n·r^{n+1}.Subtract the second equation from the first:S - rS = (r + r² + r³ + ... + r^n) - n·r^{n+1}.The left side is S(1 - r). The right side is the sum of a geometric series minus n·r^{n+1}.Sum from k=1 to n of r^k = r(1 - r^n)/(1 - r). Therefore,S(1 - r) = [r(1 - r^n)/(1 - r)] - n·r^{n+1}.Then,S = [r(1 - r^n)/(1 - r)^2] - [n·r^{n+1}/(1 - r)].Plugging in r = 3:Σ_{k=1}^n k·3^{k} = [3(1 - 3^n)/(1 - 3)^2] - [n·3^{n+1}/(1 - 3)].Compute denominators:(1 - 3)^2 = 4, and (1 - 3) = -2.So,[3(1 - 3^n)/4] - [n·3^{n+1}/(-2)] = [3(1 - 3^n)/4] + [n·3^{n+1}/2].Simplify:First term: 3(1 - 3^n)/4 = (3 - 3^{n+1})/4.Second term: n·3^{n+1}/2.Combine them:(3 - 3^{n+1})/4 + (2n·3^{n+1})/4 = [3 - 3^{n+1} + 2n·3^{n+1}]/4 = [3 + 3^{n+1}(2n - 1)]/4.Therefore, the sum Σ_{k=1}^n k·3^{k} = [3^{n+1}(2n - 1) + 3]/4.Wait, let's check the algebra again:Starting from:3(1 - 3^n)/4 + n·3^{n+1}/2Convert to quarters:= [3(1 - 3^n) + 2n·3^{n+1}]/4Factor out 3^n:= [3 - 3^{n+1} + 2n·3^{n+1}]/4= 3/4 + 3^{n+1}(2n - 1)/4So, Σk·3^k = (3^{n+1}(2n - 1) + 3)/4.Therefore, going back to Tₙ:Tₙ = (1/2)[Σk·3^k - Σk] = (1/2)[(3^{n+1}(2n - 1) + 3)/4 - n(n + 1)/2].Let's compute each part:First term inside the brackets: (3^{n+1}(2n - 1) + 3)/4Second term: -n(n + 1)/2So,Tₙ = (1/2)[(3^{n+1}(2n - 1) + 3)/4 - (2n(n + 1))/4]Because to combine the terms, we need a common denominator of 4. So, rewrite -n(n +1)/2 as -2n(n +1)/4.Therefore,Tₙ = (1/2)[ (3^{n+1}(2n -1) + 3 - 2n(n +1)) /4 ]Multiply the 1/2 and 1/4 to get 1/8:Tₙ = [3^{n+1}(2n -1) + 3 - 2n(n +1)] /8Simplify numerator:3^{n+1}(2n -1) + 3 - 2n(n +1)We can factor out the 3^{n+1} term and leave the rest as is.Alternatively, write the entire expression:Tₙ = [3^{n+1}(2n -1) - 2n(n +1) + 3] /8Check if this can be simplified further. Let's see.Alternatively, maybe factor the constants:But probably this is the simplest form. Let me check with n=1 to see if this formula works.For n=1:c₁ = a₁·b₁ = 1·1 =1. So T₁=1.Using the formula:[3^{2}(2*1 -1) - 2*1*(1 +1) +3]/8 = [9*(1) -4 +3]/8 = (9 -4 +3)/8=8/8=1. Correct.Another test with n=2:Compute a₁=1, a₂=4 (from a₂=3a₁ +1=4). b₁=1, b₂=2. So c₁=1*1=1, c₂=4*2=8. T₂=1+8=9.Using the formula:[3^{3}(2*2 -1) -2*2*3 +3]/8 = [27*(3) -12 +3]/8 = (81 -12 +3)/8=72/8=9. Correct.Another test with n=3:a₃=3a₂ +1=3*4 +1=13. b₃=3. c₃=13*3=39. T₃=1+8+39=48.Formula:[3^{4}(2*3 -1) -2*3*4 +3]/8 = [81*(5) -24 +3]/8=(405 -24 +3)/8=384/8=48. Correct. So formula works.Therefore, the sum Tₙ is [3^{n+1}(2n -1) -2n(n +1) +3]/8. To write this in a more simplified form, maybe factor the terms:Let me see:3^{n+1}(2n -1) +3 -2n(n +1). Perhaps factor 3 from 3^{n+1} and 3?But 3^{n+1} is 3*3^n, and 3 is 3*1. So:3[3^n(2n -1) +1] -2n(n +1). But not sure if that's helpful. Alternatively, leave it as is.Alternatively, write numerator as:3^{n+1}(2n -1) -2n(n +1) +3. That's acceptable.Therefore, the final answer is Tₙ = [3^{n+1}(2n -1) -2n(n +1) +3]/8.Alternatively, factor numerator:Let me check if -2n(n +1) +3 can be written differently. Maybe not. So that's as simplified as it gets.So summarizing:1. Proved that {aₙ +1/2} is geometric with ratio 3.2. Found Tₙ = [3^{n+1}(2n -1) -2n(n +1) +3]/8.Final Answer(1) boxed{left{a_n + frac{1}{2}right}} is a geometric sequence. (2) The sum of the first ( n ) terms ( T_n ) is boxed{dfrac{3^{n+1}(2n - 1) - 2n(n + 1) + 3}{8}}.</think>Given the problem, we need to address two parts:(1) Prove that (left{a_n + frac{1}{2}right}) forms a geometric sequence.We start with the recurrence relation for the sequence ({a_n}): (a_{n+1} - 3a_n - 1 = 0). Rearranging this, we get (a_{n+1} = 3a_n + 1). Adding (frac{1}{2}) to both sides, we have:[a_{n+1} + frac{1}{2} = 3a_n + 1 + frac{1}{2} = 3left(a_n + frac{1}{2}right)]This shows that (a_{n+1} + frac{1}{2}) is three times (a_n + frac{1}{2}), indicating that (left{a_n + frac{1}{2}right}) is a geometric sequence with the first term (a_1 + frac{1}{2} = frac{3}{2}) and common ratio 3.(2) Find the sum of the first (n) terms of the sequence ({c_n}) where (c_n = a_n cdot b_n).First, we determine the explicit form of (a_n). Since (left{a_n + frac{1}{2}right}) is a geometric sequence, we have:[a_n + frac{1}{2} = frac{3}{2} cdot 3^{n-1} = frac{3^n}{2}]Thus, (a_n = frac{3^n - 1}{2}).The sequence ({b_n}) is an arithmetic sequence with the first term 1 and common difference 1 (since (b_3 + 1 = a_2) gives (b_3 = 3), leading to a common difference of 1). Therefore, (b_n = n).Now, (c_n = a_n cdot b_n = left(frac{3^n - 1}{2}right) cdot n = frac{n cdot 3^n - n}{2}).To find the sum (T_n), we split it into two parts:[T_n = frac{1}{2} left( sum_{k=1}^n k cdot 3^k - sum_{k=1}^n k right)]The sum (sum_{k=1}^n k) is (frac{n(n+1)}{2}). For the sum (sum_{k=1}^n k cdot 3^k), we use the formula for the sum of (k cdot r^k):[sum_{k=1}^n k cdot 3^k = frac{3(1 - 3^n)}{(1 - 3)^2} - frac{n cdot 3^{n+1}}{1 - 3} = frac{3(1 - 3^n)}{4} + frac{n cdot 3^{n+1}}{2}]Simplifying this, we get:[sum_{k=1}^n k cdot 3^k = frac{3^{n+1}(2n - 1) + 3}{4}]Combining the results, we have:[T_n = frac{1}{2} left( frac{3^{n+1}(2n - 1) + 3}{4} - frac{n(n+1)}{2} right)]Converting to a common denominator and simplifying, we get:[T_n = frac{3^{n+1}(2n - 1) - 2n(n + 1) + 3}{8}]Final answers:(1) (boxed{left{a_n + frac{1}{2}right}}) is a geometric sequence.(2) The sum of the first (n) terms (T_n) is (boxed{dfrac{3^{n+1}(2n - 1) - 2n(n + 1) + 3}{8}}).,<think>Okay, so I need to solve this problem about arithmetic and geometric sequences. Let me start by understanding what is given and what needs to be proven and found.First, the problem says that {bₙ} is an arithmetic sequence with the first term being 1. So, arithmetic sequences have a common difference, right? So, the general formula for an arithmetic sequence is bₙ = b₁ + (n-1)d, where d is the common difference. Since the first term b₁ is 1, so bₙ = 1 + (n-1)d. But we don't know what d is yet. Maybe we can find that later.Then, the sequence {aₙ} satisfies the recurrence relation aₙ₊₁ - 3aₙ - 1 = 0. So, that means aₙ₊₁ = 3aₙ + 1. And they also tell us that a₁ = 1. So, maybe we can write out the first few terms of {aₙ} to see the pattern. Let's see:a₁ = 1a₂ = 3a₁ + 1 = 3*1 + 1 = 4a₃ = 3a₂ + 1 = 3*4 + 1 = 13a₄ = 3a₃ + 1 = 3*13 + 1 = 40Hmm, seems like each term is triple the previous plus 1. This looks like a linear nonhomogeneous recurrence relation. I remember that for such recursions, the solution can be found by finding the homogeneous solution and a particular solution.But maybe part (1) wants us to show that {aₙ + 1/2} is a geometric sequence. If that's the case, maybe transforming the recursion into a geometric sequence by adding a constant.Let me try that. Suppose we let xₙ = aₙ + c, where c is a constant to be determined. Then, substituting into the recurrence:xₙ₊₁ - c = 3(xₙ - c) + 1So, xₙ₊₁ = 3xₙ - 3c + 1 + c = 3xₙ - 2c + 1If we can choose c such that -2c + 1 = 0, then xₙ₊₁ = 3xₙ, which is a geometric sequence with ratio 3.Solving -2c + 1 = 0 gives c = 1/2. So, xₙ = aₙ + 1/2, and xₙ₊₁ = 3xₙ. Therefore, {xₙ} is a geometric sequence with ratio 3. That's exactly part (1) to prove. So, I think this approach works. Let me check with the initial term. Since a₁ = 1, then x₁ = 1 + 1/2 = 3/2. Then x₂ should be 3x₁ = 9/2. But a₂ is 4, so x₂ = 4 + 1/2 = 9/2. Correct. Then x₃ = 3x₂ = 27/2, which is 13 + 1/2 = 13.5 = 27/2. Correct. So yes, this works. Therefore, part (1) is proved by defining xₙ = aₙ + 1/2, showing that the recurrence becomes xₙ₊₁ = 3xₙ, hence a geometric sequence with ratio 3.So, part (1) is done.Now, part (2) asks to find the sum of the first n terms of the sequence {cₙ}, where cₙ = aₙ · bₙ. So, Tₙ = Σₖ=1ⁿ cₖ = Σₖ=1ⁿ aₖbₖ.First, we need expressions for aₙ and bₙ.From part (1), we know that {aₙ + 1/2} is a geometric sequence with ratio 3. The first term x₁ = a₁ + 1/2 = 1 + 1/2 = 3/2. Therefore, the general term is xₙ = (3/2) * 3^{n-1} = (3/2) * 3^{n-1} = 3^{n}/2. Therefore, aₙ = xₙ - 1/2 = (3^{n}/2) - 1/2 = (3^{n} - 1)/2.Let me verify this with the earlier terms:For n=1: (3^1 -1)/2 = (3-1)/2 = 1. Correct.For n=2: (9 -1)/2 = 8/2 = 4. Correct.For n=3: (27 -1)/2 = 26/2 =13. Correct. So yes, this formula works.Now, for the arithmetic sequence {bₙ}. It's given that b₁ = 1, and it's arithmetic. But we need to find the common difference d. The problem also gives that b₃ + 1 = a₂. Let's use that.First, find b₃. For arithmetic sequence, b₃ = b₁ + 2d = 1 + 2d. Then, b₃ +1 = (1 + 2d) +1 = 2 + 2d. According to the problem, this equals a₂. From above, a₂ =4. Therefore:2 + 2d =4Subtract 2: 2d =2Divide by 2: d=1Therefore, the common difference d is 1. Therefore, the arithmetic sequence {bₙ} is bₙ =1 + (n-1)*1 =n.Wait, that's nice. So, bₙ =n. Let me confirm:b₁ =1, b₂=2, b₃=3. Then, b₃ +1 =4, which is a₂. Correct. So, yes, bₙ =n.So, now cₙ =aₙ * bₙ = [(3^{n} -1)/2] *n = [n*3^{n} -n]/2.Therefore, Tₙ = Σₖ=1ⁿ [k*3^{k} -k]/2 = (1/2)[Σₖ=1ⁿ k*3^{k} - Σₖ=1ⁿk]So, Tₙ = (1/2)[S - Σk], where S = Σₖ=1ⁿk*3^{k} and Σk = n(n+1)/2.Therefore, we need to compute S = Σₖ=1ⁿk*3^{k}.I remember that there's a formula for the sum Σₖ=1ⁿk*r^{k}. Let me recall. The formula is r*(1 - (n+1)*r^{n} +n*r^{n+1}) ) / (1 -r)^2 when r ≠1. Let me verify this.Alternatively, let's derive it. Let S = Σₖ=1ⁿk*r^{k}Consider S = r + 2r² + 3r³ + ... +n rⁿMultiply both sides by r:rS = r² + 2r³ + ... + (n-1)rⁿ +n r^{n+1}Subtract:S - rS = (r + r² + r³ + ... +rⁿ) -n r^{n+1}Left side: S(1 - r)Right side: Σₖ=1ⁿ r^{k} -n r^{n+1} = [r(rⁿ -1)/(r-1)] -n r^{n+1}Wait, but if r ≠1, the sum Σₖ=1ⁿ r^{k} = r*(rⁿ -1)/(r -1). So, substituting:S(1 - r) = [r(rⁿ -1)/(r -1)] -n r^{n+1}Therefore,S = [r(rⁿ -1)/(r -1)² -n r^{n+1}/(1 - r)]Hmm, let me check again.Wait, starting over:S = Σₖ=1ⁿk*r^{k}rS = Σₖ=1ⁿk*r^{k+1} = Σₖ=2}^{n+1} (k-1) r^{k}Subtracting:S - rS = S(1 - r) = Σₖ=1ⁿk r^{k} - Σₖ=2}^{n+1} (k-1) r^{k}= Σₖ=1ⁿk r^{k} - Σₖ=2}^{n} (k-1) r^{k} -n r^{n+1}= [r + Σₖ=2ⁿk r^{k}] - [Σₖ=2ⁿ (k-1) r^{k}] -n r^{n+1}= r + Σₖ=2ⁿ [k - (k -1)] r^{k} -n r^{n+1}= r + Σₖ=2ⁿ r^{k} -n r^{n+1}= Σₖ=1ⁿ r^{k} -n r^{n+1}Therefore, S(1 - r) = Σₖ=1ⁿ r^{k} -n r^{n+1}So, S = [Σₖ=1ⁿ r^{k} -n r^{n+1}]/(1 - r)Since Σₖ=1ⁿ r^{k} = r*(rⁿ -1)/(r -1)Thus,S = [r(rⁿ -1)/(r -1) -n r^{n+1}]/(1 - r)Factor out r^{n+1} in the numerator? Wait, maybe simplify step by step.First, compute Σₖ=1ⁿ r^{k} = r(rⁿ -1)/(r -1). So,S = [r(rⁿ -1)/(r -1) -n r^{n+1}]/(1 - r)Note that 1 - r = -(r -1), so denominator becomes -(r -1). Therefore,S = - [r(rⁿ -1)/(r -1) -n r^{n+1}]/(r -1)= - [r(rⁿ -1) -n r^{n+1}(r -1)] / (r -1)^2Wait, maybe that's complicating. Let me just plug r=3.Since in our problem, r=3. Let's compute S for r=3.Thus, S = [3*(3ⁿ -1)/(3 -1) -n*3^{n+1}]/(1 -3)= [3*(3ⁿ -1)/2 -n*3^{n+1}]/(-2)Multiply numerator and denominator:= [3*(3ⁿ -1)/2 -n*3^{n+1}]/(-2)Let me compute numerator:First term: 3*(3ⁿ -1)/2 = (3^{n+1} -3)/2Second term: -n*3^{n+1}So, numerator: (3^{n+1} -3)/2 -n*3^{n+1} = [3^{n+1} -3 -2n*3^{n+1}]/2 = [ (1 -2n)3^{n+1} -3 ] /2Then, divide by -2:S = [ (1 -2n)3^{n+1} -3 ] / (2*(-2)) = [ (1 -2n)3^{n+1} -3 ] / (-4)Multiply numerator and denominator by -1:S = [ (2n -1)3^{n+1} +3 ] /4Hmm, let's check that again. Maybe there's a miscalculation. Let me re-express.Alternatively, perhaps using the formula from the beginning. Let me check with n=1,2,3.For n=1: S=1*3^1=3. Let's plug into the formula:[ (2*1 -1)3^{1+1} +3 ] /4 = [ (1)9 +3 ] /4 =12/4=3. Correct.For n=2: S=1*3 +2*9=3+18=21. Formula:[ (2*2 -1)3^{3} +3 ] /4= [3*27 +3]/4=(81+3)/4=84/4=21. Correct.For n=3: S=1*3 +2*9 +3*27=3+18+81=102. Formula:[ (2*3 -1)3^4 +3 ] /4= [5*81 +3]/4=(405 +3)/4=408/4=102. Correct. So, the formula works.Therefore, S=Σₖ=1ⁿk*3^{k}= [ (2n -1)3^{n+1} +3 ] /4.Therefore, going back to Tₙ.Tₙ=(1/2)[S - Σk]=(1/2)[ [ (2n -1)3^{n+1} +3 ] /4 - n(n+1)/2 ]Let me compute this step by step.First, compute S = [ (2n -1)3^{n+1} +3 ] /4.Then, compute Σk =n(n+1)/2.Therefore,Tₙ=(1/2)[ ( (2n -1)3^{n+1} +3 ) /4 - (n(n+1))/2 ]To combine the terms, let's get a common denominator of 4.So,= (1/2)[ ( (2n -1)3^{n+1} +3 -2n(n+1) ) /4 ]Multiply 1/2:= [ (2n -1)3^{n+1} +3 -2n(n+1) ] /8Let me expand and simplify:First term: (2n -1)3^{n+1}Second term: +3Third term: -2n(n+1)So, expanding the third term: -2n² -2n.But perhaps factor 3^{n+1}?Alternatively, let's see:The expression is [ (2n -1)3^{n+1} -2n² -2n +3 ] /8Maybe factor 3^{n} from the first term:= [3^{n}(2n -1)3 -2n² -2n +3 ] /8= [3^{n+1}(2n -1) -2n² -2n +3 ] /8But I don't think that helps much. Maybe leave it as is.Alternatively, check if we can factor something else. Let me test for n=1,2,3.For n=1:T₁ = c₁ =a₁b₁=1*1=1.Using the formula:[ (2*1 -1)3^{2} +3 -2*1*2 ] /8= [1*9 +3 -4]/8= (9+3-4)/8=8/8=1. Correct.For n=2:T₂ =c₁ +c₂=1 + ( (3² -1)/2 )*2=1 + (9-1)/2 *2=1 +8=9.Using formula:[ (2*2 -1)3^{3} +3 -2*2*3 ] /8= [3*27 +3 -12]/8=(81 +3 -12)/8=72/8=9. Correct.For n=3:T₃=1 +8 + ( (3³ -1)/2 )*3=1 +8 + (27 -1)/2 *3=9 +26/2 *3=9 +13*3=9 +39=48.Using formula:[ (2*3 -1)3^{4} +3 -2*3*4 ] /8= [5*81 +3 -24]/8=(405 +3 -24)/8=384/8=48. Correct.So, the formula works. Therefore, the sum Tₙ is [ (2n -1)3^{n+1} -2n² -2n +3 ] /8.But perhaps we can write it in a more factored form or combine terms.Let me see:Original expression:Tₙ = [ (2n -1)3^{n+1} +3 -2n(n+1) ] /8We can write this as:= [ (2n -1)3^{n+1} -2n² -2n +3 ] /8Alternatively, factor 3 from the first two terms? Not sure. Alternatively, factor 3^{n+1}:But maybe that's not helpful. Alternatively, write it as:= [ (2n -1)3^{n+1} -2n² -2n +3 ] /8Alternatively, factor 3^{n+1} as follows:But perhaps leave it in this form. Alternatively, notice that 3^{n+1} =3*3^{n}, so:= [3*3^{n}(2n -1) -2n² -2n +3 ] /8But still, unless there's a common factor, which I don't see.Alternatively, check if 2n² +2n -3 can be related to the other terms? Not directly.Alternatively, perhaps leave the answer as it is. Since the problem says "Find the sum...", and we have a closed-form expression, even if it's a bit complex, that's acceptable. But maybe simplify it a bit more.Let me check for possible factorization. Let me compute:Numerator:(2n -1)3^{n+1} -2n² -2n +3= 3^{n+1}(2n -1) -2n(n +1) +3Perhaps try to factor terms with 3^{n+1} and polynomial terms.Alternatively, maybe factor 3^{n+1}:But it's a sum of exponential and quadratic terms, so likely can't be factored further. Therefore, the answer is:Tₙ = [ (2n -1)3^{n+1} -2n² -2n +3 ] /8Alternatively, write the numerator as (2n -1)3^{n+1} +3 -2n(n +1). Maybe that's as simplified as it can get. Let me check with the examples again to see if this form is acceptable.For n=1:Numerator: (2*1 -1)3^{2} +3 -2*1*(1 +1) =1*9 +3 -4=8. 8/8=1. Correct.For n=2:(3*27 +3 -12)=81+3-12=72. 72/8=9. Correct.For n=3:(5*81 +3 -24)=405 +3 -24=384. 384/8=48. Correct.Therefore, the expression is correct. So, the final answer is this fraction. However, perhaps we can write it in another way. Let me see:Let me expand (2n -1)3^{n+1}:= (2n -1)*3*3^{n} =3(2n -1)3^{n}So, numerator:=3(2n -1)3^{n} -2n² -2n +3Alternatively, factor 3 from the first term and the last term:=3[(2n -1)3^{n} +1] -2n(n +1)But I don't know if that's helpful. Alternatively, maybe factor 3^{n}:But 3^{n} is multiplied by 3(2n -1), so:=3^{n}*3(2n -1) -2n(n +1) +3But again, not much better. Therefore, perhaps the original form is the simplest.Alternatively, if we factor numerator:But I don't see an obvious factorization. Let's see:Let me write the numerator:(2n -1)3^{n+1} -2n² -2n +3= (2n -1)3^{n+1} -2n² -2n +3Is there a way to factor this? Let me check for n=1,2,3:For n=1: 1*9 -2 -2 +3=9-2-2+3=8For n=2: 3*27 -8 -4 +3=81-8-4+3=72For n=3:5*81 -18 -6 +3=405-18-6+3=384No obvious common factors except 8,72,384 which are 8*1,8*9,8*48. Not helpful.Alternatively, perhaps we can write the numerator as (2n -1)3^{n+1} +3 -2n(n +1). Since 2n(n +1) is the quadratic part, and 3^{n+1} is the exponential part. Since these are different types of terms, they can't be combined further. Therefore, the expression is as simplified as possible.Therefore, the answer is Tₙ = [ (2n -1)3^{n+1} -2n² -2n +3 ] /8.Alternatively, we can factor out a 3 from the first term:= [ 3*(2n -1)3^{n} -2n² -2n +3 ] /8But that might not help. Alternatively, split the numerator into two parts:= [ (2n -1)3^{n+1} +3 ] /8 - [2n(n +1)] /8= [ (2n -1)3^{n+1} +3 ] /8 - [n(n +1)/4]But that might not help either. Since the problem just asks for the sum, and we have a closed-form expression, even if it's a bit complicated, that's acceptable.Alternatively, maybe there's a different approach to find Tₙ. Let me think. Since cₙ = aₙ * bₙ = [(3ⁿ -1)/2] *n = (n*3ⁿ -n)/2. Therefore, Tₙ= (1/2)(Σn*3ⁿ -Σn). We already found Σn*3ⁿ = S = [ (2n -1)3^{n+1} +3 ] /4, and Σn =n(n+1)/2. Therefore, substituting:Tₙ = (1/2)( [ (2n -1)3^{n+1} +3 ] /4 -n(n+1)/2 )Multiply 1/2:= [ (2n -1)3^{n+1} +3 ] /8 -n(n+1)/4Combine terms over a common denominator of 8:= [ (2n -1)3^{n+1} +3 -2n(n+1) ] /8Which is the same as before. Therefore, that's the simplest form.So, summarizing:For part (1), we showed that by defining xₙ =aₙ +1/2, the recurrence becomes xₙ₊₁=3xₙ, hence a geometric sequence with ratio 3.For part (2), after finding expressions for aₙ and bₙ, we computed cₙ =aₙbₙ, and found the sum Tₙ by using the formula for Σk*3^k, leading to the closed-form expression [ (2n -1)3^{n+1} -2n² -2n +3 ] /8.Therefore, the final answers are:(1) Proven that {aₙ +1/2} is geometric.(2) Tₙ = [ (2n -1)3^{n+1} -2n² -2n +3 ] /8Final Answer(1) boxed{left{a_n + dfrac{1}{2}right} text{ is a geometric sequence with common ratio } 3}(2) boxed{T_n = dfrac{(2n - 1) cdot 3^{n+1} - 2n^2 - 2n + 3}{8}}</think>Given that the sequence ({b_{n}}) is an arithmetic sequence with the first term being 1, and the sequence ({a_{n}}) satisfies (a_{n+1} - 3a_{n} - 1 = 0) with (a_{1} = 1). We also know that (b_{3} + 1 = a_{2}).(1) To prove that (left{a_n + frac{1}{2}right}) forms a geometric sequence:- Define (x_n = a_n + frac{1}{2}). Substituting into the recurrence relation (a_{n+1} = 3a_n + 1), we get: [ x_{n+1} - frac{1}{2} = 3left(x_n - frac{1}{2}right) + 1 ] Simplifying, we find: [ x_{n+1} = 3x_n ] This shows that ({x_n}) is a geometric sequence with ratio 3. The first term (x_1 = a_1 + frac{1}{2} = frac{3}{2}). Therefore, (left{a_n + frac{1}{2}right}) is a geometric sequence with common ratio 3.(2) To find the sum of the first (n) terms of the sequence ({c_n}) where (c_n = a_n cdot b_n):- First, determine the general terms for (a_n) and (b_n): - Since (left{a_n + frac{1}{2}right}) is a geometric sequence, (a_n = frac{3^n - 1}{2}). - The arithmetic sequence ({b_n}) has (b_1 = 1) and common difference (d = 1) (found using (b_3 + 1 = a_2)), so (b_n = n).- Then, (c_n = a_n cdot b_n = frac{n(3^n - 1)}{2}).- The sum (T_n) is calculated as: [ T_n = sum_{k=1}^n c_k = frac{1}{2} left( sum_{k=1}^n k cdot 3^k - sum_{k=1}^n k right) ] - Using the formula for (sum_{k=1}^n k cdot 3^k): [ sum_{k=1}^n k cdot 3^k = frac{(2n - 1) cdot 3^{n+1} + 3}{4} ] - The sum (sum_{k=1}^n k = frac{n(n+1)}{2}).- Combining these results: [ T_n = frac{1}{2} left( frac{(2n - 1) cdot 3^{n+1} + 3}{4} - frac{n(n+1)}{2} right) ] Simplifying, we get: [ T_n = frac{(2n - 1) cdot 3^{n+1} - 2n^2 - 2n + 3}{8} ] Final Answer(1) boxed{left{a_n + dfrac{1}{2}right} text{ is a geometric sequence with common ratio } 3}(2) boxed{T_n = dfrac{(2n - 1) cdot 3^{n+1} - 2n^2 - 2n + 3}{8}}]

🔑: Part (1) Proof:Given the recursive formula for the sequence {a_n} is a_{n+1}-3a_{n}-1=0, we can rearrange this to find an explicit formula for a_{n+1} in terms of a_n:begin{align*}a_{n+1} &= 3a_{n} + 1 implies a_{n+1} + frac{1}{2} &= 3a_{n} + 1 + frac{1}{2} &= 3left(a_{n} + frac{1}{2}right).end{align*}This shows that the sequence left{a_n + frac{1}{2}right} is a geometric sequence with the first term a_1 + frac{1}{2} = 1 + frac{1}{2} = frac{3}{2} and common ratio 3. Therefore, we have proven that left{a_n + frac{1}{2}right} forms a geometric sequence. Part (2) Finding T_n:Given a_n = frac{3^n - 1}{2}, and knowing that the sequence {b_n} is arithmetic with b_3 + 1 = a_2 and a_1 = 1, we find the common difference d of {b_n}:begin{align*}b_3 + 1 &= a_2 1 + 2d + 1 &= frac{3^2 - 1}{2} 2d &= frac{8}{2} - 2 d &= 1.end{align*}Thus, b_n = 1 + (n-1)d = n. The sequence {c_n} is defined as c_n = a_n cdot b_n = frac{3^n - 1}{2} cdot n.To find T_n, the sum of the first n terms of {c_n}:begin{align*}T_n &= sum_{k=1}^{n} c_k &= frac{1}{2} sum_{k=1}^{n} (k cdot 3^k - k) &= frac{1}{2} left[sum_{k=1}^{n} k cdot 3^k - sum_{k=1}^{n} kright].end{align*}Let M_n = sum_{k=1}^{n} k cdot 3^k, and calculate 3M_n - M_n to find a formula for M_n:begin{align*}3M_n - M_n &= sum_{k=1}^{n} k cdot 3^{k+1} - sum_{k=1}^{n} k cdot 3^k &= 3^{n+1} + 3^2 + cdots + n cdot 3^{n+1} - (3 + 3^2 + cdots + n cdot 3^n) &= -frac{2n-1}{2} cdot 3^{n+1} - frac{3}{2}.end{align*}Solving for M_n, we get M_n = frac{2n-1}{4} cdot 3^{n+1} + frac{3}{4}. Substituting M_n and the sum of the first n natural numbers into the formula for T_n:begin{align*}T_n &= frac{1}{2} left[frac{2n-1}{4} cdot 3^{n+1} + frac{3}{4} - frac{n(n+1)}{2}right] &= frac{2n-1}{8} cdot 3^{n+1} + frac{3}{8} - frac{n(n+1)}{4}.end{align*}Thus, the sum of the first n terms of the sequence {c_n}, denoted as T_n, is boxed{frac{2n-1}{8} cdot 3^{n+1} + frac{3}{8} - frac{n(n+1)}{4}}.

🧠:[<think>Okay, so I need to solve this inequality:9 cdot frac{|x+4| - |x-2|}{|3x+14| - |3x-8|} + 11 cdot frac{|x+4| + |x-2|}{|3x+14| + |3x-8|} leqslant 6And then find the sum of all integer solutions where |x| < 110. Alright, let's try to break this down step by step.First, this looks complicated with absolute values in both numerator and denominator. Maybe I should consider different cases based on the critical points where the expressions inside the absolute values change their sign. The critical points occur where the arguments of the absolute values are zero. Let's find those.For |x + 4| and |x - 2|, the critical points are at x = -4 and x = 2.For the denominators |3x +14| and |3x -8|, set 3x +14 = 0 and 3x -8 = 0:3x +14 = 0 ⇒ x = -14/3 ≈ -4.66673x -8 = 0 ⇒ x = 8/3 ≈ 2.6667So the critical points are x = -14/3, -4, 2, 8/3.These divide the real line into intervals where the expressions inside the absolute values have consistent signs. Let's list the intervals:1. x < -14/3 ≈ -4.66672. -14/3 ≤ x < -43. -4 ≤ x < 24. 2 ≤ x < 8/3 ≈ 2.66675. x ≥ 8/3Wait, but x = -4 and x = 2 are also critical points from the numerators. So these intervals are split by all critical points. Let me check again.The critical points in order from left to right are:x = -14/3 ≈ -4.6667,x = -4,x = 2,x = 8/3 ≈ 2.6667.Therefore, the intervals are:1. x < -14/32. -14/3 ≤ x < -43. -4 ≤ x < 24. 2 ≤ x < 8/35. x ≥ 8/3In each interval, the absolute value expressions can be simplified by considering their sign. Let's handle each interval one by one.But before diving into each interval, maybe I can see if there's a substitution or simplification possible. The structure of the inequality has two fractions with similar numerators and denominators, multiplied by coefficients 9 and 11. Maybe there's a way to combine them or find a substitution. Let me denote:Let A = |x + 4| - |x - 2|B = |3x +14| - |3x -8|C = |x + 4| + |x - 2|D = |3x +14| + |3x -8|Then the inequality is:9*(A/B) + 11*(C/D) ≤ 6Hmm. Not sure if that helps directly. Alternatively, maybe if I can find expressions for A/B and C/D in each interval, then compute 9*(A/B) + 11*(C/D) and see where it's ≤6.Alternatively, note that A = |x +4| - |x -2| and C = |x +4| + |x -2|. Similarly, B = |3x +14| - |3x -8| and D = |3x +14| + |3x -8|. So maybe C and D are always positive, while A and B could be positive or negative depending on x. However, denominators B and D: since B is in the denominator, we must ensure that B ≠ 0, and similarly D is a denominator, so D ≠ 0. Let's check when D can be zero. |3x +14| + |3x -8| = 0 only when both |3x +14| = 0 and |3x -8| = 0, which is impossible since 3x +14 =0 ⇒ x = -14/3, and 3x -8=0 ⇒ x =8/3. So D is always positive. Similarly, B = |3x +14| - |3x -8|. When can B = 0? When |3x +14| = |3x -8|. Let's solve that:|3x +14| = |3x -8|Which implies 3x +14 = 3x -8 or 3x +14 = - (3x -8)First equation: 3x +14 = 3x -8 ⇒ 14 = -8, which is impossible.Second equation: 3x +14 = -3x +8 ⇒ 6x = -6 ⇒ x = -1.Therefore, B = 0 when x = -1. So in the original inequality, the term 9*(A/B) would be undefined at x = -1. Therefore, x = -1 is excluded from the domain.Similarly, check if x = -1 is in any of our intervals. The critical points are at -14/3 ≈ -4.6667, -4, 2, 8/3 ≈ 2.6667. So x = -1 is in the interval -4 ≤ x < 2. So in that interval, we need to exclude x = -1.So when handling the interval -4 ≤ x < 2, we have to consider x ≠ -1.Similarly, in other intervals, B might be positive or negative. Let's check in each interval.But perhaps let's proceed with each interval one by one.Interval 1: x < -14/3 ≈ -4.6667In this interval, let's evaluate each absolute value expression:|x +4|: Since x < -14/3 ≈ -4.6667, x +4 < -14/3 +4 = (-14 +12)/3 = -2/3 ≈ -0.6667. So x +4 < 0, so |x +4| = -(x +4) = -x -4.|x -2|: x < -14/3, so x -2 < -14/3 -6/3 = -20/3 ≈ -6.6667 < 0. So |x -2| = -(x -2) = -x +2.Similarly, denominators:|3x +14|: 3x +14. Since x < -14/3, 3x < -14, so 3x +14 < 0. Therefore, |3x +14| = -(3x +14) = -3x -14.|3x -8|: 3x -8. Since x < -14/3, 3x < -14, so 3x -8 < -14 -8 = -22 < 0. So |3x -8| = -(3x -8) = -3x +8.Therefore, compute A, B, C, D:A = |x +4| - |x -2| = (-x -4) - (-x +2) = -x -4 +x -2 = -6B = |3x +14| - |3x -8| = (-3x -14) - (-3x +8) = -3x -14 +3x -8 = -22C = |x +4| + |x -2| = (-x -4) + (-x +2) = -2x -2D = |3x +14| + |3x -8| = (-3x -14) + (-3x +8) = -6x -6So substituting into the inequality:9*(A/B) + 11*(C/D) ≤ 6Plug in the values:9*(-6 / (-22)) + 11*((-2x -2)/(-6x -6)) ≤ 6Simplify each term:First term: 9*(6/22) = 9*(3/11) = 27/11 ≈ 2.4545Second term: 11*((-2x -2)/(-6x -6)) = 11*((2x +2)/(6x +6)) = 11*((2(x +1))/(6(x +1))) = 11*(1/3) = 11/3 ≈ 3.6667Wait, let's check that:(-2x -2)/(-6x -6) = [ -2(x +1) ] / [ -6(x +1) ] = (2(x +1))/(6(x +1)) = 1/3, provided that x +1 ≠ 0, i.e., x ≠ -1.But in this interval, x < -14/3 ≈ -4.6667, so x is less than -4.6667, so x +1 is definitely less than -3.6667, so x +1 ≠ 0. So the expression simplifies to 1/3.Therefore, the second term is 11*(1/3) = 11/3 ≈ 3.6667Therefore, total left-hand side (LHS) is 27/11 + 11/3 ≈ 2.4545 + 3.6667 ≈ 6.1212So 6.1212 ≤ 6? That's approximately 6.12 ≤ 6, which is false. Therefore, in this interval, the inequality does not hold. So no solutions in interval 1.Wait, but maybe we need to check exact fractions rather than approximate.Compute 27/11 + 11/3 = (27*3 + 11*11)/33 = (81 + 121)/33 = 202/33 ≈ 6.1212, which is indeed greater than 6. So the inequality 202/33 ≤ 6 is false. So no solutions here.Interval 2: -14/3 ≤ x < -4 (approximately -4.6667 ≤ x < -4)In this interval, evaluate the absolute values:|x +4|: Since x < -4, x +4 < 0, so |x +4| = -x -4.|x -2|: x < -4, so x -2 < -6 < 0. So |x -2| = -x +2.Denominators:|3x +14|: 3x +14. Since x ≥ -14/3 in this interval, 3x ≥ -14, so 3x +14 ≥ 0. So |3x +14| = 3x +14.|3x -8|: x < -4, so 3x < -12, so 3x -8 < -20 < 0. So |3x -8| = -3x +8.So compute A, B, C, D:A = |x +4| - |x -2| = (-x -4) - (-x +2) = -x -4 +x -2 = -6B = |3x +14| - |3x -8| = (3x +14) - (-3x +8) = 3x +14 +3x -8 = 6x +6C = |x +4| + |x -2| = (-x -4) + (-x +2) = -2x -2D = |3x +14| + |3x -8| = (3x +14) + (-3x +8) = 22Wait, 3x +14 -3x +8 = 22? Yes, 14 +8 =22. So D =22.Therefore, substitute into the inequality:9*(A/B) + 11*(C/D) ≤6Plugging in:9*(-6/(6x +6)) + 11*(-2x -2)/22 ≤6Simplify each term:First term: 9*(-6)/(6x +6) = 9*(-6)/(6(x +1)) = 9*(-6)/(6(x +1)) = 9*(-1)/(x +1) = -9/(x +1)Second term: 11*(-2x -2)/22 = 11*(-2)(x +1)/22 = (-22(x +1))/22 = -(x +1)So the inequality becomes:-9/(x +1) - (x +1) ≤6Let me write that as:- (x +1) - 9/(x +1) ≤6Multiply both sides by -1 (which reverses the inequality):(x +1) + 9/(x +1) ≥ -6But let's handle this step by step. Let's denote t = x +1. Then the inequality is:- t - 9/t ≤6Multiply both sides by -1 (remembering to reverse the inequality):t + 9/t ≥ -6But since x is in [-14/3, -4), then x +1 is in [-14/3 +1, -4 +1) = [-11/3, -3). So t is in [-11/3, -3). So t is negative.So t + 9/t ≥ -6Let’s let t = -s where s >0, because t is negative. So s = -t, which is in (3, 11/3]. Then:Left-hand side (LHS) becomes -s + 9/(-s) = -s -9/sSo inequality:-s -9/s ≥ -6Multiply both sides by -1 (reverse inequality):s +9/s ≤6So s +9/s ≤6, where s ∈ (3, 11/3]Wait, 11/3 is approximately 3.6667. So s is between 3 and 3.6667.We can solve s +9/s ≤6.Multiply both sides by s (since s >0):s² +9 ≤6sBring all terms to left:s² -6s +9 ≤0Which factors as (s -3)² ≤0But (s -3)² is always non-negative, and equals 0 only when s =3. Therefore, the inequality (s -3)² ≤0 holds only when s=3.But s is in (3, 11/3]. Therefore, there are no solutions here, except s=3, which is not in the interval. So in this interval, there are no solutions. Hence, in interval 2, no solutions.Interval 3: -4 ≤ x < 2In this interval, evaluate absolute values:|x +4|: Since x ≥ -4, x +4 ≥0, so |x +4| = x +4.|x -2|: x <2, so x -2 <0, so |x -2| = -x +2.Denominators:|3x +14|: 3x +14. Since x ≥ -4, 3x ≥ -12, so 3x +14 ≥ -12 +14 =2>0. So |3x +14| =3x +14.|3x -8|: x <2, so 3x <6, so 3x -8 <6 -8 =-2<0. So |3x -8| = -3x +8.Compute A, B, C, D:A = |x +4| - |x -2| = (x +4) - (-x +2) =x +4 +x -2=2x +2B = |3x +14| - |3x -8| = (3x +14) - (-3x +8) =3x +14 +3x -8=6x +6C = |x +4| + |x -2| = (x +4) + (-x +2) =6D = |3x +14| + |3x -8| = (3x +14) + (-3x +8) =22So substituting into the inequality:9*(A/B) +11*(C/D) ≤6Plug in the values:9*( (2x +2)/(6x +6) ) +11*(6/22) ≤6Simplify each term:First term: 9*(2x +2)/(6x +6) =9*(2(x +1))/(6(x +1)) =9*(2/6)=9*(1/3)=3. But note that x +1 is in the denominator. However, in this interval, x is between -4 and 2. So x +1 is between -3 and 3. But x ≠ -1, because when x =-1, the denominator B =6x +6=6*(-1)+6=0, which makes the first term undefined. So we need to exclude x = -1.Therefore, for x ≠ -1 in this interval, the first term simplifies to 3. The second term is 11*(6/22)=11*(3/11)=3.Therefore, LHS =3 +3=6.So the inequality becomes 6 ≤6, which holds as equality. Therefore, all x in this interval except x =-1 satisfy the inequality.But wait, we need to check if there are any restrictions from the denominators. The denominator B is 6x +6. So x ≠ -1. So in interval 3 (-4 ≤x <2), x ≠ -1. Therefore, the solution here is x ∈ [-4, 2) {-1}.But wait, in interval 3, x is from -4 to 2. So the solutions here are all x between -4 and 2 except x=-1.But we need to check if x=-1 is in the interval. Yes, x=-1 is between -4 and 2, so we exclude it. Therefore, solutions in interval 3 are [-4, -1) ∪ (-1, 2).But wait, the interval is -4 ≤x <2, so when we remove x=-1, it's split into two intervals: [-4, -1) and (-1, 2).So all x in [-4, -1) ∪ (-1, 2) satisfy the inequality.But since the original inequality is ≤6, and for these x values, the LHS is exactly 6. So equality holds. Therefore, the solution is x in [-4,2) with x ≠ -1.So in this interval, the solutions are x ∈ [-4, -1) ∪ (-1, 2).Interval 4: 2 ≤x <8/3 ≈2.6667In this interval, evaluate absolute values:|x +4|: x ≥2, so x +4 ≥6>0. So |x +4|=x +4.|x -2|: x ≥2, so x -2 ≥0. So |x -2| =x -2.Denominators:|3x +14|: 3x +14. Since x ≥2, 3x ≥6, so 3x +14 ≥20>0. So |3x +14|=3x +14.|3x -8|: x <8/3≈2.6667, so 3x <8. So 3x -8 <0. Hence, |3x -8|=-3x +8.Compute A, B, C, D:A = |x +4| - |x -2| = (x +4) - (x -2)=x +4 -x +2=6B = |3x +14| - |3x -8|=(3x +14)-(-3x +8)=3x +14 +3x -8=6x +6C = |x +4| + |x -2|=x +4 +x -2=2x +2D = |3x +14| + |3x -8|=(3x +14) + (-3x +8)=22So substitute into inequality:9*(6/(6x +6)) +11*((2x +2)/22) ≤6Simplify each term:First term: 9*6/(6x +6)=54/(6(x +1))=9/(x +1)Second term:11*(2x +2)/22= (22x +22)/22= x +1Therefore, inequality becomes:9/(x +1) + (x +1) ≤6Let t = x +1. Then x = t -1. Since x ∈ [2,8/3), t ∈ [3,8/3 +1)= [3, 11/3≈3.6667)So the inequality is:9/t + t ≤6Multiply both sides by t (since t >0):9 + t² ≤6tBring all terms to left:t² -6t +9 ≤0Which factors as (t -3)² ≤0Again, (t -3)^2 is always non-negative, and equal to 0 only when t=3. So the inequality holds only when t=3. Thus, t=3 ⇒x +1=3⇒x=2.But x=2 is the left endpoint of this interval. The interval is 2 ≤x <8/3, so x=2 is included. Therefore, x=2 is a solution.Therefore, in interval 4, the only solution is x=2.Interval 5: x ≥8/3≈2.6667In this interval, evaluate absolute values:|x +4|: x ≥8/3, so x +4 >8/3 +12/3=20/3>0. So |x +4|=x +4.|x -2|: x ≥8/3>2, so x -2≥8/3 -6/3=2/3>0. So |x -2|=x -2.Denominators:|3x +14|: 3x +14. Since x ≥8/3, 3x ≥8, so 3x +14 ≥8 +14=22>0. So |3x +14|=3x +14.|3x -8|: x ≥8/3, so 3x ≥8, so |3x -8|=3x -8.Compute A, B, C, D:A = |x +4| - |x -2|=x +4 -x +2=6B = |3x +14| - |3x -8|=(3x +14) - (3x -8)=3x +14 -3x +8=22C = |x +4| + |x -2|=x +4 +x -2=2x +2D = |3x +14| + |3x -8|=(3x +14) + (3x -8)=6x +6Substitute into the inequality:9*(6/22) +11*((2x +2)/(6x +6)) ≤6Simplify each term:First term: 9*6/22=54/22=27/11≈2.4545Second term:11*(2x +2)/(6x +6)=11*(2(x +1))/(6(x +1))=11*(2/6)=11*(1/3)=11/3≈3.6667But note that (2x +2)/(6x +6)= (2(x +1))/(6(x +1))=1/3, provided x +1 ≠0. Since x ≥8/3, x +1 ≥11/3>0, so no problem. Therefore, the second term is 11/3.Therefore, total LHS=27/11 +11/3≈2.4545 +3.6667≈6.1212>6. So the inequality 6.1212 ≤6 is false. Hence, no solutions in interval 5.So summarizing the solutions from all intervals:- Interval 1: No solutions.- Interval 2: No solutions.- Interval 3: All x in [-4, -1) ∪ (-1, 2).- Interval 4: x=2.- Interval 5: No solutions.But wait, in interval 3, the solution is x ∈ [-4, -1) ∪ (-1, 2). Then in interval 4, x=2 is included. So overall, the solution set is [-4,2], but excluding x=-1. Wait, but in interval 3, x <2, and in interval 4, x=2. So combining, x ∈ [-4,2], x ≠-1.Wait, interval 3 is -4 ≤x <2, and interval 4 gives x=2, so combined, x ∈ [-4,2], with x ≠-1. So x=-1 is excluded.Therefore, the solution set is [-4, -1) ∪ (-1, 2]. But wait, interval 4 adds x=2, which is not included in interval 3. So the full solution is [-4, -1) ∪ (-1, 2], since x=2 is allowed. Wait, interval 4 is 2 ≤x <8/3, but the only solution there is x=2. So yes, the solution set is [-4,2], excluding x=-1.But need to check if x=2 is included. From interval 4, yes, x=2 is included, as the solution. So the solution is x ∈ [-4,2], x ≠-1.But wait, in interval 3, x <2, and in interval 4, x=2 is allowed. So the combined solution is [-4,2], excluding x=-1.Therefore, the integer solutions within |x| <110 would be all integers from -4 up to 2, excluding -1.But wait, let's list the integer points in [-4,2]:-4, -3, -2, -1, 0, 1, 2.But exclude -1. So the integer solutions are: -4, -3, -2, 0, 1, 2.Now sum these integers:-4 + (-3) + (-2) +0 +1 +2 = (-4 -3 -2) + (0 +1 +2) = (-9) +3 = -6But wait, hold on. Wait, the solution set is [-4,2], excluding x=-1. So in integers, x can be -4, -3, -2, 0,1,2. Because:-4 is included (as -4 is in [-4,2])-3 is in [-4,2]-2 is in [-4,2]-1 is excluded.0,1,2 are included.So sum: (-4) + (-3) + (-2) +0 +1 +2.Let's compute:Start from the left: (-4) + (-3) = -7-7 + (-2) = -9-9 +0 = -9-9 +1= -8-8 +2= -6So total sum is -6.But wait, is x=2 included? From interval 4, yes, x=2 is a solution. So yes, included. So that's correct.But wait, need to check if all these integers are within |x| <110. Since |x| <110 means x is in (-110, 110). All the integers -4, -3, -2, 0,1,2 are within that range. So sum is -6.Wait, but let me double-check the solution set.Original solution set:x ∈ [-4,2] {-1}Therefore, integer solutions in [-4,2] excluding -1: integers -4, -3, -2, 0,1,2. Sum is -6.Therefore, the answer is boxed{-6}Final Answerboxed{-6}</think>To solve the inequality9 cdot frac{|x+4| - |x-2|}{|3x+14| - |3x-8|} + 11 cdot frac{|x+4| + |x-2|}{|3x+14| + |3x-8|} leqslant 6we consider the critical points where the expressions inside the absolute values change their sign: ( x = -14/3 approx -4.6667 ), ( x = -4 ), ( x = 2 ), and ( x = 8/3 approx 2.6667 ). These points divide the real line into intervals. We analyze each interval:1. Interval ( x < -14/3 ): - The left-hand side (LHS) of the inequality evaluates to approximately 6.1212, which is greater than 6. No solutions here.2. Interval ( -14/3 leq x < -4 ): - The LHS simplifies to ( -9/(x + 1) - (x + 1) ). Solving this, we find no solutions in this interval.3. Interval ( -4 leq x < 2 ): - The LHS simplifies to 6, which holds as equality. However, ( x = -1 ) is excluded due to division by zero. Solutions are ( x in [-4, -1) cup (-1, 2) ).4. Interval ( 2 leq x < 8/3 ): - The LHS evaluates to 6 only at ( x = 2 ). Thus, ( x = 2 ) is a solution.5. Interval ( x geq 8/3 ): - The LHS evaluates to approximately 6.1212, which is greater than 6. No solutions here.Combining the solutions from all intervals, the solution set is ( [-4, 2] setminus {-1} ). The integer solutions within ( |x| < 110 ) are ( -4, -3, -2, 0, 1, 2 ). Summing these integers:[-4 + (-3) + (-2) + 0 + 1 + 2 = -6]Thus, the sum of the integer solutions is (boxed{-6}).,<think>Okay, so I need to solve this inequality:9 cdot frac{|x+4|-|x-2|}{|3x+14|-|3x-8|} + 11 cdot frac{|x+4|+|x-2|}{|3x+14|+|3x-8|} leqslant 6And then find the sum of integer solutions where |x| < 110. Hmm, this looks a bit intimidating with all the absolute values. Let me try to break it down step by step.First, I remember that absolute value expressions can change their behavior depending on whether the inside is positive or negative. So, I need to find the critical points where the expressions inside the absolute values change sign. That will help me divide the number line into intervals where I can handle the absolute values without the modulus.Looking at the numerator and denominator expressions:For |x + 4| and |x - 2| in the numerators:- x + 4 = 0 ⇒ x = -4- x - 2 = 0 ⇒ x = 2For the denominators |3x + 14| and |3x - 8|:- 3x + 14 = 0 ⇒ x = -14/3 ≈ -4.666...- 3x - 8 = 0 ⇒ x = 8/3 ≈ 2.666...So, the critical points are at x = -14/3, -4, 2, 8/3. These divide the real line into intervals where the expressions inside the absolute values have consistent signs. Let me list these intervals:1. x < -14/3 (≈ -4.666...)2. -14/3 ≤ x < -43. -4 ≤ x < 24. 2 ≤ x < 8/3 (≈ 2.666...)5. x ≥ 8/3But wait, actually, since -14/3 ≈ -4.666 is less than -4, the order is correct. So these are the intervals where the expressions inside each absolute value will have consistent signs, so we can remove the absolute values by considering their signs in each interval.Therefore, I need to analyze each interval separately and rewrite the original inequality without absolute values, then solve the inequality in each interval. Finally, combine the solutions and sum the integer solutions within |x| < 110.Let me start with the first interval: x < -14/3 ≈ -4.666...Interval 1: x < -14/3In this interval, let's determine the sign of each expression inside the absolute values.- x + 4: Since x < -14/3 ≈ -4.666, x + 4 < -14/3 + 4 = (-14 + 12)/3 = -2/3 < 0. So |x + 4| = -(x + 4)- x - 2: x < -4.666, so x - 2 < -4.666 - 2 = -6.666 < 0. So |x - 2| = -(x - 2)- 3x + 14: Let's check. 3x + 14 when x < -14/3: 3x + 14 < 3*(-14/3) + 14 = -14 +14 = 0. So |3x +14| = -(3x +14)- 3x - 8: 3x -8. Since x < -4.666, 3x < -14, so 3x -8 < -14 -8 = -22 <0. Thus, |3x -8| = -(3x -8)Now substitute these into the original inequality:First term numerator: |x+4| - |x-2| = (-x -4) - (-x +2) = -x -4 +x -2 = -6Denominator of first term: |3x +14| - |3x -8| = (-3x -14) - (-3x +8) = -3x -14 +3x -8 = -22So the first fraction becomes (-6)/(-22) = 6/22 = 3/11Second term numerator: |x+4| + |x-2| = (-x -4) + (-x +2) = -2x -2Denominator of second term: |3x +14| + |3x -8| = (-3x -14) + (-3x +8) = -6x -6So the second fraction is (-2x -2)/(-6x -6) = (2x + 2)/(6x +6) = (2(x +1))/(6(x +1)) = (x +1)/(3(x +1)) = 1/3, provided that x +1 ≠ 0. But in this interval x < -14/3 ≈ -4.666, so x +1 < -14/3 +1 = -11/3 ≈ -3.666 < 0, so x +1 ≠ 0 here.Therefore, the second term simplifies to 1/3.Therefore, substituting back into the original inequality:9*(3/11) + 11*(1/3) ≤ 6Calculate left-hand side (LHS):9*(3/11) = 27/11 ≈ 2.454511*(1/3) ≈ 3.6667Sum ≈ 2.4545 + 3.6667 ≈ 6.1212So, approximately 6.1212 ≤ 6? That's not true. So in this interval, the inequality does not hold.Therefore, no solutions in interval 1.Interval 2: -14/3 ≤ x < -4Let's check the expressions here:- x +4: Since x < -4, x +4 <0. So |x +4| = -x -4- x -2: x < -4 < 2, so x -2 <0. So |x -2| = -x +2- 3x +14: At x ≥ -14/3, 3x +14 ≥0. So |3x +14| = 3x +14- 3x -8: Since x < -4, 3x < -12, so 3x -8 < -12 -8 = -20 <0. So |3x -8| = -3x +8Now compute the fractions:First term numerator: |x+4| - |x-2| = (-x -4) - (-x +2) = -x -4 +x -2 = -6Denominator: |3x +14| - |3x -8| = (3x +14) - (-3x +8) = 3x +14 +3x -8 = 6x +6So first fraction is -6/(6x +6) = -6/(6(x +1)) = -1/(x +1)Second term numerator: |x+4| + |x-2| = (-x -4) + (-x +2) = -2x -2Denominator: |3x +14| + |3x -8| = (3x +14) + (-3x +8) = 22So second fraction is (-2x -2)/22 = (-2(x +1))/22 = -(x +1)/11Therefore, the original inequality becomes:9*(-1/(x +1)) + 11*(-(x +1)/11) ≤6Simplify:-9/(x +1) - (x +1) ≤6Multiply through by -1 (remember to flip inequality sign):9/(x +1) + (x +1) ≥ -6But wait, maybe better to rearrange terms first:Let me write the inequality as:-9/(x +1) - (x +1) ≤6Let me move all terms to left-hand side:-9/(x +1) - (x +1) -6 ≤0But perhaps it's better to let y = x +1. Let me set y = x +1. Then x = y -1. Since -14/3 ≤x < -4, substituting:-14/3 ≤ y -1 < -4 ⇒ -14/3 +1 ≤ y < -4 +1 ⇒ -11/3 ≤ y < -3So y ∈ [-11/3, -3). Then the inequality becomes:-9/y - y ≤6Multiply both sides by y. But here we have to be careful because y is negative in this interval (since y < -3), so multiplying by y reverses the inequality sign.Starting with:-9/y - y ≤6Let me rearrange:-9/y - y -6 ≤0Multiply both sides by y (negative, so inequality flips):-9 - y^2 -6y ≥0Which is equivalent to:-y^2 -6y -9 ≥0Multiply both sides by -1 (inequality flips again):y^2 +6y +9 ≤0Factor left side:(y +3)^2 ≤0Since a square is always non-negative, the only solution is y +3 =0 ⇒ y = -3But in our interval for y: y ∈ [-11/3, -3). So y = -3 is the upper bound, but it's not included (since y < -3). Therefore, there's no solution in this interval.Wait, but y = -3 is not in the interval. So, the inequality (y +3)^2 ≤0 has no solution here. Therefore, in interval 2, there are no solutions.Interval 3: -4 ≤ x <2Now check the expressions:- x +4: Since x ≥-4, x +4 ≥0. So |x +4| = x +4- x -2: x <2, so x -2 <0. So |x -2| = -x +2- 3x +14: x ≥-4, so 3x ≥-12, 3x +14 ≥-12 +14 =2 >0. So |3x +14| =3x +14- 3x -8: x <2, so 3x <6, 3x -8 <6 -8 = -2 <0. So |3x -8| = -3x +8Compute the fractions:First term numerator: |x+4| - |x -2| = (x +4) - (-x +2) =x +4 +x -2 =2x +2Denominator: |3x +14| - |3x -8| = (3x +14) - (-3x +8) =3x +14 +3x -8 =6x +6First fraction: (2x +2)/(6x +6) = (2(x +1))/(6(x +1)) =1/3, provided that x +1 ≠0. Since -4 ≤x <2, x +1 ranges from -3 to 3, so x +1=0 ⇒x=-1 is in this interval. So we need to check if x=-1 is a problem.Second term numerator: |x+4| + |x -2| = (x +4) + (-x +2) =6Denominator: |3x +14| + |3x -8| = (3x +14) + (-3x +8) =22So second fraction is 6/22 =3/11Therefore, substituting into the original inequality:9*(1/3) +11*(3/11) ≤6Simplify:9*(1/3)=311*(3/11)=3So total LHS=3+3=6 ≤6. Equality holds.But wait, we need to check if x +1 ≠0 in the first fraction. Because when x=-1, the denominator in the first fraction becomes 6*(-1)+6=0, but we already substituted x=-1 into the first fraction's denominator. Wait, no:Wait, first term's denominator is 6x +6, which is 6(x +1). So at x=-1, denominator becomes 0. Therefore, x=-1 would make the first fraction undefined. Therefore, we have to exclude x=-1 from this interval.Therefore, the original inequality in interval 3 is valid for x ∈ [-4,2) except x=-1. And when x ≠-1, the inequality becomes 6 ≤6, which is equality. Therefore, the inequality holds for all x in [-4,2) except x=-1.But wait, the original inequality is ≤6. Since we get equality (6 ≤6) for all x in the interval except x=-1. Therefore, all x in [-4,2) except x=-1 are solutions. But x=-1 is excluded because it makes the first fraction undefined.Therefore, solutions in interval 3: [-4,2) {-1}But we need to check if x=-1 is part of the interval. Since interval 3 is -4 ≤x <2, and x=-1 is within this interval, so we exclude it.Therefore, solutions here are x ∈ [-4, -1) ∪ (-1, 2)But since x is real, we need to consider these intervals. However, we need to check whether the original expression is defined. At x=-1, denominator of first term is zero, so undefined. So x ≠-1.But when we solved the inequality, we found that for all x in the interval except x=-1, the expression equals 6, so 6 ≤6 is true. Hence, all x in [-4,2) except x=-1 satisfy the inequality.Therefore, in interval 3, solutions are x ∈ [-4, -1) ∪ (-1, 2)Interval 4: 2 ≤x <8/3 (≈2.666...)Check the expressions:- x +4: x ≥2, so x +4 ≥6 >0. |x +4|=x +4- x -2: x ≥2, so x -2 ≥0. |x -2|=x -2- 3x +14: x ≥2, so 3x ≥6, 3x +14 ≥20 >0. |3x +14|=3x +14- 3x -8: x <8/3 ≈2.666, so 3x -8 <8 -8=0. So |3x -8|=-3x +8Compute the fractions:First term numerator: |x+4| - |x -2| = (x +4) - (x -2) =x +4 -x +2=6Denominator: |3x +14| - |3x -8| = (3x +14) - (-3x +8) =3x +14 +3x -8=6x +6First fraction:6/(6x +6)=6/(6(x +1))=1/(x +1)Second term numerator: |x+4| + |x -2| = (x +4)+(x -2)=2x +2Denominator: |3x +14| + |3x -8|=(3x +14)+(-3x +8)=22Second fraction: (2x +2)/22=(2(x +1))/22=(x +1)/11Therefore, original inequality becomes:9*(1/(x +1)) +11*((x +1)/11) ≤6Simplify:9/(x +1) + (x +1) ≤6Let me set y =x +1. Then x = y -1. Since 2 ≤x <8/3, substituting:2 ≤y -1 <8/3 ⇒3 ≤y <8/3 +1=11/3≈3.666...So y ∈ [3,11/3). Note that y is positive here.Original inequality in terms of y:9/y + y ≤6Multiply both sides by y (positive, so inequality direction remains):9 + y^2 ≤6yBring all terms to left:y^2 -6y +9 ≤0Factor:(y -3)^2 ≤0Again, a square is non-negative, so equality holds only when y=3.Therefore, y=3 is the only solution. Since y ∈ [3,11/3), y=3 is included.Therefore, y=3 ⇒x +1=3 ⇒x=2.But check x=2 is in interval 4: 2 ≤x <8/3. x=2 is the lower bound of interval 4. So x=2 is included.Therefore, x=2 is the only solution in interval 4.Interval 5: x ≥8/3Now check expressions:- x +4: x ≥8/3 ≈2.666, so x +4 ≥8/3 +4= 8/3 +12/3=20/3 >0. |x +4|=x +4- x -2: x ≥8/3 >2, so x -2 ≥8/3 -6/3=2/3 >0. |x -2|=x -2- 3x +14: x ≥8/3, so 3x ≥8, so 3x +14 ≥22 >0. |3x +14|=3x +14- 3x -8: x ≥8/3, so 3x -8 ≥0. |3x -8|=3x -8Compute the fractions:First term numerator: |x+4| - |x -2| = (x +4) - (x -2)=6Denominator: |3x +14| - |3x -8|=(3x +14)-(3x -8)=22First fraction:6/22=3/11Second term numerator: |x+4| + |x -2|= (x +4)+(x -2)=2x +2Denominator: |3x +14| + |3x -8|=(3x +14)+(3x -8)=6x +6Second fraction: (2x +2)/(6x +6)= (2(x +1))/(6(x +1))=1/3, provided that x +1 ≠0, which is true here since x ≥8/3.Therefore, original inequality becomes:9*(3/11) +11*(1/3) ≤6Calculate LHS:27/11 +11/3 ≈2.4545 +3.6667≈6.1212, which is greater than 6. Therefore, inequality does not hold in this interval. So no solutions in interval 5.Summary of solutions from all intervals:- Interval 1: No solutions- Interval 2: No solutions- Interval 3: x ∈ [-4,2) except x=-1- Interval 4: x=2- Interval 5: No solutionsSo combining interval 3 and 4:Solutions are x ∈ [-4,2) {-1} ∪ {2}But x=2 is included from interval 4, and in interval 3, x approaches 2 but doesn't include it. But since x=2 is included in the solutions from interval 4, the total solution set is x ∈ [-4,2] {-1}Wait, hold on. In interval 3, x is in [-4,2), so up to but not including 2. Then in interval 4, x=2 is a solution. Therefore, combining them, x ∈ [-4,2] {-1}But wait, x=2 is included in interval 4. So the total solution set is x from -4 to 2, including -4 and 2, but excluding -1.But need to verify if x=-4 is included.In interval 3: -4 ≤x <2. So x=-4 is included in interval 3. At x=-4, check if the original expression is defined.Check denominators:First term denominator: |3*(-4)+14| - |3*(-4)-8| = |-12 +14| - |-12 -8| = |2| - | -20 | =2 -20=-18. Wait, so denominator is -18. But first term is [ |x+4| - |x-2| ] / [ |3x+14| - |3x-8| ]At x=-4:Numerator: | -4 +4 | - | -4 -2 | = |0| - | -6 | =0 -6=-6Denominator: |3*(-4)+14| - |3*(-4)-8| = | -12 +14 | - | -12 -8 | = |2| - | -20 | =2 -20=-18So the first fraction is (-6)/(-18)=1/3Second term numerator: | -4 +4 | + | -4 -2 | =0 +6=6Denominator: |2| + | -20 | =2 +20=22So second fraction is6/22=3/11Therefore, original expression at x=-4:9*(1/3)+11*(3/11)=3 +3=6, which is equal to 6, so satisfies the inequality. Therefore, x=-4 is included.Similarly, check x=2:From interval 4, x=2.First term numerator:6, denominator:22, first fraction 6/22=3/11Second term numerator:2x +2=6, denominator:22, second fraction6/22=3/11Wait, no. Wait, in interval 4, x=2:First term numerator is |2+4| - |2 -2|=6 -0=6Denominator |3*2 +14| - |3*2 -8|=|20| - | -2|=20 -2=18So first fraction=6/18=1/3Second term numerator |2+4| + |2 -2|=6 +0=6Denominator |20| + | -2|=20 +2=22Second fraction=6/22=3/11Therefore, original expression at x=2:9*(1/3) +11*(3/11)=3 +3=6, which is equal to 6. So x=2 is included.Therefore, the solution set is x ∈ [-4,2] {-1}Therefore, all real numbers from -4 to 2, including -4 and 2, except x=-1.Now, the problem asks for integer solutions where |x| <110. Since |x| <110 is a condition, but our solution set is between -4 and 2, which is well within |x| <110. So we can ignore the |x| <110 condition except for completeness.So integer solutions in [-4,2] {-1} are:x=-4, -3, -2, 0, 1, 2Wait, let's list integers from -4 to 2:-4, -3, -2, -1, 0,1,2But exclude -1. So integer solutions are: -4, -3, -2, 0,1,2Now sum these integers:-4 + (-3) + (-2) +0 +1 +2Compute step by step:Start with -4-4 + (-3)= -7-7 + (-2)= -9-9 +0= -9-9 +1= -8-8 +2= -6Therefore, the sum is -6.But let me verify the integer solutions:Check each integer in the solution set:- x=-4: included, as checked earlier.- x=-3: in interval 3, which is allowed (since x=-3 is in [-4,2) and not equal to -1)- x=-2: same as above.- x=-1: excluded.- x=0: included.- x=1: included.- x=2: included.Therefore, the integer solutions are -4, -3, -2, 0,1,2. Sum is -6.Therefore, the answer is boxed{-6}Final Answerboxed{-6}</think>To solve the inequality9 cdot frac{|x+4|-|x-2|}{|3x+14|-|3x-8|} + 11 cdot frac{|x+4|+|x-2|}{|3x+14|+|3x-8|} leqslant 6,we first identify the critical points where the expressions inside the absolute values change sign: (x = -14/3), (x = -4), (x = 2), and (x = 8/3). These points divide the real line into intervals. We analyze each interval separately:1. Interval (x < -14/3): - The inequality does not hold here.2. Interval (-14/3 leq x < -4): - The inequality does not hold here.3. Interval (-4 leq x < 2): - The inequality simplifies to (6 leq 6), which holds for all (x) in ([-4, 2)) except (x = -1) (where the expression is undefined).4. Interval (2 leq x < 8/3): - The inequality holds only at (x = 2).5. Interval (x geq 8/3): - The inequality does not hold here.Combining the solutions from all intervals, we find the solution set is (x in [-4, 2] setminus {-1}).Next, we identify the integer solutions within (|x| < 110):- The integers in ([-4, 2] setminus {-1}) are (-4, -3, -2, 0, 1, 2).Summing these integer solutions:[-4 + (-3) + (-2) + 0 + 1 + 2 = -6]Thus, the sum of the integer solutions is (boxed{-6}).]

🔑:Answer: -6 (solution of the inequality: [-4 ;-1) cup(-1 ; 2] ).

🧠:[<think>Okay, let's see. The problem is about finding the minimum value of ( a^2 + b^2 + c^2 ) given that ( a, b, c ) are non-zero natural numbers satisfying ( a^2 + b^2 - c = 2022 ). Then, we also need to find the values of ( a, b, c ) when this minimum is achieved. Hmm, interesting.First, let me parse the problem. So, we have three variables ( a, b, c ), all positive integers. The equation given is ( a^2 + b^2 - c = 2022 ), which we can rearrange to express ( c ) in terms of ( a ) and ( b ): ( c = a^2 + b^2 - 2022 ). Since ( c ) must be a positive integer, this tells us that ( a^2 + b^2 ) must be greater than 2022. That's a crucial point because if ( a^2 + b^2 ) is less than or equal to 2022, then ( c ) would not be a positive integer, which is required.So our goal is to minimize ( a^2 + b^2 + c^2 ). But since ( c = a^2 + b^2 - 2022 ), we can substitute that into the expression we need to minimize. Let's do that substitution:( a^2 + b^2 + c^2 = a^2 + b^2 + (a^2 + b^2 - 2022)^2 ).So now, the problem reduces to minimizing the function ( f(a, b) = a^2 + b^2 + (a^2 + b^2 - 2022)^2 ) where ( a ) and ( b ) are positive integers such that ( a^2 + b^2 > 2022 ).Hmm. So the idea is to find the smallest possible ( a ) and ( b ) such that ( a^2 + b^2 ) is just a little bit larger than 2022, which would make ( c ) as small as possible, thereby minimizing the sum ( a^2 + b^2 + c^2 ).But wait, if ( c = a^2 + b^2 - 2022 ), then ( c ) is directly dependent on ( a ) and ( b ). So, if we choose ( a ) and ( b ) such that ( a^2 + b^2 ) is as small as possible (but still greater than 2022), then ( c ) will also be as small as possible, which should lead to the minimal value of ( a^2 + b^2 + c^2 ). But I need to verify if this is indeed the case.Let me think. Suppose ( a^2 + b^2 = S ), then ( c = S - 2022 ), so the total sum is ( S + (S - 2022)^2 ). If we can minimize this expression with respect to ( S ), then we can find the optimal ( S ), and then find the corresponding ( a ) and ( b ). However, ( S ) must be an integer greater than 2022, and ( S = a^2 + b^2 ) where ( a ) and ( b ) are positive integers.Therefore, first, let's consider the problem as a continuous optimization problem. If ( S ) can be any real number greater than 2022, then the function to minimize is ( f(S) = S + (S - 2022)^2 ). Taking the derivative of this with respect to ( S ), we get ( f'(S) = 1 + 2(S - 2022) ). Setting this equal to zero gives ( 1 + 2(S - 2022) = 0 ), so ( S = 2022 - 0.5 ). But wait, this is ( S = 2021.5 ), which is less than 2022. But ( S ) must be greater than 2022, so the minimum in the continuous case would be at ( S = 2022 ), but since ( S ) must be greater than 2022, the minimum would be approached as ( S ) approaches 2022 from above.However, in our problem, ( S ) must be an integer greater than 2022, and also ( S ) must be expressible as the sum of two squares of positive integers ( a ) and ( b ). Therefore, the minimal possible ( S ) is the smallest integer greater than 2022 that can be expressed as the sum of two squares. Then, once we find that ( S ), the minimal ( a^2 + b^2 + c^2 ) would be ( S + (S - 2022)^2 ).So the problem reduces to finding the smallest integer ( S > 2022 ) such that ( S ) is a sum of two positive integer squares. Then, the minimal value is ( S + (S - 2022)^2 ), and the corresponding ( a, b, c ) are the numbers such that ( a^2 + b^2 = S ) and ( c = S - 2022 ).Therefore, step 1: Find the minimal ( S > 2022 ) such that ( S ) is a sum of two squares. Step 2: For that ( S ), find the corresponding ( a ) and ( b ). Then ( c ) is ( S - 2022 ).So the key is to find the minimal ( S ). Let's compute ( sqrt{2022} ) approximately. Since 45^2 = 2025, so ( sqrt{2022} approx 44.96 ). So, 2022 is just 3 less than 45^2. Therefore, numbers just above 2022 would be 2023, 2024, ..., 2025=45^2, etc.Now, starting from 2023 upwards, we need to check which is the first number that can be expressed as a sum of two positive integer squares.Recall that a number can be expressed as a sum of two squares if and only if in its prime factorization, every prime congruent to 3 mod 4 occurs an even number of times. This is Fermat's theorem on sums of two squares.So, let's check each number starting from 2023.2023: Let's factorize 2023.Divide by small primes:2023 ÷ 7 = 289, which is 17^2. So 2023 = 7 × 17^2.But 7 is a prime congruent to 3 mod 4 (since 7 mod 4 = 3). The exponent of 7 is 1, which is odd. Therefore, by Fermat's theorem, 2023 cannot be expressed as a sum of two squares.Therefore, 2023 is out.2024: Let's factorize 2024.2024 ÷ 2 = 10121012 ÷ 2 = 506506 ÷ 2 = 253253 ÷ 11 = 23 (since 11×23=253). So 2024 = 2^3 × 11 × 23.Now, primes congruent to 3 mod 4 in the factorization: 11 and 23 are both 3 mod 4 (11 mod 4 = 3, 23 mod 4 = 3). The exponents of both 11 and 23 are 1, which are odd. Therefore, since there are primes congruent to 3 mod 4 raised to an odd power, 2024 cannot be expressed as a sum of two squares.Moving on to 2025.2025 is 45^2. So, 45^2 + 0^2, but the problem states that a, b, c are non-zero natural numbers. So, a and b must be positive integers, so 0 is not allowed. Therefore, 2025 = 45^2 + 0^2 is invalid here because b would be zero. So 2025 cannot be expressed as a sum of two positive squares. However, maybe 2025 can be expressed as another combination. Let's check.But 2025 = 45^2. To express it as a sum of two positive squares, we need two positive integers a and b such that a^2 + b^2 = 45^2. For example, 45 is a multiple of 5, so perhaps using a Pythagorean triple. 45 = 9 × 5. The (9, 12, 15) scaled by 3 gives (27, 36, 45). Wait, 27^2 + 36^2 = 729 + 1296 = 2025. Yes! So 27^2 + 36^2 = 45^2. Therefore, 2025 can be expressed as the sum of two positive squares. Therefore, 2025 is a valid S. But since 2025 is equal to 45^2, which is greater than 2022, so c = 2025 - 2022 = 3. Therefore, c=3. Therefore, a=27, b=36, c=3. Then, the sum a^2 + b^2 + c^2 = 2025 + 9 = 2034. Wait, but is 2025 the minimal S?Wait, before 2025, we had 2023 and 2024 which couldn't be expressed as sum of two positive squares. Then 2025 can be. So S=2025 is the minimal S. Therefore, the minimal sum would be 2025 + 3^2 = 2025 + 9 = 2034.But wait, hold on. Let's check numbers between 2022 and 2025. For example, 2023, 2024, 2025. But 2023 and 2024 are both not expressible as sum of two squares. So 2025 is the next one. But 2025 is 45^2. However, perhaps there are larger numbers beyond 2025 but smaller S when combined with c might give a lower total sum? Wait, no. Because S needs to be as small as possible. Since we're adding S and (S - 2022)^2, if S is as small as possible, then (S - 2022) is as small as possible, so their sum would be minimized. Therefore, it's best to take the minimal S. Therefore, 2025 would give the minimal sum. However, let's check numbers beyond 2025 just to be sure.Wait, for example, S=2026. Let's factorize 2026. 2026 ÷ 2 = 1013. 1013 is a prime. Check 1013 mod 4: 1013 ÷ 4 = 253.25, so 1013 = 4×253 + 1, so 1013 ≡ 1 mod 4. Therefore, primes congruent to 3 mod 4 in the factorization of 2026: only 2, which is 2 mod 4, and 1013 which is 1 mod 4. Since all primes congruent to 3 mod 4 have even exponents (there are none here), so 2026 can be expressed as a sum of two squares. So 2026 is a possible S.But since S=2025 is smaller than 2026, and we already found that S=2025 is possible, then 2025 is still the minimal S. Similarly, 2025 is 45^2, which we can write as 27^2 + 36^2. So, that's valid. So, perhaps S=2025 is indeed the minimal. However, let's check if there is a smaller S between 2022 and 2025. But since S must be greater than 2022, the next integer is 2023, which we saw cannot be expressed as a sum of two squares, 2024 also cannot, 2025 can. Therefore, S=2025 is indeed the minimal.Therefore, the minimal value of ( a^2 + b^2 + c^2 ) is 2025 + 3^2 = 2025 + 9 = 2034. The corresponding values are a=27, b=36, c=3. But wait, let's confirm that 27^2 + 36^2 is indeed 2025. 27^2 is 729, 36^2 is 1296, so 729 + 1296 = 2025. Correct. Then c = 2025 - 2022 = 3. So yes, that's correct.But wait, hold on. Maybe there are other pairs (a, b) that give S=2025 but with smaller a and b? For example, 2025 can also be written as 0^2 + 45^2, but since a and b must be non-zero, that's invalid. Are there other representations?Wait, 2025 is 45^2. So any Pythagorean triple that adds up to 45. For example, multiples of smaller triples. For instance, the (3,4,5) triple scaled by 9: 3×9=27, 4×9=36, 5×9=45. So, 27, 36, 45 is one. Alternatively, maybe 15, 45-15? Wait, 15^2 + 30^2 = 225 + 900 = 1125, which is not 2025. Alternatively, 9, 42: 9^2 + 42^2 = 81 + 1764 = 1845, not 2025. 12, 39: 144 + 1521 = 1665. 18, 39: 324 + 1521 = 1845. 21, 36: 441 + 1296 = 1737. 24, 33: 576 + 1089 = 1665. So, none of these combinations. So 27, 36 is the only one? Hmm. Let me check another way.Alternatively, 45 can be expressed as the sum of two squares in different ways? Wait, 45 is itself 3^2 * 5. So, according to the sum of squares theorem, the number of representations depends on the prime factors. Since 5 is 1 mod 4, and 3 is 3 mod 4 with even exponent (since 45 = 3^2 * 5), then the number of representations is determined by the exponent of 5. Since 5 is raised to the first power, there is essentially one representation (up to order and signs). So, maybe the only primitive Pythagorean triple here is (27, 36, 45). So, that's the only way to express 45 as a hypotenuse with integer sides. Therefore, 27 and 36 are the only positive integers such that their squares add up to 45^2. Therefore, S=2025 can only be expressed as 27^2 + 36^2. So that's the only possible (a, b) pair for S=2025. Therefore, those are the values.But hold on, let me check another approach. Let's suppose that there might be another pair (a, b) with smaller a and b such that a^2 + b^2 is greater than 2022 but the total sum a^2 + b^2 + c^2 is smaller. Wait, for example, if S is larger, but c is smaller? But c = S - 2022. If S is larger, then c = S - 2022 is larger. Therefore, the total sum S + c^2 would be S + (S - 2022)^2. If S increases, then (S - 2022)^2 increases quadratically, which might dominate the linear increase of S. Therefore, it's better to have the smallest possible S.But just to confirm, let's take S=2025, sum=2025 + 3^2=2034. Suppose we take the next possible S, which is 2026. Then c=2026 -2022=4, sum=2026 + 16=2042, which is larger. Similarly, S=2027, c=5, sum=2027 +25=2052. So, the sum is increasing as S increases. Therefore, the minimal sum is indeed achieved at the minimal S, which is 2025. So the answer should be 2034, with a=27, b=36, c=3.Wait, but let me check if S=2025 is indeed the minimal possible. Suppose there's another S between 2022 and 2025 which is expressible as the sum of two squares. Wait, but 2023, 2024, 2025. We saw that 2023 and 2024 cannot be expressed as sum of two squares, so 2025 is the next. Therefore, 2025 is indeed the minimal S. Therefore, that's the answer.But just to make sure, let me check if there are numbers just above 2022 that can be expressed as sum of two squares. For example, 2022 +1=2023, which we saw can't. 2022+2=2024, also can't. 2022+3=2025, which can. So yes, 2025 is the minimal.Alternatively, maybe we can find a different S where even though S is larger, the total sum S + (S-2022)^2 is smaller. Wait, but as S increases, the term (S -2022)^2 increases quadratically, so the total sum is going to be minimized at the smallest S. Let's take a small example. Suppose S1 =2025, sum=2025 +3^2=2034. S2=2026, sum=2026 +4^2=2026 +16=2042. S3=2027, sum=2027 +25=2052. So, clearly, the sum increases as S increases. Therefore, the minimal sum is achieved at the minimal S. Therefore, 2025 is the correct S.Therefore, the minimum value is 2034, achieved at a=27, b=36, c=3.Wait, but just to check if there's another combination with a smaller sum. Suppose S=2025 is minimal, but maybe there are different a, b with S=2025 but different a, b that might give a different c? No, because c is fixed once S is fixed. Since c = S -2022. So regardless of a and b, if S=2025, then c=3. So the sum is 2025 +9=2034. So even if there are different a and b for S=2025, the sum remains the same. However, the problem asks for the values of a, b, c when the sum is minimized. So even if there are multiple (a, b) pairs for S=2025, they all result in the same minimal sum. But in our case, we have only one pair (27,36) as per the earlier check. Wait, is that the only pair? Let's see.Wait, another pair could be (36, 27), but since a and b are interchangeable, the problem doesn't specify which is which, so both are acceptable. But in the answer, we need to provide the values. So, likely, either (27,36,3) or (36,27,3) are both correct. But perhaps the problem expects the numbers in some order, maybe ascending? Let's check.Alternatively, maybe there are other pairs (a, b) such that a^2 + b^2 =2025. Let's check. For example, 2025 divided by 2 is 1012.5, so if there's a square near that. Let's try a=15, then b^2=2025 -225=1800, which is not a square. a=20, b^2=2025 -400=1625, which is 25×65, not a square. a=24, b^2=2025 -576=1449, not a square. a=25, b^2=2025 -625=1400, not a square. a=30, b^2=2025 -900=1125, not a square. a=35, b^2=2025 -1225=800, not a square. a=40, b^2=2025 -1600=425, not a square. a=5, b^2=2025 -25=2000, not a square. a=10, b^2=2025 -100=1925, not a square. a=12, b^2=2025 -144=1881, not a square. So no, seems like only (27,36) works. So, that must be the only pair.Therefore, the answer is (1) 2034; (2) a=27, b=36, c=3 (or a=36, b=27, c=3).Wait, just to confirm, let me check 27^2 + 36^2: 729 + 1296 is 2025. Then c=2025 -2022=3. Then the sum of squares is 2025 + 9=2034. Yes.Therefore, I think that's the answer.Final Answer(1) The minimum value of ( a^2 + b^2 + c^2 ) is (boxed{2034}). (2) The values of ( a, b, c ) are (boxed{27}), (boxed{36}), and (boxed{3}) (in any order).</think>Given the equation (a^2 + b^2 - c = 2022) with non-zero natural numbers (a, b, c), we need to find the minimum value of (a^2 + b^2 + c^2) and the corresponding values of (a, b, c).First, we express (c) in terms of (a) and (b):[ c = a^2 + b^2 - 2022 ]We need to minimize the expression:[ a^2 + b^2 + c^2 = a^2 + b^2 + (a^2 + b^2 - 2022)^2 ]To minimize this, we need the smallest (S = a^2 + b^2) such that (S > 2022) and (S) is a sum of two squares of positive integers. The next few numbers after 2022 are checked:- 2023: Cannot be expressed as a sum of two squares (factorization includes primes congruent to 3 mod 4 with odd exponents).- 2024: Cannot be expressed as a sum of two squares (similar factorization issues).- 2025: Can be expressed as (27^2 + 36^2) (since (45^2 = 2025) and using the Pythagorean triple (27, 36, 45)).Thus, (S = 2025) is the smallest valid value. Substituting (S) into the expression for (c):[ c = 2025 - 2022 = 3 ]The minimum value of (a^2 + b^2 + c^2) is:[ 2025 + 3^2 = 2025 + 9 = 2034 ]The corresponding values of (a, b, c) are (27, 36, 3) (or (36, 27, 3)). Final Answer(1) The minimum value of (a^2 + b^2 + c^2) is (boxed{2034}). (2) The values of (a, b, c) are (boxed{27}), (boxed{36}), and (boxed{3}) (in any order).,<think>Alright, so I need to solve this problem where we have non-zero natural numbers a, b, c satisfying the equation a² + b² - c = 2022. And we need to find the minimum value of a² + b² + c² and the corresponding values of a, b, c when that minimum is achieved. Let me think step by step.First, let me parse the problem. We have a, b, c as natural numbers (so positive integers), and they satisfy a² + b² - c = 2022. So, rearranging that equation, we can express c in terms of a and b: c = a² + b² - 2022. Since c has to be a positive integer, that means a² + b² must be greater than 2022. So, for any a and b such that a² + b² > 2022, c will be a positive integer. Our goal is to find a, b, c such that the sum a² + b² + c² is minimized.So, since c = a² + b² - 2022, substituting that into the expression we need to minimize gives us:a² + b² + c² = a² + b² + (a² + b² - 2022)²So, we can consider this as a function of a and b, where a and b are positive integers with a² + b² > 2022. Our task is to find the minimum of this function.To minimize a² + b² + c², which is a² + b² + (a² + b² - 2022)², we can consider that this expression is going to be minimized when a² + b² is as close to 2022 as possible, because the term (a² + b² - 2022)² will be minimized when a² + b² is just over 2022. However, we also have to account for the a² + b² term itself. So, there's a balance between keeping a² + b² small and keeping (a² + b² - 2022)² small. Since the latter is a square term, it might grow more rapidly if a² + b² is too large, so perhaps the optimal point is when a² + b² is just slightly larger than 2022.Therefore, the strategy would be to find the smallest possible a² + b² that is greater than 2022, and then compute c as a² + b² - 2022, and then compute the sum a² + b² + c². However, perhaps even if a² + b² is just a bit larger than 2022, the term c² could still be significant. So, maybe there's a trade-off here. For example, suppose a² + b² is 2023, then c would be 1, so the sum would be 2023 + 1 = 2024, and c² is 1, so total sum is 2023 + 1 = 2024. Wait, no: wait, a² + b² + c² would be 2023 + 1² = 2023 + 1 = 2024. If a² + b² is 2022 + k, then c = k, so the sum is (2022 + k) + k². So, to minimize this, we need to minimize k + k² where k is a positive integer (since c must be at least 1). The minimum occurs at k=1, giving 1 + 1 = 2, so total sum 2022 + 1 + 1 = 2024. But wait, but a² + b² has to be 2023 in this case. So, if we can find a and b such that a² + b² = 2023, then c=1 and the total sum is 2024, which would be the minimum. However, the problem is that 2023 might not be expressible as the sum of two squares. Therefore, we need to check if 2023 can be written as the sum of two squares. If not, then we need to check the next integer, 2024, and so on.So, the problem reduces to finding the smallest integer k ≥ 1 such that 2022 + k can be expressed as a sum of two squares a² + b², where a and b are positive integers, and then compute the sum as 2022 + k + k². The minimal such k would give the minimal sum.Therefore, first, check if 2023 can be written as the sum of two squares. Let's recall that a number can be expressed as the sum of two squares if and only if in its prime factorization, every prime congruent to 3 mod 4 appears with an even exponent.Let me factorize 2023. 2023 divided by 7 is 289, no: 7*289 is 2023? Wait, 7*289 is 2023. But 289 is 17². So, 2023 = 7 * 17². Now, 7 is a prime congruent to 3 mod 4 (since 7 mod 4 is 3). The exponent of 7 is 1, which is odd. Therefore, by the sum of two squares theorem, 2023 cannot be expressed as the sum of two squares. Therefore, k=1 is impossible.Next, check k=2: 2022 + 2 = 2024. Can 2024 be written as the sum of two squares? Let's factorize 2024. 2024 divided by 4 is 506. 506 divided by 2 is 253. 253 is 11*23. So, 2024 = 4 * 2 * 11 * 23. The primes 11 and 23 are both congruent to 3 mod 4 (11 mod 4 is 3, 23 mod 4 is 3). Each appears once, which is odd. Therefore, 2024 cannot be expressed as the sum of two squares.k=3: 2025. Let's check 2025. 2025 is 45², so that's a perfect square. So, 45² + 0², but since a and b must be non-zero natural numbers, 0 is not allowed. So, 2025 can be expressed as 45² + 0², but since a and b must be at least 1, this is invalid. However, maybe other representations. Let's see. 2025 divided by 25 is 81. So 2025 = 25 * 81. So, 2025 = (5²)(9²) = (5*9)² = 45². So, perhaps 2025 is a square, but not the sum of two non-zero squares. Wait, but 2025 can be written as 36² + 27²? Wait, let's compute 36² is 1296, 27² is 729, 1296 + 729 = 2025. Yes, 36² + 27² = 2025. So, that works. Therefore, 2025 can be expressed as the sum of two squares. Therefore, when k=3, c=3, and the sum is 2025 + 9 = 2034. However, wait, 36 and 27 are both positive integers, so that works. Therefore, with a=36, b=27, c=3, we have a² + b² + c² = 2025 + 9 = 2034.But before we accept that, let's check if there is a smaller k, like k=2 or k=1. As we saw, k=1 and k=2 don't work. Let's check k=4: 2022 +4=2026. Factorize 2026. 2026 divided by 2 is 1013. 1013 is a prime number. Let me check: 1013 divided by 2 is not, 3: 1013/3 is 337.666..., 5: ends with 3, 7: 1013/7 is approximately 144.7, 11: 1013/11 is 92.09, 13: 1013/13 is 77.92, 17: 1013/17 ≈ 59.58. So, 1013 is prime. Since 1013 ≡ 1 mod 4 (1013 divided by 4 is 253.25, so remainder 1). Therefore, 2026 = 2 * 1013. Since 1013 is prime ≡ 1 mod 4, then 2026 can be expressed as the sum of two squares. Let's find the actual decomposition.We can use the identity that if a prime p ≡ 1 mod 4, then it can be written as a sum of two squares. So, 1013 can be written as m² + n². Let's try to find m and n such that m² + n² = 1013. Let's compute sqrt(1013) ≈ 31.83, so m and n are around 31. Let's check 31² = 961, 1013 - 961 = 52. 52 is not a square. 30²=900, 1013 -900=113. 113 is 10² + 7²=100+49=149. No. 29²=841, 1013-841=172, not a square. 28²=784, 1013-784=229. 229 is 15² + 2²=225 +4=229. Yes! So, 229=15² + 2². Therefore, 1013 = 28² + 15² + 2²? Wait, no. Wait, 1013 = 28² + (sqrt(229))², but 229 is 15² + 2². Wait, perhaps I need a different approach. Alternatively, use the fact that primes ≡1 mod 4 can be expressed as sum of two squares. Maybe use an algorithm.Alternatively, perhaps using the identity that if we have primes p ≡1 mod 4, then we can find integers m and n such that m² + n² = p. For 1013, let's check possible squares:Start from 31²=961, 1013-961=52, not a square. 30²=900, 1013-900=113, not a square. 29²=841, 1013-841=172, not a square. 28²=784, 1013-784=229, which is prime. 27²=729, 1013-729=284. 284 divided by 4 is 71, so 284=4*71. 71 is prime, 71≡3 mod 4. So 284 is not a sum of two squares. 26²=676, 1013-676=337. 337 is prime, 337≡1 mod 4, so 337 can be written as sum of two squares. Let's check sqrt(337)≈18.35. 18²=324, 337-324=13, which is 3² + 2². Wait, but 337=18² + (sqrt(13))²? No, 13 isn't a square. Wait, maybe 17²=289, 337-289=48, not a square. 16²=256, 337-256=81=9². So, 16² + 9² =256 +81=337. Therefore, 337=16² +9². Therefore, 1013=26² + (16² +9²). Wait, no. Wait, 1013=26² + 337, and 337=16² +9². Therefore, 1013 can be written as 26² + 16² +9², but that's three squares. But we need two squares. Hmm. Maybe I need a better approach.Alternatively, perhaps use the fact that 2026=2*1013, and since 1013 is prime ≡1 mod 4, then 2026 can be written as (1² +1²)(m² +n²)= (m -n)² + (m +n)². Wait, the product of sums of squares formula: (a² + b²)(c² + d²) = (ac - bd)² + (ad + bc)². So, since 2=1² +1² and 1013=m² +n², then 2*1013=(1*m -1*n)² + (1*n +1*m)². Let's compute that. If 1013=m² +n², then 2*1013=(m -n)² + (m +n)². So, suppose m and n are such that m² +n²=1013. Let's say m=28 and n=15 (since 28² +15²=784 +225=1009, which is close but not 1013). Wait, that was wrong. Wait, 28²=784, 1013-784=229. 229 as we saw is 15² + 2². Wait, so maybe m=15 and n=2? Then 15² +2²=225 +4=229. No, that's not 1013. Wait, this is getting confusing. Maybe another way.Alternatively, since 1013 is a prime ≡1 mod 4, there exists a unique (up to order and sign) representation as a sum of two squares. To find it, we can compute a square root of -1 modulo 1013, then use the Euclidean algorithm.But this might be complicated. Alternatively, perhaps look up the decomposition. Alternatively, maybe 1013= 22² + 17². 22²=484, 17²=289, 484+289=773, nope. 23²=529, 1013-529=484=22². So 23² +22²=529 +484=1013. Yes! So 1013=23² +22². Therefore, 2026=2*1013=(1² +1²)(23² +22²)= (1*23 -1*22)² + (1*22 +1*23)²= (23 -22)² + (22 +23)²=1² +45²=1 +2025=2026. Therefore, 2026=1² +45². However, the problem states that a and b must be non-zero natural numbers. So 1 and 45 are both positive integers, so that's acceptable. Therefore, when k=4 (since 2022 +4=2026), c=4, so the sum a² + b² +c²=2026 +16=2042. But earlier, for k=3, we had sum 2034, which is smaller. So 2034 is better.Wait, but earlier, with k=3, a² + b²=2025=36² +27², which gives sum 2025 +9=2034. So, that is better than k=4's 2042. So, maybe k=3 is better. However, we should check if there's a k between 1 and 3 where a² + b²=2022 +k is expressible as sum of two squares.We already checked k=1:2023, which cannot be expressed as sum of two squares. k=2:2024, which also cannot be. k=3:2025=36² +27², which works. So, moving forward, k=3 gives the minimal sum so far. But let's check k=5: 2022+5=2027. Check if 2027 can be expressed as sum of two squares.Factorize 2027: 2027 divided by 2 is 1013.5, not integer. 2027 divided by 3: 2027/3≈675.666, not integer. 5: ends with 7, nope. 7: 2027/7≈289.571, nope. 11: 184.27, nope. 13: 155.92, nope. 17: 119.235, nope. 19: 106.684, nope. 23: 88.13, nope. 29: 69.896, nope. 31: 65.387, nope. So, 2027 is a prime. Check if 2027 ≡1 mod 4. 2027 divided by 4 is 506.75, remainder 3. So 2027 ≡3 mod4. Therefore, it cannot be expressed as the sum of two squares. So, k=5 is invalid.k=6:2022+6=2028. Factorize 2028: 2028 divided by 4 is 507. 507=3*13*13. So, 2028=4*3*13². The primes 3≡3 mod4, exponent 1, which is odd. So 2028 cannot be expressed as sum of two squares.k=7:2029. Check if 2029 is prime. Divided by 2,3,5,7,11,13,17,19,23,29,31, etc. 2029 divided by 2=1014.5, 3: 2029/3≈676.333, 5: ends with 9, nope. 7: 2029/7≈289.857, nope. 11: 184.454, nope. 13: 2029/13≈156.07, nope. 17≈119.35, nope. 19≈106.789, nope. 23≈88.217, nope. 29≈69.965, nope. 31≈65.45, nope. So, 2029 is prime. 2029 mod4=1 (since 2028 is divisible by 4, so 2029≡1 mod4). Therefore, 2029 can be expressed as a sum of two squares. Let's try to find them. sqrt(2029)≈45.04. So check 45²=2025, 2029-2025=4=2². Therefore, 2029=45² +2². Therefore, a=45, b=2, c=7. Then the sum a² +b² +c²=2025 +4 +49=2078. Which is larger than 2034, so not better.Therefore, k=3 is still better. Wait, but maybe there's a smaller k where a² +b²=2022 +k. Wait, we checked k=1,2,3,4,5,6,7. k=3 gives the first possible expression as sum of two squares. Let me check if there's a smaller k. Between k=1 and k=3, we saw k=1 and 2 are not possible. So k=3 is the minimal possible. Therefore, the minimal sum is 2025 +9=2034 with a=36, b=27, c=3. But wait, but 36² +27²=2025. But 36²=1296, 27²=729, 1296+729=2025. Correct. So, c=2025-2022=3. So, yes, that works. But let me check if there's a different pair (a,b) with smaller a and b that could give a sum closer to 2022. For example, perhaps a and b closer to each other. Because 36 and 27 are both quite large, but their squares sum to 2025. Maybe there's a pair with a and b smaller than 36 and 27 such that a² +b²=2025. Wait, 2025 is 45², so 45² can be written as 36² +27². Is there another way? Let's check. For example, 2025= sqrt(2025)=45, so looking for integers a,b where a² +b²=45². So, possible Pythagorean triples. The standard (45, 0), but since a,b must be non-zero, that's invalid. The primitive triples: 45 is 9*5. The primitive triples for 5 are 3,4,5. So scaled by 9: 27,36,45. So that's the 27,36,45 triple. So that's the only way to express 45² as sum of two non-zero squares. Therefore, the only representation is 27² +36². Therefore, a=27, b=36 or a=36, b=27. So, that's the only possible. So, that gives c=3. So, the sum a² +b² +c²=2025 +9=2034.But wait, let me check if there are other values of k where 2022 +k can be written as the sum of two squares in a different way, leading to a smaller sum. For example, if k=3 gives sum 2034, is there a k=3 but with a different a and b that gives a smaller sum? Well, since k=3 requires a² +b²=2025, and 2025 can only be expressed as 27² +36², which gives the sum 2025 +9=2034. So, no other representations. Therefore, that's the minimal.But let me check k=25. Wait, why 25? Because maybe c=25, then a² +b²=2022+25=2047. Then the sum would be 2047 +625=2672, which is way bigger. So, larger k will lead to larger sums.Alternatively, maybe if we take a and b such that a² +b² is larger than 2022 but not by much, but such that c is smaller. Wait, but c is equal to a² +b² -2022. So, if a² +b² is just slightly larger, c is small. But the minimal c is 1, but we saw that k=1 is not possible because 2023 can't be expressed as sum of two squares. So, c=3 is the minimal possible. Therefore, the minimal sum is 2034.Wait, but let me check another way. Suppose we fix c, and then try to find a and b such that a² +b²=2022 +c. Then, the expression we need to minimize is (2022 +c) +c². So, we can consider this as a function f(c)=c² +c +2022. To find the minimum of f(c) over positive integers c where 2022 +c is expressible as sum of two squares. The function f(c)=c² +c +2022 is a quadratic in c, which is minimized at c=-0.5, but since c must be a positive integer, the minimal value occurs at c=1. However, as we saw, c=1 is not possible. Then c=2 is not possible. c=3 is possible, giving f(3)=9 +3 +2022=2034. Next possible c=4 gives f(4)=16 +4 +2022=2042, which is higher. Then c=7 gives f(7)=49 +7 +2022=2078, which is even higher. So, the minimal value is 2034 when c=3. Therefore, the answer should be 2034, with a=27, b=36, c=3 or any permutation of a and b.But the problem says "non-zero natural numbers a, b, c", so they can be in any order. So, the values are a=27, b=36, c=3 or a=36, b=27, c=3. So, depending on which order you write them.But let me verify once again to make sure there are no mistakes. We need to check if 2025 can indeed be written as 27² +36². 27²=729, 36²=1296, sum is 729 +1296=2025. Correct. Then c=2025 -2022=3. Then a² +b² +c²=2025 +9=2034. So, yes, that seems correct. Also, 2025 is the first number after 2022 that can be expressed as a sum of two non-zero squares. The numbers 2023,2024,2026,2027,2028,2029 were checked and they cannot be expressed as such. Therefore, k=3 is indeed the minimal possible, leading to the minimal sum of 2034.Therefore, the answers are:(1) The minimum value of a² +b² +c² is 2034.(2) The values of a, b, c are 27, 36, 3 (in any order since a and b are interchangeable).Final AnswerThe minimum value of (a^2 + b^2 + c^2) is (boxed{2034}), achieved when (a = 27), (b = 36), and (c = 3) (or vice versa).</think>Given non-zero natural numbers (a, b, c) satisfying the equation (a^2 + b^2 - c = 2022), we need to find the minimum value of (a^2 + b^2 + c^2) and the corresponding values of (a, b, c).First, we express (c) in terms of (a) and (b):[ c = a^2 + b^2 - 2022 ]We need to minimize the expression:[ a^2 + b^2 + c^2 = a^2 + b^2 + (a^2 + b^2 - 2022)^2 ]To minimize this, we need (a^2 + b^2) to be as close to 2022 as possible. We check the smallest integers greater than 2022 that can be expressed as the sum of two squares:1. Check (2023): Factorizing (2023 = 7 times 17^2). Since 7 is a prime congruent to 3 mod 4 with an odd exponent, (2023) cannot be expressed as the sum of two squares.2. Check (2024): Factorizing (2024 = 2^3 times 11 times 23). Both 11 and 23 are primes congruent to 3 mod 4 with odd exponents, so (2024) cannot be expressed as the sum of two squares.3. Check (2025): (2025 = 45^2). This can be expressed as (36^2 + 27^2), both non-zero natural numbers.Thus, (a = 36), (b = 27) (or vice versa) gives (a^2 + b^2 = 2025). Substituting this into the expression for (c):[ c = 2025 - 2022 = 3 ]The minimum value of (a^2 + b^2 + c^2) is:[ 2025 + 3^2 = 2025 + 9 = 2034 ]Therefore, the minimum value of (a^2 + b^2 + c^2) is (boxed{2034}), achieved when (a = 27), (b = 36), and (c = 3) (or vice versa).]

🔑:【Answer】(1) 2034 (2) 27,36,3【Solution】Since a perfect square divided by 4 can only leave a remainder of 0 or 1, the sum of two perfect squares divided by 4 can only leave a remainder of 0, 1, or 2. From the problem, we have a^{2}+b^{2}=2022+c. (Method 1) When c=2, a^{2}+b^{2}=2024, then left(frac{a}{2}right)^{2}+left(frac{b}{2}right)^{2}=506, so frac{a}{2} div 4 cdots 1, frac{b}{2} div 4 cdots 1, trying 1,5,9,13,17, we find no solution.(Method 2) When c=2, a^{2}+b^{2}=2024, since 11 mid 2024, and a perfect square divided by 11 can only leave a remainder of 0, 1, 4, 9, 5, 3, so 11|a, 11| b, but 2024 is not a multiple of 121, so there is no solution.Similarly, using Method 2, we can conclude that a, b are both multiples of 9.When c=3, a^{2}+b^{2}=2025=45^{2}=3^{2} times 5^{2}=left(3^{2}+4^{2}right) times 3^{-1}=27^{2}+36^{2}. At this point,a^{2}+b^{2}+c^{2}=2025+9=2034, a=27, b=36, c=3 or a=36, b=27, c=3