🧠:[<think>Okay, let's try to solve this problem step by step. So, the question is about Hogwarts School of Witchcraft and Wizardry. They have a magic competition with 246 participants. The total number of boys in the school is 255, and the number of boys participating in the competition is 11 more than the number of girls not participating. We need to find the total number of students at the school. Hmm, let's parse this carefully.First, let me note down the given information:1. Total participants in the competition: 246 people.2. Total boys in the school: 255.3. The number of boys participating is 11 more than the number of girls not participating.4. We need to find the total number of students at Hogwarts.Let me define some variables to make this clearer. Let's say:- Let B_p be the number of boys participating in the competition.- Let G_p be the number of girls participating in the competition.- Let G_np be the number of girls not participating in the competition.We know that the total participants are 246, so:B_p + G_p = 246 ...(1)We also know the total number of boys in the school is 255. Since some boys are participating and others are not, the number of boys not participating would be 255 - B_p. However, the problem doesn't mention anything about boys not participating directly. Instead, it relates the number of boys participating to the number of girls not participating.The third point says: "The number of boys participating in the competition is 11 more than the number of girls not participating." So:B_p = G_np + 11 ...(2)Our goal is to find the total number of students, which would be the total number of boys plus the total number of girls. We already know the total number of boys is 255, so if we can find the total number of girls, we can add them together for the total.Let me denote the total number of girls as G. Then G = G_p + G_np. Because all girls are either participating or not participating.So, if we can find G_p and G_np, we can find G. Then total students = 255 + G.Now, let's see what equations we have.From equation (1): B_p + G_p = 246From equation (2): B_p = G_np + 11We need to relate these variables. Let's see.We can substitute B_p from equation (2) into equation (1):(G_np + 11) + G_p = 246Which simplifies to:G_np + G_p + 11 = 246But G_np + G_p is equal to the total number of girls, G. So:G + 11 = 246Therefore, G = 246 - 11 = 235Wait, hold on. If G + 11 = 246, then G = 246 - 11 = 235. That seems straightforward. But wait, let me verify.Wait, if G_p + G_np = G, then substituting into the equation:(G_np + G_p) + 11 = 246So G + 11 = 246Therefore, G = 246 - 11 = 235. So total number of girls is 235. Then total students = 255 (boys) + 235 (girls) = 490.But wait, that seems too straightforward. Let me check if there's a mistake here.Wait, maybe I oversimplified. Let me go back step by step.We have:1. Total participants: B_p + G_p = 2462. Total boys: 255, so boys not participating = 255 - B_p3. Boys participating (B_p) = Girls not participating (G_np) + 11We need to find total students = 255 (boys) + G (girls)To find G, we know G = G_p + G_npSo from equation (2): B_p = G_np + 11From equation (1): B_p = 246 - G_pSo substitute B_p into equation (2):246 - G_p = G_np + 11Therefore, G_np = 246 - G_p - 11 = 235 - G_pBut G_np is equal to total girls - G_p (since G = G_p + G_np). Therefore:G_np = G - G_pSo substituting:235 - G_p = G - G_pSubtract -G_p from both sides:235 = GTherefore, G = 235Thus, total students = 255 + 235 = 490Wait, so that's correct? Let me double-check.If total girls are 235, then G_p (girls participating) + G_np (girls not participating) = 235From equation (1): B_p + G_p = 246From equation (2): B_p = G_np + 11But since G_np = 235 - G_p, substitute into equation (2):B_p = (235 - G_p) + 11 = 246 - G_pSo from equation (1): B_p + G_p = 246But B_p = 246 - G_p, so substituting:(246 - G_p) + G_p = 246 => 246 = 246, which is an identity.Hmm, so this seems consistent, but maybe we need another way to approach it.Alternatively, let's think of it in terms of variables.Let’s denote:- Total students: S = 255 (boys) + G (girls)- Participants: 246 = B_p + G_p- Boys participating: B_p = G_np + 11But G_np = G - G_pTherefore, B_p = (G - G_p) + 11But B_p = 246 - G_p (from the participants equation)Therefore:246 - G_p = G - G_p + 11Simplify:246 = G + 11So G = 246 - 11 = 235Hence total students S = 255 + 235 = 490Yes, that's the same result. So the total number of students is 490. Therefore, the answer should be 490.But wait, let me confirm with another method.Suppose we make a table.Total participants: 246 (B_p + G_p)Total boys: 255. Therefore, boys not participating: 255 - B_pTotal girls: G = G_p + G_npGiven that B_p = G_np + 11.So, substitute B_p into participants:G_np + 11 + G_p = 246But G_np + G_p = G, so:G + 11 = 246 => G = 235Hence total students: 255 + 235 = 490Same result. So, I think that's correct.Alternatively, let's plug numbers in.If G = 235, then G_p + G_np = 235If B_p = G_np + 11Total participants: B_p + G_p = (G_np + 11) + G_p = (G_np + G_p) + 11 = G + 11 = 235 + 11 = 246, which matches.So everything checks out.Therefore, the total number of students is 255 + 235 = 490.Final AnswerThe total number of students at Hogwarts School of Witchcraft and Wizardry is boxed{490} people.</think>Given the problem, we need to determine the total number of students at Hogwarts School of Witchcraft and Wizardry. Here are the key steps to solve the problem:1. Define Variables: - Let ( B_p ) be the number of boys participating in the competition. - Let ( G_p ) be the number of girls participating in the competition. - Let ( G_{np} ) be the number of girls not participating in the competition. - The total number of boys in the school is 255. - The total number of participants in the competition is 246.2. Set Up Equations: - The total number of participants: ( B_p + G_p = 246 ) - The number of boys participating is 11 more than the number of girls not participating: ( B_p = G_{np} + 11 )3. Relate Variables: - The total number of girls ( G ) is the sum of girls participating and not participating: ( G = G_p + G_{np} )4. Substitute and Solve: - Substitute ( B_p ) from the second equation into the first equation: [ (G_{np} + 11) + G_p = 246 ] - This simplifies to: [ G_{np} + G_p + 11 = 246 ] - Recognize that ( G_{np} + G_p = G ), so: [ G + 11 = 246 ] - Solving for ( G ): [ G = 246 - 11 = 235 ]5. Calculate Total Students: - The total number of students is the sum of all boys and girls: [ 255 text{ (boys)} + 235 text{ (girls)} = 490 ]Thus, the total number of students at Hogwarts School of Witchcraft and Wizardry is (boxed{490}).,<think>Okay, let me try to figure out this problem. So, Hogwarts has a magic competition with 246 participants. The total number of boys at the school is 255. The number of boys participating is 11 more than the number of girls not participating. We need to find the total number of students at the school. Hmm, let's break this down step by step.First, let's note down what's given:1. Total participants in the competition: 2462. Total boys at the school: 2553. Number of boys participating = Number of girls not participating + 11We need to find the total number of students. Let's denote some variables to make this clearer. Let me think...Let’s let:- ( B_p ) = number of boys participating in the competition- ( G_p ) = number of girls participating in the competition- ( G_{np} ) = number of girls not participating in the competition- Total students = ( T )We know that the total participants are boys and girls who participate, so:( B_p + G_p = 246 ) ...(1)Also, the total number of boys at the school is 255. Since all boys are either participating or not participating, but we don't have information about boys not participating. Wait, but we might need to express that. Let me define:- ( B_{np} ) = number of boys not participatingThen, total boys:( B_p + B_{np} = 255 ) ...(2)But the problem doesn't mention boys not participating directly. Instead, it relates the number of boys participating to the number of girls not participating. The third piece of information is:Number of boys participating is 11 more than the number of girls not participating. So,( B_p = G_{np} + 11 ) ...(3)Our goal is to find the total number of students ( T ). The total students would be total boys plus total girls. Let me denote total girls as ( G ). So,( T = 255 + G )Therefore, if we can find the number of girls at the school, we can find ( T ). So, how to find ( G )?Total girls would be girls participating plus girls not participating:( G = G_p + G_{np} )So, if we can find ( G_p ) and ( G_{np} ), we can find ( G ). Let's see how.From equation (1): ( B_p + G_p = 246 ). If we can find ( B_p ), then we can find ( G_p ). Then, from equation (3), if we can find ( G_{np} ), we can relate it to ( B_p ). Let me try to link these.From equation (3): ( G_{np} = B_p - 11 )So, once we know ( B_p ), we can get ( G_{np} ).But how do we find ( B_p )? Let's see. From equation (2): ( B_p + B_{np} = 255 ). But we don't know ( B_{np} ). However, maybe we don't need it. Let's see if we can relate other variables.Wait, total participants are 246, which is ( B_p + G_p ). So, if we can express ( G_p ) in terms of other variables, maybe we can substitute.But ( G_p = 246 - B_p ). So, substituting into the total girls:( G = G_p + G_{np} = (246 - B_p) + (B_p - 11) )Wait, let me check that:( G = G_p + G_{np} )But ( G_p = 246 - B_p ), and ( G_{np} = B_p - 11 ). So,( G = (246 - B_p) + (B_p - 11) = 246 - B_p + B_p - 11 = 246 - 11 = 235 )Oh! Wait a minute. That simplifies to 235. So, total girls would be 235. Therefore, total students would be 255 (boys) + 235 (girls) = 490. Is that possible?Let me verify this. Let me check each step again.Starting from equation (3):( B_p = G_{np} + 11 implies G_{np} = B_p - 11 )Total girls: ( G = G_p + G_{np} )But ( G_p = 246 - B_p ), as per equation (1). So substituting,( G = (246 - B_p) + (B_p - 11) = 246 - B_p + B_p - 11 = 235 )Yes, that's right. The ( B_p ) terms cancel out, leaving 246 - 11 = 235. Therefore, the total number of girls is 235. Then total students is 255 + 235 = 490.But let's check if this makes sense with all the given information. Let's verify:Total students: 255 boys + 235 girls = 490. Okay.Number of participants: 246. So, number of non-participants would be 490 - 246 = 244.But let's check the boys participating and girls not participating.If total girls are 235, then girls not participating ( G_{np} = B_p - 11 ). Wait, we can find ( B_p ):From equation (1): ( B_p + G_p = 246 ). But ( G_p = 246 - B_p ). Also, total girls ( G = 235 = G_p + G_{np} ). Substitute ( G_p = 246 - B_p ) and ( G_{np} = B_p - 11 ):235 = (246 - B_p) + (B_p - 11) = 235, which checks out.But we can also check boys. The total number of boys is 255. If we can find ( B_p ), we can see if the numbers make sense.But from the girls side, we found that ( G_{np} = B_p - 11 ). Let's see, if total girls are 235, then girls not participating are ( G_{np} = 235 - G_p ). But ( G_p = 246 - B_p ). So:( G_{np} = 235 - (246 - B_p) = 235 - 246 + B_p = B_p - 11 ). Which matches equation (3). So that's consistent.But we still don't know ( B_p ). However, since total boys are 255, boys not participating would be ( 255 - B_p ). And total non-participants are 244 (from 490 total students - 246 participants). So:Total non-participants = boys not participating + girls not participating244 = (255 - B_p) + (B_p - 11)Simplify:244 = 255 - B_p + B_p -11 = 255 -11 = 244Which checks out. So, regardless of ( B_p ), the numbers hold. Therefore, the total number of students is indeed 490.But wait, that seems a bit too straightforward. Let me confirm again.If total students are 490, with 255 boys and 235 girls.Total participants are 246. So, participants consist of some boys and some girls.Number of boys participating is 11 more than the number of girls not participating.So, girls not participating = ( G_{np} = 235 - G_p )But ( G_p = 246 - B_p ), so:( G_{np} = 235 - (246 - B_p) = B_p - 11 )Which gives:( B_p - 11 = G_{np} implies B_p = G_{np} + 11 ). Which is exactly the given condition. So that holds.Additionally, total non-participants: 244, which is 255 - B_p (boys not participating) + G_{np} = 255 - B_p + B_p - 11 = 244. So that works.Therefore, all equations are satisfied. The answer is 490. Hmm, I think that's correct. Let me just recap the steps:1. Let total girls = G2. Participants: B_p + G_p = 2463. Total boys: B_p + B_np = 2554. Boys participating: B_p = G_np + 115. Total girls: G = G_p + G_np6. Substitute G_p = 246 - B_p into G: G = (246 - B_p) + (B_p - 11) = 2357. Total students = 255 + 235 = 490Yes, that seems to all check out. I don't see any missing steps or errors in the logic. So the total number of students is 490.Final Answerboxed{490}</think>Given the problem, we need to find the total number of students at Hogwarts School of Witchcraft and Wizardry. Here are the key steps to solve the problem:1. Define Variables: - ( B_p ): Number of boys participating in the competition. - ( G_p ): Number of girls participating in the competition. - ( G_{np} ): Number of girls not participating in the competition. - Total participants: 246 (given). - Total boys at the school: 255 (given).2. Key Equations: - Total participants: ( B_p + G_p = 246 ) - Number of boys participating: ( B_p = G_{np} + 11 ) - Total girls: ( G = G_p + G_{np} )3. Express ( G_p ) and ( G_{np} ) in terms of ( B_p ): - From total participants: ( G_p = 246 - B_p ) - From the given relationship: ( G_{np} = B_p - 11 )4. Calculate Total Number of Girls: - Substitute ( G_p ) and ( G_{np} ) into the total girls equation: [ G = (246 - B_p) + (B_p - 11) = 246 - 11 = 235 ]5. Calculate Total Number of Students: - Total students = Total boys + Total girls [ T = 255 + 235 = 490 ]6. Verification: - Total non-participants: ( 490 - 246 = 244 ) - Boys not participating: ( 255 - B_p ) - Girls not participating: ( B_p - 11 ) - Total non-participants check: [ (255 - B_p) + (B_p - 11) = 255 - 11 = 244 ]All equations are satisfied, confirming the solution is correct.[boxed{490}]]

🔑:begin{tabular}{l}mathbf{2 2} hline 490end{tabular}

🧠:[<think>Okay, so I need to solve this problem where part (1) is to find the rectangular coordinate equation and the parametric equation of curve C, which is given in polar form as ρ = 2 sin θ + 4 cos θ. Then part (2) is to find the length of the chord cut by line l on curve C. Let me start with part (1).First, I remember that converting a polar equation to rectangular coordinates involves using the relationships between polar and rectangular coordinates. The standard conversions are x = ρ cos θ, y = ρ sin θ, and ρ² = x² + y². Let me write that down:x = ρ cos θ y = ρ sin θ ρ = √(x² + y²)Given the polar equation ρ = 2 sin θ + 4 cos θ. Hmm. Let me see. I need to convert this into rectangular form. So maybe I can multiply both sides by ρ to make it easier to substitute using the above equations.If I multiply both sides by ρ, I get:ρ² = 2 ρ sin θ + 4 ρ cos θNow, substituting ρ² with x² + y², ρ sin θ with y, and ρ cos θ with x:x² + y² = 2y + 4xThat seems like the rectangular equation. Let me check. So, rearranging terms:x² - 4x + y² - 2y = 0Now, completing the square for both x and y terms.For the x terms: x² - 4x. Take half of -4, which is -2, square it to get 4. So add and subtract 4.For the y terms: y² - 2y. Take half of -2, which is -1, square it to get 1. Add and subtract 1.So:(x² - 4x + 4) - 4 + (y² - 2y + 1) - 1 = 0 Which simplifies to:(x - 2)² + (y - 1)² - 5 = 0 Therefore,(x - 2)² + (y - 1)² = 5So the rectangular coordinate equation is a circle with center at (2, 1) and radius √5. That makes sense because the original polar equation is linear in sin θ and cos θ, which usually represents a circle. Good.Now, part (1) also asks for the parametric equation of curve C. Since it's a circle, parametric equations are straightforward. The standard parametric equations for a circle with center (h, k) and radius r are:x = h + r cos θ y = k + r sin θBut since the original polar equation was given, maybe there's another parametrization. Wait, but in rectangular coordinates, the parametric equations can be written using an angle parameter. Let's confirm.Given the circle (x - 2)² + (y - 1)² = 5, so center (2,1), radius √5. So parametric equations would be:x = 2 + √5 cos θ y = 1 + √5 sin θWhere θ is the parameter varying from 0 to 2π. So that should be the parametric equation. Alternatively, since the original polar equation can be manipulated into this form, maybe there's another parametrization, but this seems straightforward.Wait, let me check if there's another way to parametrize it based on the original equation. The original equation was ρ = 2 sin θ + 4 cos θ. Maybe using θ as the parameter? But in that case, converting to parametric equations would be:From polar to parametric: x = ρ cos θ, y = ρ sin θ. So substituting ρ:x = (2 sin θ + 4 cos θ) cos θ y = (2 sin θ + 4 cos θ) sin θSimplify those:x = 2 sin θ cos θ + 4 cos² θ y = 2 sin² θ + 4 sin θ cos θBut these can be simplified using trigonometric identities. Let's see:First, 2 sin θ cos θ is sin 2θ, and 4 cos² θ can be written as 4*(1 + cos 2θ)/2 = 2 + 2 cos 2θ.Similarly, 2 sin² θ can be written as 2*(1 - cos 2θ)/2 = 1 - cos 2θ, and 4 sin θ cos θ is 2 sin 2θ.So:x = sin 2θ + 2 + 2 cos 2θ y = 1 - cos 2θ + 2 sin 2θHmm, but maybe it's more complicated than necessary. Alternatively, perhaps keeping θ as the parameter is acceptable. However, the parametric equations for a circle are typically given in terms of an angle parameter, which is similar to θ here. So maybe both parametrizations are valid. But the user probably expects the standard parametric equations for the circle, which are the ones with center (2,1) and radius √5. Let me confirm.Wait, if I use θ as the parameter in the original polar equation, converting to parametric equations would indeed give x and y in terms of θ, but with more complex expressions. But if the problem just wants a parametric equation, either form is acceptable. But since part (1) says "the parametric equation", maybe the first one is better, as it's simpler.But let me check the two parametrizations. Let me compare.If I use the standard circle parametrization:x = 2 + √5 cos φ y = 1 + √5 sin φwhere φ is the parameter. Alternatively, if I use θ from the original polar equation:x = 2 sin θ cos θ + 4 cos² θ y = 2 sin² θ + 4 sin θ cos θBut are these equivalent? Let's see.Suppose we set φ = θ + α for some angle α, perhaps. But maybe not. Alternatively, maybe there's a different parametrization. Let me check with specific angles.Take θ = 0. Then in the original parametrization:x = 2*0 + 4*1 = 4 y = 2*0 + 4*0 = 0In the standard circle parametrization, we need to find φ such that 2 + √5 cos φ = 4, which implies cos φ = (4 - 2)/√5 = 2/√5. Then sin φ = ±1/√5. So y = 1 + √5*(±1/√5) = 1 ±1. Therefore, y = 2 or 0. At θ =0, in the original parametrization, y=0. So corresponding φ would have sin φ = -1/√5, which gives y=1 -1=0. So that works. But it's a different parameter. So the two parametrizations are different but both describe the same circle.Therefore, the problem might accept either form. But the standard parametrization is more straightforward. However, since the original equation is given in polar coordinates, maybe they expect the parametric equations derived directly from the polar equation. Let me see.The problem says: "Find the rectangular coordinate equation and the parametric equation of curve C". Since the polar equation is given, perhaps they want the parametric equations in terms of θ, which is the parameter in polar coordinates. So even though θ here is the polar angle, using θ as the parameter for parametric equations is acceptable. So perhaps the parametric equations are:x = 4 cos θ + 2 sin θ cos θ Wait, no, let me recast:Wait, the original polar equation is ρ = 2 sin θ + 4 cos θ. So in parametric equations, we can write:x = ρ cos θ = (2 sin θ + 4 cos θ) cos θ y = ρ sin θ = (2 sin θ + 4 cos θ) sin θTherefore, x = 4 cos² θ + 2 sin θ cos θ y = 2 sin² θ + 4 sin θ cos θAlternatively, simplifying using double-angle identities:x = 4*(1 + cos 2θ)/2 + sin 2θ = 2 + 2 cos 2θ + sin 2θ y = 2*(1 - cos 2θ)/2 + 2 sin 2θ = 1 - cos 2θ + 2 sin 2θSo x = 2 + 2 cos 2θ + sin 2θ y = 1 - cos 2θ + 2 sin 2θBut this seems more complicated. However, it's a valid parametrization with parameter θ. Alternatively, using a different parameter. But maybe it's acceptable. However, the standard parametrization is simpler. So I need to check the problem's requirement. It says "parametric equation", so probably either is okay, but perhaps the one derived from polar coordinates is expected here.Alternatively, since the rectangular equation is a circle, maybe they expect parametric equations in terms of a parameter (like t) going around the circle. But in that case, as I mentioned earlier, the parametric equations would be:x = 2 + √5 cos t y = 1 + √5 sin tBut this is different from the parametrization using θ. Hmm. Let me check if these two parametrizations are equivalent. For instance, in the standard parametrization, when t=0, x=2 + √5, y=1. In the original parametrization, when θ=0, x=4, y=0. So they start at different points. So they are different parametrizations.Therefore, the problem might accept either, but since part (1) asks for both rectangular and parametric equations, and the original is given in polar, perhaps they want the parametric equations derived from the polar form. Let me confirm with my first calculation.Original polar equation: ρ = 2 sin θ + 4 cos θ. Converting to parametric:x = ρ cos θ = (2 sin θ + 4 cos θ) cos θ y = ρ sin θ = (2 sin θ + 4 cos θ) sin θSo expanding these:x = 2 sin θ cos θ + 4 cos² θ y = 2 sin² θ + 4 sin θ cos θAlternatively, factor terms:x = 2 sin θ cos θ + 4 cos² θ y = 2 sin² θ + 4 sin θ cos θBut these can be written as:x = 4 cos² θ + 2 sin θ cos θ y = 2 sin² θ + 4 sin θ cos θAlternatively, factoring out cos θ and sin θ:x = cos θ (4 cos θ + 2 sin θ) y = sin θ (2 sin θ + 4 cos θ)But this might not help much. Alternatively, using double-angle identities.For x:2 sin θ cos θ = sin 2θ 4 cos² θ = 4*(1 + cos 2θ)/2 = 2 + 2 cos 2θ So x = sin 2θ + 2 + 2 cos 2θFor y:2 sin² θ = 2*(1 - cos 2θ)/2 = 1 - cos 2θ 4 sin θ cos θ = 2 sin 2θ So y = 1 - cos 2θ + 2 sin 2θTherefore, the parametric equations can be written as:x = 2 + 2 cos 2θ + sin 2θ y = 1 - cos 2θ + 2 sin 2θThis is a valid parametrization, but perhaps not the most straightforward. Alternatively, maybe using t = 2θ, but that complicates it further.Alternatively, since the rectangular equation is a circle, perhaps using a parameter φ for the angle around the circle. Let's see.Center at (2,1), radius √5. So the parametric equations would be:x = 2 + √5 cos φ y = 1 + √5 sin φWhere φ is the parameter from 0 to 2π. This is much simpler. But does the problem require one or the other? The problem says "the parametric equation", so maybe either is acceptable. But since we were given a polar equation, they might expect the parametric equations in terms of θ. However, in many cases, when converting from polar to parametric, you just use θ as the parameter and express x and y in terms of θ, which is what I did first.Alternatively, maybe the problem expects both answers: the rectangular equation is the circle equation, and the parametric equations can be either the standard circle parametrization or the one derived from the polar equation. However, since the problem is structured as two separate parts, maybe part (1) is expecting both the rectangular equation and parametric equations for curve C. So perhaps they want the parametric equations derived from the polar equation, which would use θ as the parameter.In that case, the parametric equations would be:x = 4 cos θ + 2 sin θ cos θ y = 2 sin² θ + 4 sin θ cos θBut that seems messy. Alternatively, leaving it as:x = (2 sin θ + 4 cos θ) cos θ y = (2 sin θ + 4 cos θ) sin θBut perhaps simplifying:x = 4 cos² θ + 2 sin θ cos θ y = 2 sin² θ + 4 sin θ cos θBut maybe factoring:x = 2 cos θ (2 cos θ + sin θ) y = 2 sin θ (sin θ + 2 cos θ)Hmm, not sure if that helps. Alternatively, maybe using a substitution. Alternatively, since we know it's a circle, perhaps the parametrization can be expressed more cleanly. But since the original polar equation is ρ = 2 sin θ + 4 cos θ, which is equivalent to a circle, the parametric equations can also be written as the standard circle parametrization. Therefore, both forms are correct, but which one is expected here?Given that part (1) is to find both the rectangular and parametric equations, and the rectangular equation is a circle, I think the parametric equations would be expected to be the standard ones for a circle, i.e., using a parameter like φ:x = 2 + √5 cos φ y = 1 + √5 sin φBut the problem might consider θ as the parameter, even though in the polar equation θ is the polar angle, which is different from the parameter in the parametric equations. Wait, no. In parametric equations, θ can be used as a parameter, but here θ in the polar equation is the polar angle. However, when converting from polar to parametric, typically you end up with parametric equations in terms of θ, which is the same θ. So maybe they want that.Therefore, perhaps the parametric equations are:x = (2 sin θ + 4 cos θ) cos θ y = (2 sin θ + 4 cos θ) sin θBut written in simplified form. Let me compute them again:x = 2 sin θ cos θ + 4 cos² θ y = 2 sin² θ + 4 sin θ cos θYes. So these are the parametric equations directly converted from the polar equation. So maybe the problem expects this answer. Alternatively, the standard parametrization. However, since the rectangular equation is a circle, the standard parametrization is more elegant. But since the problem didn't specify, perhaps both answers are acceptable. But since part (1) is asking for both rectangular and parametric equations, and given that the rectangular equation is a circle, the standard parametrization is likely acceptable. Let me check with the original polar equation.Alternatively, since the polar equation is a circle, we can find its center and radius. Wait, we already converted it to rectangular and found the center (2,1) and radius √5. Therefore, the parametric equations are indeed the standard circle parametrization. Therefore, I think both answers are correct, but since we derived the rectangular equation as a circle, the parametric equations should correspond to that.Therefore, I will present both answers:Rectangular equation: (x - 2)² + (y - 1)² = 5 Parametric equations: x = 2 + √5 cos θ, y = 1 + √5 sin θ, where θ is the parameter.But let me verify once more. If we take the parametric equations as x = 2 + √5 cos θ, y = 1 + √5 sin θ, then converting back to polar coordinates:x = 2 + √5 cos θ y = 1 + √5 sin θBut this θ is the parameter, not the polar angle. To avoid confusion, maybe use a different symbol, like φ:x = 2 + √5 cos φ y = 1 + √5 sin φBut regardless, the parametric equations are valid. However, if we use θ as the parameter here, it's different from the polar angle θ in the original equation. To prevent confusion, maybe using t as the parameter is better.Alternatively, since the problem uses t as the parameter in line l's equation, maybe they expect a different parameter for curve C. But the problem statement just says "the parametric equation of curve C", so we can choose any parameter. Given that, the standard parametrization with a parameter t would be:x = 2 + √5 cos t y = 1 + √5 sin twhere t ∈ [0, 2π). This is the parametric equation. Therefore, I think that's acceptable.However, let's cross-verify by plugging in θ from the original polar equation. For example, in the original polar equation, when θ = 0, ρ = 2*0 + 4*1 = 4, so the point in polar coordinates is (4, 0). Converting to rectangular coordinates: x = 4*1 = 4, y = 4*0 = 0. So the point (4,0).In the rectangular equation (x - 2)² + (y - 1)² = 5, plugging in (4,0):(4 - 2)² + (0 - 1)² = 4 + 1 = 5. Correct.In the parametric equations x = 2 + √5 cos t, y = 1 + √5 sin t, to get the point (4,0):Set 2 + √5 cos t = 4 ⇒ cos t = 2/√5 1 + √5 sin t = 0 ⇒ sin t = -1/√5So t is an angle in the fourth quadrant where cos t = 2/√5 and sin t = -1/√5. That's valid, so the parametric equations can represent the point (4,0). Similarly, other points can be checked.Therefore, both the rectangular equation and parametric equations are correct.Now, moving on to part (2): Find the length of the chord cut by line l on curve C.Given line l's parametric equations:x = 2 + t y = -1 + √3 tSo line l is given parametrically with parameter t. Curve C is the circle (x - 2)² + (y - 1)² = 5. We need to find the points where line l intersects curve C, then compute the distance between those two points.First, let's substitute the parametric equations of line l into the rectangular equation of curve C.Substituting x = 2 + t and y = -1 + √3 t into (x - 2)² + (y - 1)² = 5:( (2 + t) - 2 )² + ( (-1 + √3 t) - 1 )² = 5 Simplify:(t)² + ( -2 + √3 t )² = 5Compute each term:First term: t² Second term: (-2 + √3 t)² = (-2)² + 2*(-2)*(√3 t) + (√3 t)^2 = 4 - 4√3 t + 3 t²Therefore, adding both terms:t² + 4 - 4√3 t + 3 t² = 5 Combine like terms:4 t² - 4√3 t + 4 = 5 Subtract 5 from both sides:4 t² - 4√3 t -1 = 0So we have a quadratic equation in t: 4t² - 4√3 t -1 = 0Let me solve this quadratic equation for t.The quadratic formula is t = [4√3 ± √( (4√3)^2 - 4*4*(-1) ) ] / (2*4)Compute discriminant:(4√3)^2 - 4*4*(-1) = 16*3 + 16 = 48 + 16 = 64So discriminant is √64 = 8Therefore, t = [4√3 ±8]/8 = [ (4√3 ±8) ] /8 = [ √3 ±2 ] /2Therefore, t1 = (√3 + 2)/2 and t2 = (√3 - 2)/2Therefore, the two points of intersection correspond to t1 and t2.Now, we need to find the points on line l corresponding to these t values and then compute the distance between them.Alternatively, since the chord length can be found using the distance between the two solutions t1 and t2 in the parameter t, multiplied by the magnitude of the direction vector of the line.The direction vector of line l is given by the coefficients of t in the parametric equations: (1, √3). So the direction vector is (1, √3), and its magnitude is √(1² + (√3)^2) = √(1 + 3) = √4 = 2.The chord length can be computed as |t1 - t2| * magnitude of direction vector.First, compute |t1 - t2|:t1 = (√3 + 2)/2 t2 = (√3 - 2)/2 t1 - t2 = [ (√3 + 2)/2 - (√3 - 2)/2 ] = [ (√3 + 2 - √3 + 2)/2 ] = (4)/2 = 2 Therefore, |t1 - t2| = 2Therefore, chord length = 2 * 2 = 4Wait, that seems straightforward. Let me check.Alternatively, compute the points:For t1 = (√3 + 2)/2:x = 2 + t1 = 2 + (√3 + 2)/2 = (4 + √3 + 2)/2 = (6 + √3)/2 y = -1 + √3 t1 = -1 + √3*(√3 + 2)/2 = -1 + (3 + 2√3)/2 = (-2 + 3 + 2√3)/2 = (1 + 2√3)/2For t2 = (√3 - 2)/2:x = 2 + t2 = 2 + (√3 - 2)/2 = (4 + √3 - 2)/2 = (2 + √3)/2 y = -1 + √3 t2 = -1 + √3*(√3 - 2)/2 = -1 + (3 - 2√3)/2 = (-2 + 3 - 2√3)/2 = (1 - 2√3)/2Now, compute the distance between the two points:Point 1: ( (6 + √3)/2 , (1 + 2√3)/2 )Point 2: ( (2 + √3)/2 , (1 - 2√3)/2 )Compute Δx = [ (6 + √3)/2 - (2 + √3)/2 ] = [ (6 + √3 - 2 - √3)/2 ] = (4)/2 = 2Compute Δy = [ (1 + 2√3)/2 - (1 - 2√3)/2 ] = [ (1 + 2√3 -1 + 2√3)/2 ] = (4√3)/2 = 2√3Therefore, distance = √( (Δx)^2 + (Δy)^2 ) = √(2² + (2√3)^2 ) = √(4 + 12) = √16 = 4So the length of the chord is 4. Therefore, the answer is 4.Alternatively, using the parametric approach with the direction vector's magnitude and the difference in parameters:Since the line's direction vector has magnitude 2, and |t1 - t2| = 2, then chord length is 2 * 2 = 4. Which matches. Therefore, both methods confirm the chord length is 4.Therefore, part (2)'s answer is 4.Final Answer(1) The rectangular coordinate equation is boxed{(x - 2)^2 + (y - 1)^2 = 5} and the parametric equations are begin{cases} x = 2 + sqrt{5} cos theta y = 1 + sqrt{5} sin theta end{cases} (where theta is the parameter). (2) The length of the chord is boxed{4}.</think>Given the parametric equation of line ( l ):[begin{cases}x = 2 + t y = -1 + sqrt{3}tend{cases}]and the polar equation of curve ( C ):[rho = 2 sin theta + 4 cos theta](1) Rectangular Coordinate Equation of Curve ( C ):1. Multiply both sides of the polar equation by ( rho ): [ rho^2 = 2 rho sin theta + 4 rho cos theta ]2. Substitute ( rho^2 = x^2 + y^2 ), ( rho sin theta = y ), and ( rho cos theta = x ): [ x^2 + y^2 = 2y + 4x ]3. Rearrange and complete the square: [ x^2 - 4x + y^2 - 2y = 0 implies (x - 2)^2 + (y - 1)^2 = 5 ] This is a circle with center ((2, 1)) and radius (sqrt{5}).Parametric Equations of Curve ( C ):Using the standard parametric equations for a circle:[begin{cases}x = 2 + sqrt{5} cos theta y = 1 + sqrt{5} sin thetaend{cases}]where (theta) is the parameter.(2) Length of the Chord:1. Substitute the parametric equations of line ( l ) into the rectangular equation of curve ( C ): [ (t)^2 + (-2 + sqrt{3}t)^2 = 5 ]2. Simplify and solve the quadratic equation: [ 4t^2 - 4sqrt{3}t - 1 = 0 ]3. Solve for ( t ) using the quadratic formula: [ t = frac{4sqrt{3} pm 8}{8} = frac{sqrt{3} pm 2}{2} ]4. Calculate the difference in parameters ( t_1 ) and ( t_2 ): [ |t_1 - t_2| = 2 ]5. The length of the chord is the magnitude of the direction vector of line ( l ) multiplied by the difference in parameters: [ text{Length} = 2 times 2 = 4 ]Final Answers:(1) The rectangular coordinate equation is (boxed{(x - 2)^2 + (y - 1)^2 = 5}) and the parametric equations are:[begin{cases}x = 2 + sqrt{5} cos theta y = 1 + sqrt{5} sin thetaend{cases}](2) The length of the chord is (boxed{4}).,<think>Okay, let's tackle this problem step by step. First, I need to solve part (1), which is to find the rectangular coordinate equation and the parametric equation of curve C given by the polar equation ρ = 2 sin θ + 4 cos θ. Then, part (2) asks for the length of the chord cut by line l on curve C. Let me start with part (1).For the first part, converting a polar equation to rectangular coordinates usually involves using the relationships between polar and rectangular coordinates: x = ρ cos θ, y = ρ sin θ, and ρ² = x² + y². The given equation is ρ = 2 sin θ + 4 cos θ. Hmm, let's see. Maybe I can multiply both sides by ρ to make it easier to substitute.Multiplying both sides by ρ gives: ρ² = 2 ρ sin θ + 4 ρ cos θ. Then, substituting ρ sin θ with y and ρ cos θ with x, we get:ρ² = 2y + 4x.But since ρ² = x² + y², substituting that in:x² + y² = 2y + 4x.Now, this looks like the equation of a circle, but let me rearrange it to confirm. Let's bring all terms to the left side:x² - 4x + y² - 2y = 0.To complete the squares for the x and y terms. For the x terms: x² - 4x. Take half of -4, which is -2, square it to get 4. Similarly for the y terms: y² - 2y. Take half of -2, which is -1, square it to get 1. So add and subtract these values:(x² - 4x + 4) - 4 + (y² - 2y + 1) - 1 = 0.This simplifies to:(x - 2)² + (y - 1)² - 5 = 0.Therefore:(x - 2)² + (y - 1)² = 5.So that's the rectangular coordinate equation of curve C. It's a circle with center at (2, 1) and radius √5.Now, for the parametric equation of curve C. Since it's a circle, a standard parametrization is:x = h + r cos t,y = k + r sin t,where (h, k) is the center and r is the radius. So substituting h = 2, k = 1, and r = √5, the parametric equations would be:x = 2 + √5 cos t,y = 1 + √5 sin t,where t is the parameter (angle). But let me verify if this makes sense. In polar coordinates, the original equation was ρ = 2 sin θ + 4 cos θ. If I recall, such equations can represent circles. The general form of a circle in polar coordinates is ρ = A sin θ + B cos θ, which corresponds to a circle with center at (B/2, A/2) and radius √((A/2)² + (B/2)²). Let me check that.Given ρ = 2 sin θ + 4 cos θ, so A = 2, B = 4. Then the center would be (4/2, 2/2) = (2, 1), which matches the center we found earlier. The radius would be √((2/2)^2 + (4/2)^2) = √(1 + 4) = √5, which also matches. So this seems correct.Therefore, the parametric equations are as I wrote above. So part (1) is done. The rectangular equation is (x - 2)² + (y - 1)² = 5, and the parametric equations are x = 2 + √5 cos t, y = 1 + √5 sin t.Now moving on to part (2): Find the length of the chord cut by line l on curve C. The line l is given parametrically as x = 2 + t, y = -1 + √3 t. So we need to find the points where this line intersects curve C, then compute the distance between those two points.First, perhaps substitute the parametric equations of line l into the rectangular equation of curve C, solve for t, then find the two points and compute the distance.The rectangular equation of curve C is (x - 2)² + (y - 1)² = 5. Line l is x = 2 + t, y = -1 + √3 t. Substitute x and y into the circle equation.Substituting x = 2 + t, so x - 2 = t. Similarly, y = -1 + √3 t, so y - 1 = -2 + √3 t.Therefore, substituting into the equation:(t)² + (-2 + √3 t)² = 5.Compute this:First, expand (-2 + √3 t)²:= (-2)^2 + 2*(-2)*(√3 t) + (√3 t)^2= 4 - 4√3 t + 3 t².So, the equation becomes:t² + 4 - 4√3 t + 3 t² = 5.Combine like terms:(1 t² + 3 t²) + (-4√3 t) + 4 = 54 t² - 4√3 t + 4 = 5Subtract 5 from both sides:4 t² - 4√3 t -1 = 0.So, quadratic equation in t: 4 t² -4√3 t -1 = 0.Let me solve this quadratic equation. Let's use the quadratic formula. For at² + bt + c = 0, t = [-b ± √(b² - 4ac)]/(2a).Here, a = 4, b = -4√3, c = -1.Compute discriminant D:b² -4ac = (-4√3)^2 -4*4*(-1) = 16*3 - (-16) = 48 +16 = 64.√D = 8.Therefore, t = [4√3 ±8]/(2*4) = [4√3 ±8]/8 = [√3 ±2]/2.So the solutions are t = (√3 + 2)/2 and t = (√3 - 2)/2.Now, these t values correspond to the points of intersection. Let's find the coordinates.For t1 = (√3 + 2)/2:x1 = 2 + t1 = 2 + (√3 + 2)/2 = (4 + √3 + 2)/2 = (6 + √3)/2.Wait, hold on: x = 2 + t, so substituting t1:x1 = 2 + [(√3 + 2)/2] = (4 + √3 + 2)/2? Wait, no. 2 is equal to 4/2. So:x1 = 2 + t1 = 4/2 + (√3 + 2)/2 = (4 + √3 + 2)/2 = (6 + √3)/2. Yes, that's correct.Similarly, y1 = -1 + √3 t1 = -1 + √3*(√3 + 2)/2.Compute that:√3*(√3 + 2)/2 = (3 + 2√3)/2.Therefore, y1 = -1 + (3 + 2√3)/2 = (-2/2 + 3/2) + (2√3)/2 = (1/2) + √3.So y1 = 1/2 + √3.Similarly for t2 = (√3 - 2)/2:x2 = 2 + t2 = 2 + (√3 - 2)/2 = (4 + √3 - 2)/2 = (2 + √3)/2.And y2 = -1 + √3*t2 = -1 + √3*(√3 - 2)/2.Compute that:√3*(√3 - 2)/2 = (3 - 2√3)/2.Thus, y2 = -1 + (3 - 2√3)/2 = (-2/2 + 3/2) - √3 = (1/2) - √3.So the two points of intersection are:Point 1: ( (6 + √3)/2 , 1/2 + √3 )Point 2: ( (2 + √3)/2 , 1/2 - √3 )Now, to find the distance between these two points, which is the length of the chord.Let me denote the coordinates:x1 = (6 + √3)/2, y1 = 1/2 + √3x2 = (2 + √3)/2, y2 = 1/2 - √3Compute the differences:Δx = x1 - x2 = [ (6 + √3)/2 - (2 + √3)/2 ] = (6 + √3 - 2 - √3)/2 = (4)/2 = 2.Δy = y1 - y2 = [ (1/2 + √3 ) - (1/2 - √3 ) ] = (1/2 -1/2) + (√3 + √3 ) = 2√3.Therefore, the distance between the two points is √( (Δx)^2 + (Δy)^2 ) = √(2² + (2√3)^2 ) = √(4 + 12 ) = √16 = 4.Wait, that's a nice number. So the length of the chord is 4. Hmm, let me verify my steps to be sure.First, solving the quadratic equation: 4 t² -4√3 t -1 =0. Discriminant 48 + 16=64. Solutions (4√3 ±8)/8 = (√3 ±2)/2. Correct.Then, computing the coordinates:For t1 = (√3 + 2)/2:x1 = 2 + t1. 2 is 4/2, so 4/2 + (√3 + 2)/2 = (6 + √3)/2. Correct.y1 = -1 + √3 t1 = -1 + √3*(√3 + 2)/2. √3*(√3) = 3, √3*2=2√3. So (3 + 2√3)/2. Then -1 is -2/2, so total is (-2 + 3 + 2√3)/2 = (1 + 2√3)/2. Wait, but above I had:Wait, wait a second. Wait, in my previous calculation, I had:y1 = -1 + (3 + 2√3)/2. Let me redo this:Start with y1 = -1 + [√3*(√3 + 2)/2]First compute √3*(√3 + 2) = (√3 * √3) + √3 * 2 = 3 + 2√3.So [3 + 2√3]/2. Then, add (-1):-1 is -2/2, so total y1 = (-2 + 3 + 2√3)/2 = (1 + 2√3)/2. Wait, but in my previous calculation, I said y1 = 1/2 + √3. Wait, that's different.Wait, hold on. So (1 + 2√3)/2 = 1/2 + √3. Yes, because 2√3 /2 = √3. So 1/2 + √3. Correct. So y1 is 1/2 + √3. Similarly for y2:y2 = -1 + [√3*(√3 -2)/2] = -1 + [ (3 - 2√3)/2 ] = (-2/2 + 3/2 ) - √3 = (1/2) - √3. Correct.Then, the coordinates:Point1: ( (6 + √3)/2 , 1/2 + √3 )Point2: ( (2 + √3)/2 , 1/2 - √3 )Then, Δx = (6 + √3)/2 - (2 + √3)/2 = (6 + √3 -2 -√3)/2 = 4/2 = 2. Correct.Δy = (1/2 + √3) - (1/2 - √3) = 2√3. Correct.Therefore, distance squared is 2² + (2√3)^2 = 4 + 12 = 16, square root is 4. Therefore, the length of the chord is 4.That seems straightforward, but let me check again.Alternatively, maybe there's a smarter way without computing the points. Since the line intersects the circle, the length of the chord can be found using the formula: 2√(r² - d²), where d is the distance from the center to the line, and r is the radius.Given that, let's try this method to verify.The radius r is √5, and d is the distance from the center (2, 1) to the line l. If that distance is d, then the chord length would be 2√(r² - d²).First, find the equation of line l. The parametric equations are x = 2 + t, y = -1 + √3 t. Let's convert this to standard form.From the parametric equations, direction vector is (1, √3). Therefore, the slope m is √3 /1 = √3. Then, the line can be written as y = √3 x + c. To find c, use a point on the line. When t=0, x=2, y=-1. So substituting into y = √3 x + c:-1 = √3*2 + c → c = -1 -2√3.Therefore, the equation is y = √3 x -1 -2√3.Alternatively, standard form: √3 x - y - (1 + 2√3) = 0.Now, the distance from center (2,1) to this line is:d = |√3*2 -1 - (1 + 2√3)| / √( (√3)^2 + (-1)^2 ) = |2√3 -1 -1 -2√3| / √(3 +1 ) = | -2 | / 2 = 2 / 2 = 1.Therefore, d =1. Then, chord length is 2√( (√5)^2 -1^2 ) = 2√(5 -1) = 2√4 = 4. Same result! So that confirms the chord length is indeed 4. So both methods agree. Therefore, the answer is 4.Therefore, the length of the chord is 4.Final Answer(1) The rectangular coordinate equation of curve ( C ) is (boxed{(x - 2)^2 + (y - 1)^2 = 5}) and its parametric equations are (begin{cases} x = 2 + sqrt{5} cos t y = 1 + sqrt{5} sin t end{cases}).(2) The length of the chord cut by line ( l ) on curve ( C ) is (boxed{4}).</think>Given the parametric equation of line ( l ) is (begin{cases} x = 2 + t y = -1 + sqrt{3}t end{cases}) (where ( t ) is the parameter), and the polar equation of curve ( C ) is (rho = 2sin theta + 4cos theta):(1) To find the rectangular coordinate equation and the parametric equation of curve ( C ):- Convert the polar equation to rectangular coordinates: [ rho = 2sin theta + 4cos theta ] Multiply both sides by (rho): [ rho^2 = 2rhosintheta + 4rhocostheta ] Substitute (rhosintheta = y), (rhocostheta = x), and (rho^2 = x^2 + y^2): [ x^2 + y^2 = 2y + 4x ] Rearrange and complete the square: [ x^2 - 4x + y^2 - 2y = 0 implies (x - 2)^2 + (y - 1)^2 = 5 ] The rectangular coordinate equation of curve ( C ) is (boxed{(x - 2)^2 + (y - 1)^2 = 5}).- The parametric equations for the circle with center ((2, 1)) and radius (sqrt{5}) are: [ begin{cases} x = 2 + sqrt{5} cos t y = 1 + sqrt{5} sin t end{cases} ] Thus, the parametric equations of curve ( C ) are (boxed{begin{cases} x = 2 + sqrt{5} cos t y = 1 + sqrt{5} sin t end{cases}}).(2) To find the length of the chord cut by line ( l ) on curve ( C ):- Substitute the parametric equations of line ( l ) into the rectangular equation of curve ( C ): [ x = 2 + t, quad y = -1 + sqrt{3}t ] Substitute into ((x - 2)^2 + (y - 1)^2 = 5): [ t^2 + (-2 + sqrt{3}t)^2 = 5 ] Simplify and solve the quadratic equation: [ 4t^2 - 4sqrt{3}t - 1 = 0 ] Using the quadratic formula, the solutions are ( t = frac{sqrt{3} + 2}{2} ) and ( t = frac{sqrt{3} - 2}{2} ).- Calculate the coordinates of the intersection points and the distance between them: - For ( t_1 = frac{sqrt{3} + 2}{2} ): ( left( frac{6 + sqrt{3}}{2}, frac{1}{2} + sqrt{3} right) ) - For ( t_2 = frac{sqrt{3} - 2}{2} ): ( left( frac{2 + sqrt{3}}{2}, frac{1}{2} - sqrt{3} right) )- The distance between the points is: [ sqrt{(2)^2 + (2sqrt{3})^2} = sqrt{4 + 12} = sqrt{16} = 4 ]Thus, the length of the chord cut by line ( l ) on curve ( C ) is (boxed{4}).]

🔑:Solution: (1) The polar equation of curve C can be transformed into rho^{2}=2rhosin theta+4rhocos theta, By rho^{2}=x^{2}+y^{2}, x=rhocos theta, y=rhosin theta, we get x^{2}+y^{2}=2y+4x, therefore The rectangular coordinate equation of curve C is (x-2)^{2}+(y-1)^{2}=5. The parametric equation is begin{cases} x=2+ sqrt{5}cos alpha y=1+ sqrt{5}sin alpha end{cases} (where alpha is the parameter). (2) Solution one: Since the parametric equation of line l is begin{cases} x=2+t y=-1+ sqrt{3}t end{cases}, therefore The standard equation of line l is sqrt{3}x-y-2sqrt{3}-1=0. therefore Curve C represents a circle with center at (2,1) and radius sqrt{5}, The distance from the center (2,1) to line l is dfrac{|2sqrt{3}-1-2sqrt{3}-1|}{2}=1, therefore The length of the chord cut by line l on circle C is 2sqrt{(sqrt{5})^{2}-1}=4. Solution two: Substituting begin{cases} x=2+t y=-1+ sqrt{3}t end{cases} into (x-2)^{2}+(y-1)^{2}=5 gives 4t^{2}-4sqrt{3}t-1=0. Let the parameters corresponding to the intersection points of line l and curve C be t_{1} and t_{2}, therefore t_{1}+t_{2}=sqrt{3}, t_{1}⋅t_{2}=-dfrac{1}{4}. Since the parametric equation of line l can be transformed into begin{cases} x=2+ dfrac{1}{2}(2t) y=-1+ dfrac{sqrt{3}}{2}(2t) end{cases}, therefore The length of the chord cut by line l on curve C is |2t_{1}-2t_{2}|=2sqrt{(t_{1}+t_{2})^{2}-4t_{1}t_{2}}=2×sqrt{3+1}=4. Therefore, the final answers are: (1) The rectangular coordinate equation of curve C is boxed{(x-2)^{2}+(y-1)^{2}=5} and the parametric equation is boxed{begin{cases} x=2+ sqrt{5}cos alpha y=1+ sqrt{5}sin alpha end{cases}}. (2) The length of the chord cut by line l on curve C is boxed{4}.

🧠:[<think>Okay, let's see. I need to solve the equation 9^x + 6^x = 2^(2x+1). Hmm. Alright, first, let me try to rewrite all the terms with the same base or in terms of exponents that might be related. So, 9 is 3 squared, right? So 9^x is the same as (3^2)^x, which simplifies to 3^(2x). Similarly, 6 can be written as 2*3, so 6^x is (2*3)^x, which is 2^x * 3^x. Then on the right side, 2^(2x+1) can be written as 2^(2x) * 2^1, which is 2*2^(2x). So substituting all these into the equation, we get:3^(2x) + 2^x * 3^x = 2 * 2^(2x)Hmm. Let me check that again. 9^x becomes 3^(2x), 6^x is (2*3)^x = 2^x*3^x, and 2^(2x+1) is 2*2^(2x). Yes, that's correct.Now, maybe I can divide all terms by 2^(2x) to simplify things. Let's try that. Dividing each term by 2^(2x):[3^(2x) / 2^(2x)] + [2^x*3^x / 2^(2x)] = [2*2^(2x) / 2^(2x)]Simplifying each term:First term: (3/2)^(2x)Second term: (3^x * 2^x) / 2^(2x) = (3/2)^x * (2^x / 2^(2x)) = (3/2)^x * 2^(-x) = (3/2)^x * (1/2)^x = (3/2 * 1/2)^x = (3/4)^x. Wait, maybe I did that step wrong. Let's re-examine.Wait, (2^x * 3^x) / 2^(2x) = (3^x) * (2^x / 2^(2x)) = 3^x * 2^(-x) = (3/2)^x. Wait, no. Wait, 2^(-x) is 1/(2^x), so 3^x * 2^(-x) is (3/2)^x. Wait, but 3/2 is 1.5, so that term is (1.5)^x. Hmm. Wait, but let me check again. 2^x divided by 2^(2x) is 2^(x - 2x) = 2^(-x). So 3^x * 2^(-x) is (3/2)^x. Yes, that's correct. So the second term is (3/2)^x. Wait, but that contradicts what I thought earlier. Let me verify.Original second term after division: [2^x * 3^x] / 2^(2x) = (3^x / 2^(2x - x)) = 3^x / 2^x = (3/2)^x. Yes, correct. So the second term is (3/2)^x.Wait, but then the first term is [3^(2x)/2^(2x)] = (3/2)^(2x) = [(3/2)^x]^2. So substituting back, the equation becomes:(3/2)^(2x) + (3/2)^x = 2Hmm, that seems more manageable. Let's let t = (3/2)^x. Then the equation becomes t^2 + t - 2 = 0. Wait, because substituting t in:t^2 + t = 2, so t^2 + t - 2 = 0.Now, solving this quadratic equation. The quadratic is t^2 + t - 2 = 0. Let's use the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)]/(2a). Here a=1, b=1, c=-2. So discriminant D = 1^2 - 4*1*(-2) = 1 + 8 = 9. So sqrt(9)=3. Therefore, t = [-1 ± 3]/2. So two solutions:t = (-1 + 3)/2 = 2/2 = 1t = (-1 -3)/2 = -4/2 = -2But t = (3/2)^x, and since (3/2) is a positive number, any real power of it will be positive. Therefore, t can't be negative. So we discard t = -2. Thus, the only solution is t = 1.Therefore, (3/2)^x = 1. But any number to the power of 0 is 1, so x = 0.Wait, but let me check if x=0 is a solution. Let's substitute back into the original equation.Left side: 9^0 + 6^0 = 1 + 1 = 2Right side: 2^(2*0 +1) = 2^1 = 2. So yes, 2=2. Correct. So x=0 is a solution.But wait, is there another solution? Because sometimes exponentials can cross in more than one point. Let me check. Maybe I made a mistake in assuming only t=1 is valid. Let's see.Wait, when we set t = (3/2)^x, we transformed the equation into t^2 + t - 2 = 0, which had solutions t=1 and t=-2. Since t must be positive, we discard t=-2. Therefore, only t=1, leading to x=0. Therefore, the only solution is x=0.But let's check if there might be another solution. Let me consider the function f(x) = 9^x + 6^x - 2^(2x+1). We found that f(0) = 0. Let's check at x=1:f(1) = 9 +6 - 2^(3) = 15 -8 =7 >0.At x= -1:f(-1)= 9^{-1} +6^{-1} -2^{ -2 +1 }=1/9 +1/6 -2^{-1}= (2/18 +3/18) -1/2=5/18 -9/18= -4/18= -2/9 <0.So f(-1) is negative. So between x=-1 and x=0, the function crosses zero. But we found x=0 as a solution. Wait, but when x approaches negative infinity, 9^x and 6^x approach zero, and 2^(2x+1) approaches zero, but which term dominates? Let's see:As x approaches negative infinity, 9^x = (3^2)^x = 3^{2x}, which tends to zero. Similarly, 6^x = (6)^x tends to zero. The right side is 2^{2x+1} = 2*4^x, which also tends to zero. So the behavior might be that f(x) approaches zero from some direction, but maybe not crossing again.Wait, but at x=-1, f(-1) is -2/9, and at x=0, f(0)=0. So between x=-1 and x=0, the function goes from negative to zero. Then from x=0 onwards, it's increasing because f(1)=7, so it's positive. So maybe x=0 is the only real solution. But let me check another point between x=-1 and x=0, say x=-0.5:f(-0.5)=9^{-0.5} +6^{-0.5} -2^{2*(-0.5)+1}= 1/3 + 1/sqrt(6) -2^{0}= (1/3 + approx 0.408) -1 ≈ (0.333 +0.408) -1 ≈ 0.741 -1 ≈ -0.259 <0.So at x=-0.5, f(x) is still negative. Then at x approaching 0 from the left, f(x) approaches 0 from below. So maybe x=0 is the only real solution. But wait, perhaps there's another solution when x is negative? Let me check x=-2:f(-2)=9^{-2} +6^{-2} -2^{2*(-2)+1}=1/81 +1/36 -2^{-3}= (4/324 +9/324) -1/8=13/324 -1/8≈0.0401 -0.125≈-0.0849 <0.Still negative. How about x approaching negative infinity? Let's see, as x→-infty:9^x and 6^x go to zero, while 2^{2x+1}=2*4^x also goes to zero. But which goes faster? Let's compare exponents. For example, 4^x decreases faster than 6^x as x→-infty (since 4 <6), so 6^x approaches zero slower than 4^x. So 6^x would dominate over 4^x in the limit. Wait, but in the equation f(x) =9^x +6^x -2*4^x. So as x→-infty, 9^x ~0, 6^x ~0, and 2*4^x ~0. But maybe 6^x is larger than 4^x for negative x. Let's see:Take x=-1, 6^{-1}=1/6 ≈0.166, 4^{-1}=0.25. So here, 4^{-1} is larger. Wait, so 4^x when x is negative is larger than 6^x. So at x=-1, 2*4^{-1}=0.5, and 6^{-1}=0.166, so 0.5>0.166. So as x→-infty, 4^x becomes (since x is negative) 4^x = e^{x ln4}, which approaches zero as x→-infty. Similarly, 6^x approaches zero. But the rate: 4^x vs 6^x. Let's compare ln(4) vs ln(6). Since ln(4)≈1.386, ln(6)≈1.792. So as x→-infty, the term with the smaller ln would dominate (since x is negative). So 4^x decays slower than 6^x. Therefore, 4^x >6^x for x negative. Therefore, 2*4^x >6^x for x negative. Thus, f(x)=9^x +6^x -2*4^x ≈0 +6^x -2*4^x ≈ negative because 2*4^x >6^x. Therefore, as x→-infty, f(x) approaches zero from below. Therefore, the function f(x) is negative for all x <0 except at x=0 where it's zero, and then becomes positive for x>0. Therefore, the only solution is x=0.Wait, but wait a second. At x=0, f(x)=0. When x>0, f(x) increases because all terms are increasing, but since 9^x and 6^x grow faster than 4^x? Wait, for x>0, 9^x grows as 3^(2x), which is faster than 4^x=2^(2x). So 9^x would dominate, making f(x) positive. For x=1, as checked earlier, f(1)=7>0. So there's no other crossing. Therefore, x=0 is the only solution.But wait, let me confirm by checking the behavior of the function. Let's consider f(x)=9^x +6^x -2*4^x.At x=0, f(x)=2 -2=0.For x>0: 9^x and 6^x increase faster than 4^x, so f(x) becomes positive.For x<0: Let's take x=-0.5: f(-0.5)=1/3 +1/sqrt(6) -2*4^{-0.5}=1/3 + approx0.408 -2*(1/2)=1/3+0.408-1≈-0.259<0.At x approaching negative infinity, as discussed, f(x) approaches zero from below. So the function crosses zero only once at x=0. Therefore, the only solution is x=0.But wait, let me check if there's another solution. Suppose x is a negative value where 9^x +6^x equals 2*4^x. Let's suppose x is negative. Let me try x= log base something. Wait, but maybe there's another solution. Let's try x= -1: f(-1)=1/9 +1/6 -1/2= (2/18 +3/18) -9/18=5/18 -9/18=-4/18=-2/9 <0.x=-0.5: as above, negative. x=-0.1: let's compute f(-0.1)=9^{-0.1} +6^{-0.1} -2*4^{-0.1}.Calculate each term:9^{-0.1}=1/9^{0.1}=1/(3^{0.2})= approx 1/1.2457≈0.8027.6^{-0.1}=1/6^{0.1}≈1/1.1798≈0.8473.4^{-0.1}=1/4^{0.1}=1/(2^{0.2})≈1/1.1487≈0.8706.So 2*4^{-0.1}≈2*0.8706≈1.7412.Then f(-0.1)=0.8027 +0.8473 -1.7412≈1.65 -1.7412≈-0.0912 <0.Still negative. x=-0.05:9^{-0.05}=1/3^{0.1}≈1/1.1161≈0.896.6^{-0.05}=1/6^{0.05}≈1/1.0938≈0.914.4^{-0.05}=1/4^{0.05}=1/(2^{0.1})≈1/1.0718≈0.933.So 2*4^{-0.05}≈2*0.933≈1.866.f(-0.05)=0.896+0.914 -1.866≈1.81 -1.866≈-0.056 <0.Hmm, still negative. x approaching 0 from the left, say x=-0.001:9^{-0.001}=e^{-0.001*ln9}≈1 -0.001*ln9≈0.9978.Similarly, 6^{-0.001}=e^{-0.001*ln6}≈1 -0.001*ln6≈0.9972.4^{-0.001}=e^{-0.001*ln4}≈1 -0.001*ln4≈0.9963.So 2*4^{-0.001}≈2*0.9963≈1.9926.Then f(-0.001)=0.9978 +0.9972 -1.9926≈1.995 -1.9926≈0.0024. Wait, that's positive. Wait, that contradicts earlier. Wait, did I compute correctly?Wait, when x approaches 0 from the left, x=-ε where ε is a small positive number approaching zero. So:9^{-ε} ≈1 - ε ln96^{-ε}≈1 - ε ln64^{-ε}≈1 - ε ln4Then f(-ε)=9^{-ε} +6^{-ε} -2*4^{-ε}≈[1 -ε ln9] + [1 -ε ln6] -2*[1 -ε ln4] =1 -ε ln9 +1 -ε ln6 -2 +2ε ln4Simplify: (1+1-2) + ε(-ln9 -ln6 +2 ln4)=0 + ε(-ln9 -ln6 +2 ln4)Compute the coefficient:ln9=2 ln3, ln6=ln2 +ln3, ln4=2 ln2.Thus,-2 ln3 - (ln2 + ln3) +2*(2 ln2)= -2 ln3 -ln2 -ln3 +4 ln2= (-3 ln3) +3 ln2=3(ln2 -ln3)=3 ln(2/3)=3 ln(2/3)≈3*(-0.4055)=≈-1.2165.Therefore, f(-ε)≈ ε*(-1.2165). Since ε is positive (as x approaches 0 from below), then f(-ε)≈-1.2165 ε <0. Therefore, as x approaches 0 from the left, f(x) approaches 0 from below. Therefore, near x=0 from the left, f(x) is still negative. But when I computed x=-0.001, according to the linear approximation, f(-0.001)≈-1.2165*(-0.001)=0.0012165, which is positive. But wait, there's a discrepancy here. Wait, maybe the linear approximation is not accurate enough. Let's compute x=-0.001 more accurately.Using exact exponents:9^{-0.001}=e^{-0.001*ln9}=approx 1 -0.001*ln9 + (0.001*ln9)^2/2Similarly for the others. Let's compute each term:ln9≈2.1972, so 0.001*ln9≈0.00219729^{-0.001}≈1 -0.0021972 + (0.0021972)^2/2 ≈1 -0.0021972 +0.00000241≈0.997805Similarly, ln6≈1.7918, so 0.001*ln6≈0.00179186^{-0.001}≈1 -0.0017918 + (0.0017918)^2/2≈1 -0.0017918 +0.0000016≈0.998210ln4≈1.3863, so 0.001*ln4≈0.00138634^{-0.001}≈1 -0.0013863 + (0.0013863)^2/2≈1 -0.0013863 +0.00000096≈0.998615Then 2*4^{-0.001}≈2*0.998615≈1.99723Then f(-0.001)=9^{-0.001} +6^{-0.001} -2*4^{-0.001}≈0.997805 +0.998210 -1.99723≈1.995015 -1.99723≈-0.002215 <0Ah, so the actual value is negative. Therefore, my initial calculation with linear approximation was correct in sign. Therefore, near x=0 from the left, f(x) approaches 0 from below. Thus, there's no crossing other than at x=0. Therefore, the only solution is x=0.But wait, wait, when I first approximated with x=-0.001 using the linear terms, I thought f(-0.001)≈0.0024, but with the quadratic terms included, it's actually negative. Therefore, x=0 is the only solution.So the answer is x=0. Let me confirm once more.Original equation: 9^x +6^x =2^(2x+1).At x=0: 1+1=2, which matches 2^1=2. Correct.For any x>0, left side grows faster than the right side. For x<0, left side is less than right side, so no solution. Therefore, x=0 is the only solution.</think>The problem is to solve the equation (9^{x} + 6^{x} = 2^{2x+1}).Step-by-Step Solution:1. Rewrite terms with common bases: - (9^{x} = (3^2)^x = 3^{2x}) - (6^{x} = (2 cdot 3)^x = 2^x cdot 3^x) - (2^{2x+1} = 2 cdot 2^{2x} = 2 cdot (2^2)^x = 2 cdot 4^x)2. Substitute into the equation: [ 3^{2x} + 2^x cdot 3^x = 2 cdot 4^x ]3. Divide all terms by (4^x) to simplify: [ frac{3^{2x}}{4^x} + frac{2^x cdot 3^x}{4^x} = 2 ] - Simplify each term: - (frac{3^{2x}}{4^x} = left(frac{9}{4}right)^x = left(frac{3}{2}right)^{2x}) - (frac{2^x cdot 3^x}{4^x} = left(frac{6}{4}right)^x = left(frac{3}{2}right)^x)4. Let (t = left(frac{3}{2}right)^x): - The equation becomes (t^2 + t = 2). - Rearranged: (t^2 + t - 2 = 0).5. Solve the quadratic equation: - Discriminant: (D = 1^2 - 4(1)(-2) = 9) - Roots: (t = frac{-1 pm sqrt{9}}{2} = frac{-1 pm 3}{2}) - Solutions: (t = 1) or (t = -2).6. Reject invalid solutions: - (t = left(frac{3}{2}right)^x > 0), so (t = -2) is discarded. - Thus, (t = 1).7. Solve for (x): [ left(frac{3}{2}right)^x = 1 implies x = 0. ]8. Verify (x = 0): - Left side: (9^0 + 6^0 = 1 + 1 = 2) - Right side: (2^{2(0)+1} = 2^1 = 2) - Both sides equal, so (x = 0) is valid.Conclusion: The only solution is (boxed{0}).,<think>Okay, so I need to solve this equation: 9^x + 6^x = 2^{2x+1}. Hmm, let's start by trying to rewrite everything in terms of the same base if possible. I know that 9 is 3 squared, so 9^x is (3^2)^x which simplifies to 3^{2x}. Similarly, 6 can be written as 2*3, so 6^x is (2*3)^x, which is 2^x * 3^x. On the right side, 2^{2x+1} can be written as 2^{2x} * 2^1, which is 2*2^{2x} or 2*(2^2)^x. Wait, 2^{2x} is the same as 4^x. So, the right-hand side is 2*4^x.So rewriting the equation, we have:3^{2x} + 2^x * 3^x = 2*4^xHmm, maybe we can divide both sides by 4^x to make things simpler? Let's try that. Dividing each term by 4^x gives:(3^{2x})/(4^x) + (2^x * 3^x)/(4^x) = 2Let me simplify each term. Since 4^x is (2^2)^x = 2^{2x}, so:3^{2x}/2^{2x} = (3/2)^{2x} = ( (3/2)^x )^2Similarly, (2^x * 3^x)/2^{2x} = (3^x)/(2^x) = (3/2)^xSo substituting back, the equation becomes:( (3/2)^x )^2 + (3/2)^x = 2Let me set y = (3/2)^x. Then the equation becomes:y^2 + y - 2 = 0Wait, because it's y^2 + y = 2. So moving 2 to the left, we get y^2 + y - 2 = 0.Now, solving this quadratic equation for y. Let's factor it. Looking for two numbers that multiply to -2 and add up to 1. Hmm, 2 and -1. So, (y + 2)(y - 1) = 0. Therefore, y = -2 or y = 1.But y is (3/2)^x, and since (3/2)^x is always positive for real x, y cannot be negative. So y = 1 is the only valid solution.Therefore, (3/2)^x = 1. Taking logarithm on both sides. Remember that any number to the power 0 is 1, so x must be 0.Wait, let's check that. If x = 0, then 9^0 + 6^0 = 1 + 1 = 2. On the right-hand side, 2^{2*0 +1} = 2^1 = 2. So 2 = 2. That works.But wait, is there another solution? Let's double-check. The quadratic gave only y = 1 as a valid solution, leading to x = 0. Let's check if maybe there's another solution that I missed.Suppose we graph both sides. The left side is 9^x + 6^x, which grows exponentially. The right side is 2*4^x, which also grows exponentially. Let's see the behavior as x increases. 9^x grows faster than 4^x, so for large x, left side would dominate. When x is 0, both sides are 2. When x is 1: left is 9 + 6 = 15, right is 2*4 = 8. So left is larger. When x is negative: let's try x = -1. Left side is 9^{-1} + 6^{-1} = 1/9 + 1/6 = (2 + 3)/18 = 5/18 ≈ 0.277. Right side is 2^{2*(-1)+1} = 2^{-1} = 0.5. So right side is larger here. So the left side is less than the right side at x = -1. So maybe there's a solution somewhere between x = -1 and x = 0?Wait, but according to our algebra, we only found x = 0. That seems inconsistent. Let's check x = 0.5. Left side: 9^{0.5} + 6^{0.5} = 3 + sqrt(6) ≈ 3 + 2.45 ≈ 5.45. Right side: 2^{2*0.5 +1} = 2^{2} = 4. Left side is larger. What about x = -0.5: Left side is 9^{-0.5} + 6^{-0.5} = 1/3 + 1/sqrt(6) ≈ 0.333 + 0.408 ≈ 0.741. Right side: 2^{2*(-0.5)+1} = 2^{0} = 1. So right side is still larger. So at x = -0.5, left ≈ 0.741 < 1. At x = 0, left = 2 = right. So maybe x = 0 is the only solution. But let's check if there's another crossing point.Wait, when x approaches negative infinity, 9^x approaches 0, 6^x approaches 0, so left side approaches 0. The right side is 2^{2x +1} = 2*4^x, which also approaches 0. But the rates might differ. Let me see: 9^x is (3^2)^x = 3^{2x}, 6^x = (6)^x, and 4^x. As x approaches negative infinity, 3^{2x} and 6^x decay to 0. But 4^x also decays. Which decays faster? The base with larger number decays faster when x is negative. Wait, for example, 9^x when x is negative is (1/9)^{|x|}, which decays faster than 4^x = (1/4)^{|x|}. Similarly, 6^x decays faster than 4^x. So the left side would approach 0 faster than the right side. So as x approaches negative infinity, the right side is larger. At x = 0, they are equal. For x > 0, left side is larger. For x < 0, left side is less than right side. So the equation crosses only at x = 0? But according to the algebra, x=0 is the only solution.But wait, perhaps we need to check more carefully. Let's check x = -1 again. Left side: 1/9 + 1/6 ≈ 0.111 + 0.166 ≈ 0.277. Right side: 0.5. So left is smaller. At x=0, left=2, right=2. So between x=-infty and x=0, the left side goes from 0 to 2, and the right side goes from 0 to 2. But maybe they cross only once at x=0? Let's check the derivative. Maybe the left side is always increasing and the right side is also increasing. Wait, the left side is 9^x + 6^x. The derivative is ln(9)*9^x + ln(6)*6^x. The right side's derivative is ln(4)*2^{2x+1} = ln(4)*2*4^x. At x=0, the left derivative is ln(9) + ln(6) ≈ 2.197 + 1.792 ≈ 3.989. The right derivative is ln(4)*2 ≈ 1.386*2 ≈ 2.772. So at x=0, the left side is increasing faster. So after x=0, left side overtakes the right side and stays above. For x <0, let's see the behavior. For x approaching negative infinity, both sides approach 0. But left side approaches 0 faster. At x=0, they are equal. So is there another crossing point between x=-infty and x=0?Wait, at x=-1, left ≈0.277, right=0.5. So left < right. At x=0, left=right=2. So maybe the left side crosses the right side once at x=0. But how? Because when x approaches from the left (x approaching 0 from the negative side), what are the values? Let's pick x=-0.1. Left: 9^{-0.1} +6^{-0.1} ≈ (1/9^{0.1}) + (1/6^{0.1}). 9^{0.1} is e^{0.1*ln9} ≈ e^{0.1*2.197} ≈ e^{0.2197} ≈ 1.2457. So 1/1.2457 ≈0.803. Similarly, 6^{0.1} ≈ e^{0.1*1.7918} ≈ e^{0.17918} ≈1.196, so 1/1.196 ≈0.836. So total left≈0.803 +0.836≈1.639. Right side: 2^{2*(-0.1)+1} =2^{-0.2 +1}=2^{0.8}≈1.741. So left≈1.639 < right≈1.741. So at x=-0.1, left is less than right. At x=-0.05: Left:9^{-0.05} +6^{-0.05}. 9^{-0.05}=1/9^{0.05}=1/e^{0.05*ln9}≈1/e^{0.05*2.197}=1/e^{0.10985}≈1/1.116≈0.896. 6^{-0.05}=1/6^{0.05}=1/e^{0.05*1.7918}=1/e^{0.0896}≈1/1.094≈0.914. So left≈0.896+0.914≈1.810. Right side:2^{2*(-0.05)+1}=2^{-0.1 +1}=2^{0.9}≈1.866. So left≈1.810 < right≈1.866.At x=-0.02: Left:9^{-0.02}=1/9^{0.02}=1/e^{0.02*2.197}=1/e^{0.04394}=≈0.957. 6^{-0.02}=1/6^{0.02}=1/e^{0.02*1.7918}=1/e^{0.0358}=≈0.965. Left≈0.957+0.965≈1.922. Right:2^{2*(-0.02)+1}=2^{-0.04 +1}=2^{0.96}≈1.933. Still left < right.At x=-0.01: Left:9^{-0.01}≈1/e^{0.01*2.197}=≈1/1.022≈0.979. 6^{-0.01}≈1/e^{0.01*1.792}=≈1/1.018≈0.982. Sum≈0.979+0.982≈1.961. Right:2^{2*(-0.01)+1}=2^{-0.02+1}=2^{0.98}≈≈1.933. Wait, 2^{0.98}=e^{0.98 ln2}≈e^{0.98*0.693}≈e^{0.679}≈1.972. So left≈1.961 < right≈1.972.At x=0, left=2, right=2. So between x=-0.01 and x=0, left increases from ~1.961 to 2, and right increases from ~1.972 to 2. So there's a crossing point at x=0 where they meet. Wait, but does that mean there's another solution very close to x=0? Because at x=-0.01, left is 1.961 and right is ~1.972. At x=0, both are 2. So the left side is increasing faster here. Wait, but the left side's rate of increase near x=0 is higher. The derivative at x=0 for left is ln9 + ln6 ≈2.197+1.792≈3.989, and right derivative is ln4*2≈1.386*2≈2.772. So the left side is increasing faster. Therefore, just to the left of x=0, left side is catching up to the right side. Since at x=-0.01, left is 1.961 and right is ~1.972. So left is less than right. At x=0, they are equal. So the only crossing is at x=0. So there's only one solution: x=0.But wait, maybe there's another solution at x=1? Let's check. x=1: left=9+6=15, right=2^{3}=8. No. x=2: left=81 +36=117, right=2^{5}=32. No. How about x=-2: left=9^{-2} +6^{-2}=1/81 +1/36≈0.0123+0.0278≈0.0401. Right=2^{2*(-2)+1}=2^{-3}=1/8=0.125. Left < right. So no.Therefore, the only solution is x=0. But wait, in the quadratic equation, we had y=1, leading to x=0, which works. So maybe that's the only solution. But why does the graph seem to approach each other near x=0?Wait, maybe I need to re-examine the substitution. The equation after substitution was y^2 + y =2, leading to y=1 and y=-2. Since y=(3/2)^x can't be negative, only y=1. So x=0. Therefore, the only solution is x=0.So the answer is x=0. But just to confirm, let's plug x=0 into the original equation: 9^0 +6^0=1+1=2, and 2^{2*0+1}=2^1=2. Correct. So yes, x=0 is the solution. Therefore, the solution is x=0.</think>7.214. 9^{x}+6^{x}=2^{2 x+1}.Step-by-Step Solution:1. Rewrite terms using common bases: - (9^x = (3^2)^x = 3^{2x}) - (6^x = (2 cdot 3)^x = 2^x cdot 3^x) - (2^{2x+1} = 2 cdot (2^{2x}) = 2 cdot 4^x)2. Substitute into the equation: [ 3^{2x} + 2^x cdot 3^x = 2 cdot 4^x ]3. Divide all terms by (4^x) to simplify: [ frac{3^{2x}}{4^x} + frac{2^x cdot 3^x}{4^x} = 2 ] - Simplify fractions: - (frac{3^{2x}}{4^x} = left(frac{3}{2}right)^{2x} = left(left(frac{3}{2}right)^xright)^2) - (frac{2^x cdot 3^x}{4^x} = left(frac{3}{2}right)^x)4. Let (y = left(frac{3}{2}right)^x). The equation becomes: [ y^2 + y = 2 ] Rearrange: [ y^2 + y - 2 = 0 ]5. Solve the quadratic equation (y^2 + y - 2 = 0): - Factor: ((y + 2)(y - 1) = 0) - Solutions: (y = -2) or (y = 1)6. Reject invalid solutions: - (y = left(frac{3}{2}right)^x) is always positive, so (y = -2) is discarded. - Valid solution: (y = 1).7. Solve for (x): [ left(frac{3}{2}right)^x = 1 implies x = 0 ]8. Verify (x = 0) in the original equation: - Left side: (9^0 + 6^0 = 1 + 1 = 2) - Right side: (2^{2(0)+1} = 2^1 = 2) - Both sides equal (2), confirming the solution.Final Answer: The solution is (boxed{0}).]

🔑:Solution.Rewrite the equation as 3^{2 x}+2^{x} cdot 3^{x}-2 cdot 2^{2 x}=0 and divide it by 2^{2 x} neq 0. Then left(frac{3}{2}right)^{2 x}+left(frac{3}{2}right)^{x}-2=0 Rightarrowleft(left(frac{3}{2}right)^{x}right)=-2 (no solutions) or left(left(frac{3}{2}right)^{x}right)_{2}=1 Rightarrow x=0.Answer: 0.

🧠:[<think>Okay, so I need to find the range of values for ( x + y ) given that ( x^3 + y^3 = a^3 ) where ( a > 0 ). Hmm, let's think about how to approach this. Maybe I can use some algebraic identities or inequalities. First, I remember that ( x^3 + y^3 ) can be factored as ( (x + y)(x^2 - xy + y^2) ). Let me write that down:[x^3 + y^3 = (x + y)(x^2 - xy + y^2) = a^3]So, ( (x + y)(x^2 - xy + y^2) = a^3 ). Let me denote ( s = x + y ) and ( p = xy ). Then, ( x^2 - xy + y^2 ) can be expressed in terms of ( s ) and ( p ). Let's see:We know that ( x^2 + y^2 = (x + y)^2 - 2xy = s^2 - 2p ). Therefore, ( x^2 - xy + y^2 = (x^2 + y^2) - xy = (s^2 - 2p) - p = s^2 - 3p ). So, substituting back into the original equation:[s(s^2 - 3p) = a^3]So now we have ( s^3 - 3sp = a^3 ). Hmm, but we still have two variables here, ( s ) and ( p ). Maybe we can relate ( s ) and ( p ) using another equation or inequality. I remember that for real numbers ( x ) and ( y ), the discriminant of the quadratic equation they satisfy must be non-negative. If ( s = x + y ) and ( p = xy ), then ( x ) and ( y ) are roots of the equation ( t^2 - st + p = 0 ). For real roots, the discriminant must be non-negative:[s^2 - 4p geq 0 implies p leq frac{s^2}{4}]So that gives us an upper bound for ( p ). Let me note that down: ( p leq frac{s^2}{4} ).Going back to the equation ( s^3 - 3sp = a^3 ), let's solve for ( p ):[s^3 - 3sp = a^3 implies 3sp = s^3 - a^3 implies p = frac{s^3 - a^3}{3s}]But ( s ) cannot be zero here because if ( s = 0 ), then ( x + y = 0 ), which would imply ( x^3 + y^3 = 0 ), but we have ( x^3 + y^3 = a^3 neq 0 ). So ( s neq 0 ). Therefore, ( p = frac{s^3 - a^3}{3s} ).Now, substituting this expression for ( p ) into the inequality ( p leq frac{s^2}{4} ):[frac{s^3 - a^3}{3s} leq frac{s^2}{4}]Let me multiply both sides by ( 3s ). But wait, since ( s ) can be positive or negative, I need to be careful about the direction of the inequality. If ( s > 0 ), multiplying both sides by ( 3s ) preserves the inequality. If ( s < 0 ), multiplying by ( 3s ) (which is negative) would reverse the inequality. Hmm, this complicates things. Maybe I should consider cases for ( s > 0 ) and ( s < 0 ).But let me first see if there's another way. Alternatively, perhaps I can use the AM-GM inequality. Wait, but ( x ) and ( y ) are real numbers, not necessarily positive. So AM-GM might not directly apply here. Alternatively, maybe I can parametrize ( x ) and ( y ). Let me set ( x = u + v ) and ( y = u - v ). Then ( x + y = 2u ) and ( x - y = 2v ). Then, let's express ( x^3 + y^3 ) in terms of ( u ) and ( v ):[x^3 + y^3 = (u + v)^3 + (u - v)^3 = 2u^3 + 6uv^2 = 2u(u^2 + 3v^2) = a^3]So, ( 2u(u^2 + 3v^2) = a^3 ). Since ( u^2 + 3v^2 geq 0 ), and ( a^3 > 0 ), then ( u ) must be positive. So, ( u > 0 ). Therefore, ( x + y = 2u ), which is positive. Wait, does that mean ( x + y ) can only be positive? Hmm, but if ( x ) and ( y ) are real numbers, maybe ( x + y ) can also be negative? Let me check with an example. Suppose ( x = -y ), then ( x + y = 0 ), but ( x^3 + y^3 = 0 ), which is not equal to ( a^3 ). So, maybe ( x + y ) cannot be zero. But can it be negative? Let me see. Suppose ( x ) and ( y ) are both negative. Then ( x^3 + y^3 ) would be negative, but ( a^3 ) is positive, so that's not possible. If one is negative and the other is positive, then maybe ( x^3 + y^3 ) could be positive. For instance, if ( x ) is a large positive number and ( y ) is a small negative number, such that ( x^3 + y^3 = a^3 ). So, in that case, ( x + y ) could still be positive or negative depending on the values. Hmm, so maybe ( s = x + y ) can be both positive and negative. But in the parametrization above, I got ( u > 0 ), so ( x + y = 2u > 0 ). Maybe that parametrization restricts ( x + y ) to be positive. Is that correct?Wait, let's see. If ( x + y = 2u ), and we set ( x = u + v ), ( y = u - v ), then ( u = frac{x + y}{2} ). If ( x + y ) is negative, then ( u ) is negative. However, in the equation ( 2u(u^2 + 3v^2) = a^3 ), since ( a^3 > 0 ), then ( 2u(u^2 + 3v^2) ) must be positive. Therefore, ( u ) must be positive because ( u^2 + 3v^2 geq 0 ). So, ( u > 0 ), which implies ( x + y = 2u > 0 ). Therefore, this parametrization suggests that ( x + y ) must be positive. So, maybe ( x + y ) can't be negative? But wait, that contradicts my previous thought. Let me verify with an example. Suppose ( a = 1 ), so ( x^3 + y^3 = 1 ). Can we have ( x + y ) negative? Let's try. Let me pick ( x = -2 ) and solve for ( y ). Then ( (-2)^3 + y^3 = 1 implies -8 + y^3 = 1 implies y^3 = 9 implies y = sqrt[3]{9} approx 2.08 ). Then ( x + y = -2 + 2.08 approx 0.08 ), which is positive. Hmm. What if I pick ( x = -3 )? Then ( (-3)^3 + y^3 = 1 implies -27 + y^3 = 1 implies y^3 = 28 implies y ≈ 3.036 ). Then ( x + y ≈ -3 + 3.036 ≈ 0.036 ), still positive. Wait a second, maybe even if ( x ) is negative, ( y ) has to be sufficiently positive to compensate, making ( x + y ) positive? Let me check another case. Suppose ( x ) is negative and ( y ) is positive but smaller. Let's see: suppose ( x = -1 ), then ( (-1)^3 + y^3 = 1 implies -1 + y^3 = 1 implies y^3 = 2 implies y = sqrt[3]{2} ≈ 1.26 ). Then ( x + y ≈ -1 + 1.26 ≈ 0.26 ), still positive. Hmm. If ( x ) approaches negative infinity, then ( y^3 = a^3 - x^3 ), so as ( x to -infty ), ( y^3 to infty ), so ( y to infty ). Then ( x + y to -infty + infty ). Hmm, indeterminate form. Let me compute ( x + y ) as ( x to -infty ). Let ( x = -k ), then ( y^3 = a^3 + k^3 implies y = sqrt[3]{a^3 + k^3} approx k ) for large ( k ). So ( x + y ≈ -k + k = 0 ). But more accurately, ( y = sqrt[3]{k^3 + a^3} = k sqrt[3]{1 + frac{a^3}{k^3}} ≈ k left(1 + frac{a^3}{3k^3}right) = k + frac{a^3}{3k^2} ). Therefore, ( x + y ≈ -k + k + frac{a^3}{3k^2} = frac{a^3}{3k^2} ), which approaches 0 from the positive side as ( k to infty ). So even if ( x ) is very large negative, ( y ) is slightly larger positive, making ( x + y ) approach 0 from above. Therefore, ( x + y ) can approach 0 but can't be negative. Therefore, ( x + y ) must be positive. So the range of ( x + y ) is from some minimum value up to some maximum value, but all positive. So, going back, since ( x + y = s > 0 ), then in our parametrization, ( u > 0 ). Then, from the equation ( 2u(u^2 + 3v^2) = a^3 ). Since ( u^2 + 3v^2 geq u^2 ), then ( 2u(u^2 + 3v^2) geq 2u^3 ). Therefore:[2u^3 leq a^3 implies u^3 leq frac{a^3}{2} implies u leq frac{a}{sqrt[3]{2}}]Since ( u > 0 ), then ( 0 < u leq frac{a}{sqrt[3]{2}} ). But ( s = 2u ), so ( 0 < s leq 2 cdot frac{a}{sqrt[3]{2}} = 2^{1 - 1/3} a = 2^{2/3} a approx 1.5874a ). Hmm, but is this the maximum value of ( s )?Alternatively, if ( v = 0 ), then ( x = y = u ). Then ( x^3 + y^3 = 2u^3 = a^3 implies u = frac{a}{sqrt[3]{2}} ), so ( s = 2u = 2 cdot frac{a}{sqrt[3]{2}} = 2^{2/3}a ). So that's the maximum value when ( x = y ). What's the minimum value? If we take ( v ) to be as large as possible, how does that affect ( u )? Wait, but in the equation ( 2u(u^2 + 3v^2) = a^3 ), if ( v ) increases, then ( u ) must decrease to keep the product constant. So, as ( v to infty ), ( u to 0 ). But when ( u to 0 ), the left-hand side ( 2u(u^2 + 3v^2) ) tends to 0, which contradicts ( a^3 > 0 ). Wait, so maybe there's a lower bound on ( u ). Wait, no. If ( u ) approaches zero, then ( v^2 ) must approach infinity such that ( 2u cdot 3v^2 approx a^3 ). So ( 6uv^2 approx a^3 implies v approx sqrt{frac{a^3}{6u}} ). Therefore, as ( u to 0^+ ), ( v to infty ), but ( s = 2u to 0 ). So ( x + y ) can approach 0 from the positive side. Therefore, the minimum value is 0, but not including 0. However, we need to check if ( x + y ) can actually be as small as approaching 0. Wait, let's see. If ( x + y = s ) is approaching 0, then from the equation ( s(s^2 - 3p) = a^3 ), as ( s to 0^+ ), ( s^2 - 3p approx -3p ), so ( -3p approx a^3 / s ). Therefore, ( p approx -a^3 / (3s) ). But since ( p = xy ), and ( s = x + y ), if ( s ) is approaching 0, then ( x ) is approaching ( -y ), but ( p = xy ) is approaching ( -y^2 ). But according to ( p approx -a^3/(3s) ), as ( s to 0 ), ( p to -infty ). But ( p = xy = (-y + s)y = -y^2 + sy ). Wait, if ( s to 0 ), then ( xy approx -y^2 ), so ( -y^2 approx -a^3/(3s) implies y^2 approx a^3/(3s) implies y approx pm sqrt{a^3/(3s)} ). But ( s = x + y approx -y + y = 0 ). Wait, this seems conflicting. Let me check with specific values. Suppose ( s = 0.1 ), then ( p approx -a^3 / (3*0.1) = -10a^3/3 ). But ( x + y = 0.1 ), ( xy = -10a^3/3 ). Then, the quadratic equation is ( t^2 - 0.1t - 10a^3/3 = 0 ). The discriminant is ( (0.1)^2 + 40a^3/3 ), which is positive, so real roots exist. Therefore, such ( x ) and ( y ) exist. Therefore, as ( s to 0^+ ), we can have solutions. Therefore, ( s ) can approach 0 from above. However, when ( s = 0 ), there's no solution because ( x^3 + y^3 = 0 neq a^3 ). Therefore, the range of ( s = x + y ) is ( 0 < s leq 2^{2/3}a ).But wait, the problem states that ( x ) and ( y ) are real numbers. So, we need to confirm that the maximum value of ( s ) is indeed ( 2^{2/3}a ) and the minimum approaches 0. So the range is ( (0, 2^{2/3}a] ). But the problem asks for the range of ( x + y ). However, in the problem statement, it's possible that ( x ) and ( y ) can be such that ( x + y ) can actually reach down to a negative value? Wait, but earlier analysis suggested that ( x + y ) must be positive. Let me verify again. If ( x + y ) is negative, then ( x ) and ( y ) are both negative or one is negative and the other is positive. But if both are negative, then ( x^3 + y^3 ) is negative, which can't equal ( a^3 > 0 ). If one is negative and the other is positive, say ( x > 0 ), ( y < 0 ), then ( x^3 + y^3 = a^3 ). Let's see if ( x + y ) can be negative here. For example, take ( x = 1 ), ( y = -2 ). Then ( x^3 + y^3 = 1 + (-8) = -7 neq a^3 ). To have ( x^3 + y^3 = a^3 ), suppose ( x ) is positive and ( y ) is negative. Let's set ( y = -k ), ( k > 0 ). Then, ( x^3 - k^3 = a^3 implies x^3 = a^3 + k^3 implies x = sqrt[3]{a^3 + k^3} ). Then, ( x + y = sqrt[3]{a^3 + k^3} - k ). Let's see if this can be negative. Let's compute when ( sqrt[3]{a^3 + k^3} < k ). Then, ( a^3 + k^3 < k^3 implies a^3 < 0 ), which contradicts ( a > 0 ). Therefore, ( x + y = sqrt[3]{a^3 + k^3} - k ) is always positive. Therefore, even if one variable is negative, ( x + y ) must remain positive. Therefore, ( x + y ) is always positive, and the range is ( (0, 2^{2/3}a] ). But let me check this with another approach. Let's consider using the method of Lagrange multipliers. Let ( f(x, y) = x + y ) and the constraint ( g(x, y) = x^3 + y^3 - a^3 = 0 ). Then, setting up the Lagrangian:[mathcal{L}(x, y, lambda) = x + y - lambda(x^3 + y^3 - a^3)]Taking partial derivatives:[frac{partial mathcal{L}}{partial x} = 1 - 3lambda x^2 = 0 implies 1 = 3lambda x^2][frac{partial mathcal{L}}{partial y} = 1 - 3lambda y^2 = 0 implies 1 = 3lambda y^2][frac{partial mathcal{L}}{partial lambda} = -(x^3 + y^3 - a^3) = 0 implies x^3 + y^3 = a^3]From the first two equations, we have ( 3lambda x^2 = 1 ) and ( 3lambda y^2 = 1 ). Therefore, ( x^2 = y^2 implies y = pm x ). Case 1: ( y = x ). Then, substituting into the constraint: ( 2x^3 = a^3 implies x = left( frac{a^3}{2} right)^{1/3} = frac{a}{2^{1/3}} ). Therefore, ( x + y = 2x = 2 cdot frac{a}{2^{1/3}} = 2^{2/3}a ). This is the maximum value.Case 2: ( y = -x ). Then, substituting into the constraint: ( x^3 + (-x)^3 = 0 = a^3 ), which implies ( a = 0 ), contradicting ( a > 0 ). Therefore, this case is invalid. Therefore, the only critical point is when ( y = x ), giving the maximum ( x + y = 2^{2/3}a ).Now, for the minimum value. Since ( x + y ) can approach 0 (as shown in previous examples), but cannot be negative, the minimum is 0. However, 0 is not attainable because ( x^3 + y^3 = a^3 > 0 ). Therefore, the range of ( x + y ) is ( (0, 2^{2/3}a] ).But the problem is from a competition, so they might expect an answer in terms of radicals or fractions. Let me confirm the maximum value. ( 2^{2/3} ) is the same as ( sqrt[3]{4} ). So ( 2^{2/3}a = sqrt[3]{4}a ). Therefore, the range is ( 0 < x + y leq sqrt[3]{4}a ).But maybe there's another way to write it. Alternatively, using the earlier substitution. Let's recall that in the equation ( s^3 - 3sp = a^3 ), and ( p leq s^2/4 ). So substituting ( p leq s^2/4 ) into the equation:[s^3 - 3s cdot frac{s^2}{4} geq a^3 implies s^3 - frac{3s^3}{4} geq a^3 implies frac{s^3}{4} geq a^3 implies s^3 geq 4a^3 implies s geq sqrt[3]{4}a]Wait, but this is the opposite of our previous conclusion. Wait, this suggests that ( s geq sqrt[3]{4}a ), but we know from the Lagrange multiplier method that the maximum is ( sqrt[3]{4}a ). There must be an error here. Let me check the substitution. Wait, originally, we had ( p = frac{s^3 - a^3}{3s} leq frac{s^2}{4} ). Then:[frac{s^3 - a^3}{3s} leq frac{s^2}{4}]Multiply both sides by ( 3s ). Since ( s > 0 ), the inequality direction remains:[s^3 - a^3 leq frac{3s^3}{4}][s^3 - frac{3s^3}{4} leq a^3][frac{s^3}{4} leq a^3][s^3 leq 4a^3][s leq sqrt[3]{4}a]Ah, okay, so this gives the upper bound ( s leq sqrt[3]{4}a ), which matches our previous conclusion. Then, the lower bound? For the lower bound, we can consider the minimal value of ( s ). Since ( s ) can approach 0, as shown earlier, the lower bound is 0, but it's not included. Therefore, the range is ( 0 < s leq sqrt[3]{4}a ).But let's see if there's a way to express this more rigorously. If we consider ( s > 0 ), then we can write the range as ( (0, sqrt[3]{4}a] ).But in the problem statement, they might expect the answer written in interval notation. So, the range of ( x + y ) is ( (0, sqrt[3]{4}a] ). However, since it's a competition problem, perhaps they expect the answer in terms of ( 2^{2/3}a ), which is the same as ( sqrt[3]{4}a ). Both are correct. Alternatively, let's verify with another method. Suppose we set ( t = x + y ). We want to express the equation ( x^3 + y^3 = a^3 ) in terms of ( t ). As we did before, ( x^3 + y^3 = t(t^2 - 3xy) = a^3 ). So, ( t^3 - 3t xy = a^3 ). Also, we know that ( xy leq frac{t^2}{4} ) from AM-GM (since ( xy leq left( frac{x + y}{2} right)^2 )). Therefore:[t^3 - 3t cdot frac{t^2}{4} geq a^3 implies t^3 - frac{3t^3}{4} geq a^3 implies frac{t^3}{4} geq a^3 implies t geq sqrt[3]{4}a]Wait, but this is again the upper bound. Wait, no, this gives ( t geq sqrt[3]{4}a ), which contradicts our previous conclusion. Wait, this is confusing. Let me check the direction of the inequality. If we have ( xy leq frac{t^2}{4} ), then ( -3t xy geq -3t cdot frac{t^2}{4} ). Therefore, substituting into ( t^3 - 3t xy geq t^3 - frac{3t^3}{4} = frac{t^3}{4} ). So, ( frac{t^3}{4} leq a^3 implies t leq sqrt[3]{4}a ). Wait, no, wait:Wait, ( t^3 - 3t xy = a^3 ). If ( xy leq frac{t^2}{4} ), then ( -3t xy geq -3t cdot frac{t^2}{4} ). Therefore, ( t^3 - 3t xy geq t^3 - frac{3t^3}{4} = frac{t^3}{4} ). So, ( frac{t^3}{4} leq a^3 implies t leq sqrt[3]{4}a ). Therefore, this gives the upper bound. To find the lower bound, note that ( xy ) can be made arbitrarily negative. Because if ( x ) approaches positive infinity and ( y ) approaches negative infinity such that ( x^3 + y^3 = a^3 ), then ( xy ) can be made very negative. However, in reality, if ( x ) is large positive, ( y ) must be such that ( y^3 = a^3 - x^3 ). So, as ( x to infty ), ( y^3 = a^3 - x^3 to -infty implies y to -infty ). Then, ( xy approx x(-|y|) approx -x|y| approx -x^{4/3} to -infty ). So, ( xy ) can be made arbitrarily negative. Then, in the equation ( t^3 - 3t xy = a^3 ), if ( xy to -infty ), then ( -3t xy to infty ). So, ( t^3 + infty = a^3 ). This is not possible unless ( t ) is approaching some limit. Wait, this is contradictory. Let me think again. If ( x to infty ), ( y to -infty ), but such that ( x^3 + y^3 = a^3 ). Let me parametrize ( y = -k ), so ( x^3 - k^3 = a^3 implies x^3 = k^3 + a^3 implies x = sqrt[3]{k^3 + a^3} ). Then, ( t = x + y = sqrt[3]{k^3 + a^3} - k ). Let's analyze the behavior as ( k to infty ). Using the approximation for large ( k ):[sqrt[3]{k^3 + a^3} = k sqrt[3]{1 + frac{a^3}{k^3}} approx k left(1 + frac{a^3}{3k^3}right) = k + frac{a^3}{3k^2}]Therefore,[t = x + y approx k + frac{a^3}{3k^2} - k = frac{a^3}{3k^2} to 0 quad text{as} quad k to infty]So, as ( k ) increases, ( t ) approaches 0 from the positive side. Therefore, ( t ) can get arbitrarily close to 0 but cannot be negative or zero. Hence, the minimal value is 0 (exclusive), and the maximum value is ( sqrt[3]{4}a ). Therefore, the range of ( x + y ) is ( (0, sqrt[3]{4}a] ). Translating ( sqrt[3]{4} ) into another form, since ( 4 = 2^2 ), then ( sqrt[3]{4} = 2^{2/3} ). So, the range is ( (0, 2^{2/3}a] ). But in the competition problem, they might prefer writing it as ( (0, 2^{2/3}a] ) or using a rational exponent. Also, depending on the answer format, they might need to write it in interval notation. However, in Chinese competitions, sometimes the answer is presented using boxed notation with inequalities. Wait, but maybe there's a different approach. Let me think again. We can consider the equation ( x^3 + y^3 = a^3 ). Let me set ( t = x + y ). Then, ( x^3 + y^3 = (x + y)^3 - 3xy(x + y) = t^3 - 3txy = a^3 ). So,[t^3 - 3txy = a^3 implies xy = frac{t^3 - a^3}{3t}]But ( x ) and ( y ) are real numbers, so the discriminant of the quadratic equation ( z^2 - tz + xy = 0 ) must be non-negative:[t^2 - 4xy geq 0]Substituting ( xy = frac{t^3 - a^3}{3t} ):[t^2 - 4 cdot frac{t^3 - a^3}{3t} geq 0][t^2 - frac{4(t^3 - a^3)}{3t} geq 0]Multiply both sides by ( 3t ) (since ( t > 0 )):[3t^3 - 4(t^3 - a^3) geq 0][3t^3 - 4t^3 + 4a^3 geq 0][-t^3 + 4a^3 geq 0][t^3 leq 4a^3][t leq sqrt[3]{4}a]So, combining with the fact that ( t > 0 ), we have ( 0 < t leq sqrt[3]{4}a ). Therefore, the range of ( x + y ) is ( (0, sqrt[3]{4}a] ).Thus, the final answer should be ( boxed{(0, 2^{2/3}a]} ) or ( boxed{(0, sqrt[3]{4}a]} ). Since ( 2^{2/3} = sqrt[3]{4} ), both are equivalent. In the context of a competition, perhaps they would prefer ( 2^{2/3}a ), so I'll present it as ( boxed{(0, 2^{2/3}a]} ).Final AnswerThe range of values for ( x + y ) is boxed{(0, 2^{frac{2}{3}}a]}.</think>Given the equation ( x^3 + y^3 = a^3 ) where ( a > 0 ), we need to find the range of values for ( x + y ).First, we factorize the given equation using the identity for the sum of cubes:[x^3 + y^3 = (x + y)(x^2 - xy + y^2) = a^3]Let ( s = x + y ) and ( p = xy ). Then, we can express ( x^2 - xy + y^2 ) in terms of ( s ) and ( p ):[x^2 - xy + y^2 = s^2 - 3p]Substituting this into the equation, we get:[s(s^2 - 3p) = a^3]Solving for ( p ), we find:[p = frac{s^3 - a^3}{3s}]Next, we use the discriminant condition for the quadratic equation ( t^2 - st + p = 0 ) to ensure real roots:[s^2 - 4p geq 0]Substituting ( p ) into this inequality:[s^2 - 4 left( frac{s^3 - a^3}{3s} right) geq 0]Multiplying through by ( 3s ) (since ( s > 0 )):[3s^3 - 4(s^3 - a^3) geq 0]Simplifying this inequality:[3s^3 - 4s^3 + 4a^3 geq 0 implies -s^3 + 4a^3 geq 0 implies s^3 leq 4a^3 implies s leq sqrt[3]{4}a]We also need to consider the lower bound of ( s ). By analyzing the behavior of ( x ) and ( y ) as they approach infinity and negative infinity, we find that ( s = x + y ) can approach 0 from the positive side but cannot be zero or negative. Therefore, the lower bound is 0 (exclusive).Thus, the range of values for ( x + y ) is:[boxed{(0, 2^{frac{2}{3}}a]}],<think>Okay, so I have this problem here: given real numbers x and y that satisfy the equation x³ + y³ = a³ where a is a positive real number. I need to find the range of possible values for x + y. Hmm, let's see. I remember there are some formulas related to cubes that might be helpful here. First, maybe I can recall the identity for the sum of cubes. The equation x³ + y³ can be factored as (x + y)(x² - xy + y²). So, substituting that into the given equation: (x + y)(x² - xy + y²) = a³. Let me denote s = x + y and p = xy. Then, the equation becomes s(s² - 3p) = a³. Because x² - xy + y² is equal to (x + y)² - 3xy, which is s² - 3p. So, now we have s(s² - 3p) = a³. But I need to relate this to the possible values of s. The problem is that there are two variables here, s and p, so maybe I need another equation or a way to eliminate p. Alternatively, perhaps I can use inequalities. Since x and y are real numbers, there might be some constraints on s and p. For example, the discriminant of the quadratic equation t² - st + p = 0 must be non-negative, because x and y are real roots of this equation. The discriminant D = s² - 4p ≥ 0. So, p ≤ s²/4. So, combining this with the previous equation. We have s(s² - 3p) = a³. Since p ≤ s²/4, substituting the maximum value of p into the equation would give me a minimum or maximum value for s. Let's try substituting p = s²/4 into the equation. Then, s(s² - 3*(s²/4)) = s(s² - (3/4)s²) = s*( (1/4)s² ) = (1/4)s³. So, this equals a³. Therefore, (1/4)s³ = a³ implies s³ = 4a³, so s = cube root of 4 * a. But wait, cube root of 4 is about 1.587, which is greater than 1. But is this a maximum or a minimum?Wait, if p ≤ s²/4, then when p is at its maximum (s²/4), then the term (s² - 3p) is minimized. Therefore, s*(s² - 3p) is minimized when p is maximized. So, (1/4)s³ = a³ would give the minimum value of s? Hmm, maybe not. Let's think.Wait, if p is as large as possible, then (s² - 3p) is as small as possible. Therefore, s*(s² - 3p) would be as small as possible, but we have s*(s² - 3p) = a³. So, if (s² - 3p) is smaller, then s must be larger to compensate. So, if p is maximized, then s must be maximized? Hmm, perhaps this is the case. Let me check with an example.Suppose a is 1. Then, the equation becomes x³ + y³ = 1. Suppose x = y. Then, 2x³ = 1, so x = (1/2)^(1/3), so x + y = 2*(1/2)^(1/3) = 2^(1 - 1/3) = 2^(2/3) ≈ 1.587, which is the cube root of 4. So, when x = y, we get s = cube root of 4 * a. But is this the maximum or minimum?Alternatively, suppose y = 0. Then x³ = a³, so x = a, and x + y = a + 0 = a. So, s = a. But this is less than 2^(2/3)*a. So, when one variable is 0, the sum is a, which is smaller than when both variables are equal. So, maybe the maximum value of s is when x = y, and the minimum is when one of them is 0? But is that the case?Wait, but when you set y = 0, you get x = a, so s = a. But can s be smaller than a? Let's see. Suppose x and y are both negative. For example, let x be a large negative number and y adjust accordingly. Wait, but x³ + y³ = a³. If x approaches negative infinity, then y³ would have to approach positive infinity to compensate. However, x + y would be negative infinity plus positive infinity, which is indeterminate. But maybe there's a lower bound?Wait, let's think about this more carefully. Let's consider that x and y can be any real numbers such that x³ + y³ = a³. So, if x is very large negative, then y³ must be a³ - x³. If x is negative, then x³ is negative, so y³ = a³ - x³ would be a³ + |x³|, which is larger than a³. Therefore, y would be the cube root of a number larger than a³, so y is larger than a. But if x is very large negative, then y is very large positive, so x + y would be y + x = (large positive) + (large negative). Depending on the rates, which one dominates? Since x³ + y³ = a³, so as x approaches negative infinity, y³ ≈ -x³, so y ≈ -x. Therefore, x + y ≈ x - x = 0. Wait, but this is only approximate. Let me check with specific numbers.Let’s take a = 1. Suppose x = -n, where n is a large positive number. Then, y³ = 1 - (-n)³ = 1 + n³. So, y = (1 + n³)^(1/3). Then, x + y = -n + (1 + n³)^(1/3). Let's approximate (1 + n³)^(1/3). For large n, 1 is negligible, so it's approximately n. Therefore, x + y ≈ -n + n = 0. So, as n increases, x + y approaches 0 from above? Because (1 + n³)^(1/3) is slightly larger than n. Let's compute (n³ + 1)^(1/3) = n*(1 + 1/n³)^(1/3) ≈ n*(1 + 1/(3n³)) by the binomial approximation. Therefore, x + y ≈ -n + n + 1/(3n²) = 1/(3n²). So, as n approaches infinity, x + y approaches 0 from the positive side. So, in this case, x + y can approach 0. But can it be negative?Wait, if x and y are both negative, then x³ + y³ would be negative + negative = negative, which cannot be equal to a³ (which is positive). So, at least one of x or y has to be positive. If one is positive and the other is negative, their sum could be positive or negative. But if we can have x + y approaching 0 from the positive side, but can it actually reach 0?Suppose x + y = 0. Then y = -x. Then x³ + y³ = x³ + (-x)³ = x³ - x³ = 0. But we need x³ + y³ = a³ > 0. Therefore, x + y cannot be 0. So, the sum x + y can approach 0 but cannot be 0. So, is there a lower bound?Wait, let's suppose x + y = k. Then, we can use the equation s(s² - 3p) = a³. If k is positive, then maybe we can find p in terms of k. But we also have the discriminant condition: s² - 4p ≥ 0, so p ≤ s²/4. Therefore, in the equation s(s² - 3p) = a³, since p ≤ s²/4, then s² - 3p ≥ s² - 3*(s²/4) = s²/4. Therefore, s(s² - 3p) ≥ s*(s²/4) = s³/4. So, a³ = s(s² - 3p) ≥ s³/4. Therefore, a³ ≥ s³/4 ⇒ s³ ≤ 4a³ ⇒ s ≤ cube root(4) * a. Which gives us the upper bound for s, which matches the case when x = y. So, s ≤ cube root(4) * a.For the lower bound, we need to see how small s can be. If s approaches 0, then the left-hand side s(s² - 3p) approaches 0, but a³ is a positive constant, so s cannot be too small. Wait, but earlier when we considered x approaching negative infinity and y approaching positive infinity, x + y approached 0. But in that case, the product s(s² - 3p) must equal a³. If s approaches 0, then s(s² - 3p) approaches 0, which cannot equal a³. Therefore, there must be a contradiction here. So perhaps my previous thought experiment was wrong.Wait, let's re-examine. If x approaches negative infinity, then y³ = a³ - x³. So, as x approaches negative infinity, y³ ≈ -x³, so y ≈ -x. Therefore, x + y ≈ x - x = 0. But x³ + y³ ≈ x³ - x³ = 0, which is not equal to a³. Therefore, in reality, even though y is approximately -x, there is a small correction term. Let me compute this more precisely.Let x = -n, then y³ = a³ - (-n)³ = a³ + n³. So, y = (n³ + a³)^(1/3). Let's expand this for large n:y = n*(1 + (a³)/n³)^(1/3) ≈ n*(1 + (a³)/(3n³)) = n + a³/(3n²). Therefore, x + y ≈ -n + n + a³/(3n²) = a³/(3n²). So as n approaches infinity, x + y approaches 0. Therefore, the sum x + y can get arbitrarily close to 0 from the positive side, but cannot be negative. Because if x + y were negative, then since y must be positive (as x is negative and y³ = a³ - x³ > a³), so y is positive, but x is negative. Then, x + y could be positive or negative depending on the magnitudes. Wait, but earlier example shows that x + y approaches 0 from the positive side. Let me check with actual numbers.Take a = 1, x = -2. Then y³ = 1 - (-8) = 9, so y = ∛9 ≈ 2.08. Then x + y ≈ -2 + 2.08 ≈ 0.08. If x = -3, y³ = 1 - (-27) = 28, so y ≈ 3.036, so x + y ≈ -3 + 3.036 ≈ 0.036. If x = -10, y ≈ ∛(1001) ≈ 10.00333, so x + y ≈ -10 + 10.00333 ≈ 0.00333. So, as x becomes more negative, x + y approaches 0 from the positive side. Therefore, the sum x + y can be made arbitrarily close to 0, but not negative. So, the lower bound is approaching 0, but not including 0. However, the problem states that x and y are real numbers. So, technically, the range of x + y is (0, ∛4 * a]. Wait, but maybe even when x and y are both positive, the sum x + y can be as small as a, when one of them is 0. Wait, but in the case when one is 0, like y = 0, then x = a, so x + y = a + 0 = a. So, if we set y = 0, x = a, sum is a. So, is a the minimal value?Wait, but earlier examples show that we can get x + y smaller than a. For example, when x = -2 and y ≈ 2.08, sum ≈ 0.08, which is less than a=1. So, actually, the minimal value is approaching 0, but the maximum is ∛4 *a. So, the range is (0, ∛4 *a]. But wait, when we take y =0, x =a, sum is a. But in that case, can we get sums between a and ∛4 a?Wait, let's check. If x and y are both positive, then by AM ≥ GM, x³ + y³ ≥ 2 (x³ y³)^(1/2) = 2 (xy)^(3/2). But maybe that's not helpful. Alternatively, if we set y = k - x, then maybe we can analyze it as a function. Let me try.Let s = x + y. Then, y = s - x. Substitute into x³ + y³ = a³:x³ + (s - x)³ = a³Expanding (s - x)³: s³ - 3s²x + 3s x² - x³.Therefore, x³ + s³ - 3s²x + 3s x² - x³ = a³.Simplify: s³ - 3s²x + 3s x² = a³.So, 3s x² - 3s²x + s³ - a³ = 0.Divide both sides by 3s (assuming s ≠ 0):x² - s x + (s³ - a³)/(3s) = 0.This is a quadratic in x. For real solutions to exist, the discriminant must be non-negative.Discriminant D = [ -s ]² - 4 *1* [(s³ - a³)/(3s)] = s² - (4/3)(s³ - a³)/s = s² - (4/3)(s² - a³/s).Simplify:s² - (4/3)s² + (4/3)(a³)/s = (-1/3)s² + (4 a³)/(3 s).Multiply both sides by 3s to eliminate denominators:3s * D = (-1/3)s² * 3s + (4 a³)/(3 s) * 3s = -s³ + 4 a³.But since D ≥ 0,(-1/3)s² + (4 a³)/(3 s) ≥ 0Multiply both sides by 3s (since s > 0, as we saw earlier that s approaches 0 from the positive side):-s³ + 4a³ ≥ 0 ⇒ 4a³ ≥ s³ ⇒ s³ ≤ 4a³ ⇒ s ≤ ∛(4) a.Which confirms the upper bound. For the lower bound, since s can approach 0, but s must be positive. Wait, but in the discriminant approach, we assumed s ≠ 0, but when s approaches 0, what happens?If s approaches 0 from the positive side, then the equation x³ + y³ = a³ requires that y ≈ (a³ - x³)^(1/3). If s = x + y is approaching 0, then y ≈ -x, so x ≈ -y. But then x³ + y³ ≈ 0, which contradicts x³ + y³ = a³. So, there must be a minimal positive value of s where such real solutions exist. Wait, but our previous analysis suggested that as x approaches negative infinity, s approaches 0. So, maybe there is no lower bound, and s can approach 0, but cannot be negative.Wait, but according to the discriminant approach, the discriminant must be non-negative:-s³ + 4a³ ≥ 0 ⇒ s³ ≤ 4a³ ⇒ s ≤ ∛4 a. But this is the upper bound. The discriminant condition doesn't give a lower bound. So, perhaps the lower bound is indeed 0, but s must be positive? Because if s were negative, then x + y = s < 0. But x and y are real numbers such that x³ + y³ = a³. If s is negative, then one of x or y is more negative, but the other would have to compensate. Let's test this.Suppose s = x + y = -k, where k > 0. Then, we can write the equation s(s² - 3p) = a³. Since s is negative, and a³ is positive, then (s² - 3p) must be negative. But s² is positive, so (s² - 3p) < 0 ⇒ 3p > s² ⇒ p > s²/3. However, from the discriminant condition, we have p ≤ s²/4. But if s is negative, then s² is positive, and p ≤ s²/4. But 3p > s² would require p > s²/3. But s²/4 < s²/3 (since 1/4 < 1/3), so p ≤ s²/4 < s²/3. Therefore, 3p ≤ 3*(s²/4) = (3/4)s² < s². Therefore, (s² - 3p) ≥ s² - (3/4)s² = (1/4)s² > 0. But this contradicts the requirement that (s² - 3p) < 0. Therefore, there are no real solutions when s is negative. Therefore, s must be positive. Therefore, the range of s is (0, ∛4 a]. But wait, when s approaches 0, does the equation x³ + y³ = a³ hold? Let's check. If s approaches 0, then as x approaches -infty and y approaches +infty such that their sum approaches 0. However, in reality, as x becomes very negative, y must be slightly larger than -x to compensate for the a³ term. Therefore, as x approaches -infty, y approaches +infty such that y ≈ -x + δ, where δ is a small positive number. Then, x + y ≈ δ, which approaches 0. But in this case, the equation x³ + y³ = a³ becomes (-n)³ + (n + δ)³ ≈ a³. Let's compute (n + δ)^3:(n + δ)^3 = n³ + 3n² δ + 3n δ² + δ³.So, (-n)^3 + (n + δ)^3 = -n³ + n³ + 3n² δ + 3n δ² + δ³ = 3n² δ + 3n δ² + δ³.Set this equal to a³:3n² δ + 3n δ² + δ³ = a³.For large n, the dominant term is 3n² δ. So, approximately, 3n² δ ≈ a³ ⇒ δ ≈ a³/(3n²). Therefore, s = x + y ≈ -n + (n + δ) = δ ≈ a³/(3n²). As n approaches infinity, δ approaches 0, so s approaches 0. Therefore, s can get arbitrarily close to 0, but remains positive. Therefore, the lower bound is 0 (exclusive) and the upper bound is ∛4 a (inclusive). But wait, when one variable is 0, like y = 0, then x = a, so s = a + 0 = a, which is between 0 and ∛4 a. So, the range is all real numbers from just above 0 up to ∛4 a. So, the answer should be that x + y can take any value in (0, ∛4 a]. But in the original problem statement, it's from a competition, so maybe they expect a different form. Let me check with calculus.Alternatively, we can parametrize x and y. Let's set x = t, then y = (a³ - t³)^(1/3). Then, x + y = t + (a³ - t³)^(1/3). Let’s define this as a function f(t) = t + (a³ - t³)^(1/3). We need to find the range of f(t) for real t.To find the maximum and minimum of f(t), take the derivative and set it to 0. Let's compute f'(t):f'(t) = 1 + (1/3)(a³ - t³)^(-2/3)(-3t²) = 1 - t²/(a³ - t³)^(2/3).Set f'(t) = 0:1 - t²/(a³ - t³)^(2/3) = 0 ⇒ t² = (a³ - t³)^(2/3).Raise both sides to the power of 3/2:t³ = (a³ - t³).Therefore, t³ = a³ - t³ ⇒ 2t³ = a³ ⇒ t³ = a³/2 ⇒ t = a/(2^{1/3}).Therefore, the critical point is at t = a/(2^{1/3}). Then, compute f(t) at this point:f(a/(2^{1/3})) = a/(2^{1/3}) + (a³ - (a/(2^{1/3}))³ )^(1/3)Compute (a/(2^{1/3}))³ = a³ / 2. Therefore,(a³ - a³/2)^(1/3) = (a³/2)^(1/3) = a/(2^{1/3}).Thus, f(a/(2^{1/3})) = a/(2^{1/3}) + a/(2^{1/3}) = 2a/(2^{1/3}) = a * 2^{1 - 1/3} = a * 2^{2/3} = a * ∛4. So, that's the maximum value.For the minimum, as t approaches -infty, f(t) = t + (a³ - t³)^(1/3). Let's see, as t approaches -infty, (a³ - t³) ≈ -t³, so (a³ - t³)^(1/3) ≈ -t. Therefore, f(t) ≈ t - t = 0. But since t is approaching -infty, (a³ - t³)^(1/3) is slightly larger than -t. Let's write t = -n where n approaches +infty. Then,f(-n) = -n + (a³ + n³)^(1/3) ≈ -n + n*(1 + a³/(n³))^(1/3) ≈ -n + n + (a³)/(3n²) ≈ a³/(3n²). Therefore, as n approaches infinity, f(-n) approaches 0 from the positive side. Hence, the infimum is 0, but it's never actually reached. Therefore, the range of x + y is (0, ∛4 a]. So, the possible values of x + y are all real numbers greater than 0 and up to ∛4 times a. But let's check when t = a. Then, f(a) = a + (a³ - a³)^(1/3) = a + 0 = a. So, a is attainable. Similarly, when t =0, f(0) =0 + (a³)^(1/3) = a. So, a is attainable. Wait, but earlier when we set x = a, y =0, sum is a. But according to the calculus, the maximum is ∛4 a, which is approximately 1.587 a. So, there are values higher than a. For example, when x = y, we saw that x + y = ∛4 a. So, that's the maximum. The minimum is approaching 0. Therefore, the range is (0, ∛4 a]. However, when t is between 0 and a, the sum x + y ranges between a and ∛4 a? Wait, no. Wait, if t is between -infty and a, since if t > a, then y = (a³ - t³)^(1/3) becomes imaginary. Wait, no, t can be any real number. Wait, but if t > a, then a³ - t³ is negative, so (a³ - t³)^(1/3) is the cube root of a negative number, which is negative. So, for t > a, y would be negative. So, x = t > a, y negative. Then x + y could be greater or less than a. Let's take t = 2a. Then, y = (a³ - (2a)^3)^(1/3) = (a³ - 8a³)^(1/3) = (-7a³)^(1/3) = -∛7 a. Then x + y = 2a - ∛7 a ≈ 2a - 1.913a ≈ 0.087a. Which is positive but less than a. So, in this case, t =2a gives x + y ≈0.087a, which is still positive, but close to 0. So, even when t >a, the sum x + y is positive but approaching 0 as t approaches infinity. So, actually, the maximum occurs at t = a/∛2, and the minimum is still approaching 0.Therefore, the range is (0, ∛4 a]. But in the case when t = a, the sum is a. But when t = a/∛2, sum is ∛4 a. So, between t = a/∛2 and t approaches infinity or negative infinity, the sum decreases from ∛4 a to 0. Therefore, the possible values of x + y are all positive real numbers less than or equal to ∛4 a. Therefore, the answer should be that the range of x + y is (0, ∛4 a]. However, in the problem statement, it's from a competition, and usually, competitions expect simplified forms. Since ∛4 can be written as 2^{2/3}, but perhaps they want it in the cube root form. Also, note that in the problem, it's specified that a >0, so the range is from greater than 0 up to ∛4 a. Wait, but in the case when x and y are both positive, then x + y would be at least a. Wait, no. Wait, if x and y are positive, then by Power Mean inequality, (x³ + y³)/2 ≥ ((x + y)/2)^3. Therefore, a³ /2 ≥ ((x + y)/2)^3 ⇒ (x + y)/2 ≤ (a³ /2)^{1/3} = a / 2^{1/3} ⇒ x + y ≤ 2a / 2^{1/3} = a * 2^{2/3} = ∛4 a. Which is the same upper bound. But for the lower bound when x and y are both positive, then by AM ≥ GM: (x + y)/2 ≥ (xy)^{1/2}. But not sure if that helps. However, if x and y are both positive, then x + y is at least ... Wait, if x and y are positive, then x³ + y³ =a³. What's the minimal value of x + y? For positive x and y, we can use the inequality that (x + y)^3 ≤ 4(x³ + y³). Wait, let me check:We have for positive numbers, by Holder's inequality: (x + y)^3 ≤ (1 + 1)^2 (x³ + y³) = 4(x³ + y³). So, (x + y)^3 ≤ 4a³ ⇒ x + y ≤ ∛4 a, which is the upper bound again. But for the lower bound with positive x and y, when does x + y become minimal? If one variable is 0, then x + y =a. But if both variables are positive, can x + y be less than a? Suppose x approaches 0, then y approaches a. Then, x + y approaches a. Similarly, if y approaches 0, x approaches a, sum approaches a. Therefore, if x and y are both positive, the minimal sum is a, achieved when one of them is 0. But earlier analysis shows that if one variable is negative, the sum can be less than a, approaching 0. So, in the problem, since x and y are real numbers (not necessarily positive), the minimal value is approaching 0, but not including 0. Therefore, combining all cases, the range of x + y is all positive real numbers from just above 0 up to ∛4 a. Therefore, the answer is that x + y can take any value in the interval (0, ∛4 a]. But in the competition problem, they might expect the answer written as 0 < x + y ≤ ∛4 a, or using boxed notation. Since the problem says "find the range of values for x + y," and in Chinese competitions, often they write the range using inequalities. Therefore, the range is 0 < x + y ≤ a∛4. Translating this into LaTeX, it's (0, asqrt[3]{4}]. So, boxed answer would be boxed{(0, asqrt[3]{4}]}.Wait, but in some textbooks, cube roots of 4 are written as 2^{2/3}, so maybe they prefer that. But in the problem statement, since a is given, it's better to write in terms of a and the cube root. So, final answer is boxed{(0, asqrt[3]{4}]}Final AnswerThe range of values for ( x + y ) is boxed{(0, asqrt[3]{4}]}.</think>Given real numbers ( x ) and ( y ) that satisfy the equation ( x^3 + y^3 = a^3 ) where ( a > 0 ), we need to find the range of possible values for ( x + y ).First, we use the identity for the sum of cubes:[x^3 + y^3 = (x + y)(x^2 - xy + y^2)]Let ( s = x + y ) and ( p = xy ). Substituting these into the equation, we get:[s(s^2 - 3p) = a^3]We also know that for real numbers ( x ) and ( y ), the discriminant of the quadratic equation ( t^2 - st + p = 0 ) must be non-negative:[s^2 - 4p geq 0 implies p leq frac{s^2}{4}]Substituting the maximum value of ( p ) into the equation ( s(s^2 - 3p) = a^3 ), we get:[s left( s^2 - 3 cdot frac{s^2}{4} right) = a^3 implies s left( frac{s^2}{4} right) = a^3 implies frac{s^3}{4} = a^3 implies s^3 = 4a^3 implies s = sqrt[3]{4}a]This gives the upper bound for ( s ).For the lower bound, consider the case where ( x ) approaches negative infinity and ( y ) approaches positive infinity such that ( x + y ) approaches 0. However, since ( x^3 + y^3 = a^3 ), the sum ( x + y ) must remain positive. By analyzing the behavior of the function ( f(t) = t + sqrt[3]{a^3 - t^3} ), we find that as ( t ) approaches negative infinity, ( x + y ) approaches 0 from the positive side. Thus, the lower bound is 0 (exclusive).Therefore, the range of values for ( x + y ) is:[boxed{(0, asqrt[3]{4}]}]]

🔑:Let x+y=t. Thenbegin{aligned}x^{3}+y^{3} & =(x+y)^{3}-3(x+y) x y & =t^{3}-3 t x y=a^{3} . therefore quad x y & =frac{t^{3}-a^{3}}{3 t} .end{aligned}From this, we know that x, y are the two real roots of the equation about zz^{2}-t z+frac{t^{3}-a^{3}}{3 t}=0By the discriminant theorem, we getDelta=t^{2}-4 cdot frac{t^{3}-a^{3}}{3 t} geqslant 0 .Solving this, we get 0<t leqslant sqrt[3]{4} a.That is, 0<x+y leqslant sqrt[3]{4} a.

🧠:[<think>Okay, so I need to solve this problem about the graph of a logarithmic function passing through a fixed point, and then find the minimum value of some expression involving m and n. Let me try to break this down step by step.First, the function given is y = log_a (x + 3) - 1, where a > 0 and a ≠ 1. The problem says that this graph always passes through a fixed point A, regardless of the value of a. Fixed points on logarithmic graphs usually occur where the argument of the logarithm becomes 1 because log_a(1) = 0 for any base a. Let me check that.The argument here is (x + 3). If I set x + 3 = 1, then x = 1 - 3 = -2. So, when x = -2, the logarithm becomes log_a(1) = 0. Then, y = 0 - 1 = -1. Therefore, the fixed point A should be (-2, -1). Let me verify if this is indeed a fixed point.If x = -2 and y = -1, then substituting into the original equation: -1 = log_a(-2 + 3) - 1 → -1 = log_a(1) - 1. Since log_a(1) is 0, we get -1 = 0 - 1, which is true. So, no matter what the base a is (as long as a > 0 and a ≠ 1), the point (-2, -1) will always lie on the graph. Therefore, point A is (-2, -1).Next, the problem states that this point A lies on the line mx + ny + 2 = 0. So, substituting the coordinates of A into this line equation:m*(-2) + n*(-1) + 2 = 0Simplifying this:-2m - n + 2 = 0Which can be rewritten as:2m + n = 2So, we have the equation 2m + n = 2. Now, the problem mentions that mn > 0. This means that m and n must be both positive or both negative. However, since we need to find the minimum value of 2/m + 1/n, we should consider the cases where m and n are positive because if they were both negative, m and n would be negative numbers, and dividing by them could lead to negative values, which might complicate finding a minimum (and the expression 2/m +1/n would be negative if m and n are negative, but mn >0). However, let's think more carefully.If mn > 0, then m and n are both positive or both negative. Let's check both possibilities.Case 1: m > 0 and n > 0.Case 2: m < 0 and n < 0.But in the expression 2/m +1/n, if m and n are both negative, then 2/m and 1/n are negative numbers. However, the problem asks for the minimum value. If we can have negative values, then the minimum would be unbounded below, which doesn't make sense. Therefore, maybe we can assume that m and n are positive. Alternatively, maybe the problem implicitly assumes m and n are positive, but the given condition is mn >0. Hmm. Let's check.Wait, if m and n are both negative, then 2/m +1/n would be negative, but the problem is asking for the minimum value. If negative values are allowed, then there is no minimum because as m and n become more negative, 2/m +1/n approaches zero from below, but since m and n can be any negative numbers satisfying 2m +n =2, but if m and n are negative, then 2m +n =2 would require m and n to be positive. Wait, wait, let's think again.Wait, if m and n are both negative, then 2m + n would be negative (since 2m is negative and n is negative, their sum is negative), but the equation is 2m +n =2, which is positive. Therefore, 2m +n =2 cannot be satisfied if both m and n are negative. Therefore, m and n must both be positive. That's the only possibility. Because if one is positive and the other is negative, mn would be negative, which contradicts mn >0. Therefore, since 2m +n =2 is positive, and mn >0, then m and n must be positive. Therefore, m >0 and n >0.Therefore, we can proceed under the assumption that m and n are positive real numbers satisfying 2m +n =2. Then, we need to find the minimum value of 2/m +1/n.This seems like an optimization problem with constraint 2m +n =2, and m, n >0. So, the problem reduces to minimizing 2/m +1/n subject to 2m +n =2. This is a typical problem that can be solved using the method of Lagrange multipliers, but since it's likely intended for a calculus student or someone familiar with AM-GM inequality, maybe we can use the Cauchy-Schwarz inequality or AM-GM.Let me recall that for optimization problems with linear constraints, the method of substitution is straightforward. Let's solve the constraint for n: n = 2 - 2m. Since n >0, 2 -2m >0 → 2m <2 → m <1. Also, m >0. So, m is in (0,1).Then, substitute n =2 -2m into the expression 2/m +1/n:Expression = 2/m + 1/(2 - 2m)We need to find the minimum of this expression for m ∈ (0,1).Let me denote f(m) = 2/m + 1/(2 - 2m). To find the minimum, take derivative f’(m), set to zero.First, compute derivative:f’(m) = -2/m² + (2)/(2 -2m)² * ( -2 )Wait, wait:Wait, the derivative of 2/m is -2/m².The derivative of 1/(2 -2m) is derivative of (2 -2m)^{-1} is (-1)*(2 -2m)^{-2}*(-2) = 2/(2 -2m)^2.Wait, let me check again.If f(m) = 2/m +1/(2 -2m)Then f’(m) = -2/m² + derivative of [1/(2 -2m)]Let me compute derivative of 1/(2 -2m):Let’s write it as (2 -2m)^{-1}Derivative is (-1)*(2 -2m)^{-2} * (-2) = 2/(2 -2m)^2Therefore, f’(m) = -2/m² + 2/(2 -2m)^2Set derivative equal to zero:-2/m² + 2/(2 -2m)^2 =0Move one term to the other side:2/(2 -2m)^2 = 2/m²Divide both sides by 2:1/(2 -2m)^2 =1/m²Take reciprocal:(2 -2m)^2 = m²Take square roots:|2 -2m| = |m|Since m ∈ (0,1), 2 -2m is positive because when m is in (0,1), 2m is in (0,2), so 2 -2m is in (0,2). So, 2 -2m is positive, and m is positive, so absolute values can be dropped:2 -2m = mSolve:2 =3m → m =2/3Check if this is within (0,1): Yes, m=2/3 is between 0 and1.Then, n=2 -2m=2 -2*(2/3)=2 -4/3=2/3.Therefore, the critical point is at m=2/3, n=2/3.Check second derivative or test intervals to ensure it's a minimum.Alternatively, since the function tends to infinity as m approaches 0 or 1, the critical point must be a minimum.Therefore, the minimum value of 2/m +1/n is 2/(2/3) +1/(2/3)= 3 + 3/2= 4.5=9/2.Wait, let me compute:If m=2/3 and n=2/3,Then 2/m =2/(2/3)=3,And 1/n=1/(2/3)=3/2,So sum is 3 + 3/2=9/2=4.5.Therefore, the minimum value is 9/2.Alternatively, maybe there's a way to use the Cauchy-Schwarz inequality or AM-GM.Let me try AM-GM.We have the expression 2/m +1/n with constraint 2m +n=2.Let me consider variables u = m, v =n.But maybe use substitution.Let me set variables such that the constraint is incorporated.Let me let m and n be positive numbers with 2m +n=2.Let me denote t = m. Then n=2 -2t, with t ∈ (0,1).But maybe another substitution.Alternatively, use Cauchy-Schwarz.Let me write the expression 2/m +1/n as (2/m +1/n). Let's consider vectors.But another approach: Let’s use Lagrange multipliers.Define function F(m,n) = 2/m +1/n with constraint G(m,n)=2m +n -2=0.The Lagrangian is L=2/m +1/n +λ(2m +n -2).Take partial derivatives:∂L/∂m = -2/m² +2λ=0∂L/∂n = -1/n² +λ=0∂L/∂λ=2m +n -2=0From the first equation: -2/m² +2λ=0 → λ=1/m²From the second equation: -1/n² +λ=0 → λ=1/n²Therefore, 1/m²=1/n² → n²=m² → n=m or n=-m.But m,n>0, so n=m.Thus, n=m.But the constraint is 2m +n=2. If n=m, then 2m +m=3m=2 → m=2/3, n=2/3. Which gives the same result as before.Therefore, the minimum value is 9/2.Alternatively, using AM-GM inequality.We can consider the expression 2/m +1/n. Let's think in terms of weighted AM-GM.But let me think of variables x and y such that 2/m and 1/n can be related.Alternatively, let's set variables a and b such that 2/m = a and 1/n = b. Then, the expression to minimize is a + b.But we need to relate a and b via the constraint.But 2/m =a → m=2/a1/n =b → n=1/bThen, the constraint 2m +n=2 becomes 2*(2/a) +1/b=2 → 4/a +1/b=2So, we need to minimize a + b subject to 4/a +1/b=2, with a>0 and b>0.This might be more complicated, but maybe we can use AM-GM here.Alternatively, use substitution.Let’s express 1/b=2 -4/a → b=1/(2 -4/a)But then a + b = a +1/(2 -4/a)This seems messy. Maybe not helpful.Alternatively, use Cauchy-Schwarz on the terms 4/a +1/b=2.Let’s consider vectors:Let’s write 4/a +1/b=2 as (2/sqrt(a))^2 + (1/sqrt(b))^2=2.But I don’t see a direct application.Alternatively, use Hölder's inequality.Alternatively, use Lagrange multipliers again.Alternatively, note that from the earlier result, the minimum is 9/2, but I need to confirm using AM-GM.Wait, let me try writing 2/m +1/n as a sum involving terms related to 2m and n.Since 2m +n=2, let's denote 2m = t, so n=2 -t, with t ∈ (0,2). Then, m=t/2.Thus, 2/m = 2/(t/2)=4/t and 1/n=1/(2 -t).So, the expression becomes 4/t +1/(2 -t). Need to minimize this for t ∈(0,2).This is similar to the previous substitution.Alternatively, let's set t=2m, so t=2m, so m=t/2, and n=2 -t.Then, the expression becomes 2/(t/2) +1/(2 -t) =4/t +1/(2 -t). Same as above.So, to minimize 4/t +1/(2 -t) where t ∈ (0,2).Take derivative:d/dt [4/t +1/(2 -t)] = -4/t² +1/(2 -t)^2Set equal to zero:-4/t² +1/(2 -t)^2=0 → 1/(2 -t)^2=4/t² → 1/(2 -t)=2/t → t=2*(2 -t)→ t=4 -2t → 3t=4 → t=4/3Thus, t=4/3, so m= t/2= (4/3)/2=2/3, and n=2 -t=2 -4/3=2/3. So, same result.Therefore, the minimum is 4/(4/3) +1/(2 -4/3)=3 +1/(2/3)=3 +3/2=9/2. So, same answer.Alternatively, using AM-GM inequality.Let’s see, suppose I try to apply AM-GM to the terms 4/t and 1/(2 -t). Let’s set them as two terms.But 4/t and 1/(2 -t). To apply AM-GM, we need positive numbers. Let's think about how to split them.Alternatively, use Cauchy-Schwarz:(4/t +1/(2 -t))*(t + (2 -t)) ≥ (sqrt(4/t *t) + sqrt(1/(2 -t)*(2 -t)))^2Wait, that would be:(4/t +1/(2 -t))*(t + (2 -t)) ≥ (sqrt(4) + sqrt(1))^2 = (2 +1)^2=9But t + (2 -t)=2, so left side is (4/t +1/(2 -t))*2 ≥9 → 4/t +1/(2 -t) ≥9/2.Therefore, the minimum is 9/2, achieved when equality in Cauchy-Schwarz, which requires that 4/t / t =1/(2 -t) / (2 -t), i.e., 4/t²=1/(2 -t)² → 2/t=1/(2 -t) → 2*(2 -t)=t →4 -2t =t →4=3t →t=4/3. Which is the same as before. Thus, equality holds when t=4/3, so minimum is 9/2.Therefore, regardless of the method, the minimum value is 9/2.Hence, the answer should be boxed{dfrac{9}{2}}.Final Answerboxed{dfrac{9}{2}}</think>The given function is ( y = log_a(x+3) - 1 ). To find the fixed point ( A ), we set the argument of the logarithm to 1, since ( log_a(1) = 0 ) for any base ( a ). This gives ( x + 3 = 1 ), so ( x = -2 ). Substituting ( x = -2 ) into the function, we get ( y = 0 - 1 = -1 ). Thus, the fixed point ( A ) is ( (-2, -1) ).Next, we substitute the coordinates of ( A ) into the line equation ( mx + ny + 2 = 0 ):[-2m - n + 2 = 0 implies 2m + n = 2]Given ( mn > 0 ), both ( m ) and ( n ) must be positive. We need to minimize the expression ( frac{2}{m} + frac{1}{n} ) subject to the constraint ( 2m + n = 2 ).Express ( n ) in terms of ( m ):[n = 2 - 2m]Substitute ( n ) into the expression:[frac{2}{m} + frac{1}{2 - 2m}]To find the critical points, take the derivative and set it to zero:[f(m) = frac{2}{m} + frac{1}{2 - 2m}][f'(m) = -frac{2}{m^2} + frac{2}{(2 - 2m)^2}]Setting ( f'(m) = 0 ):[-frac{2}{m^2} + frac{2}{(2 - 2m)^2} = 0 implies frac{1}{(2 - 2m)^2} = frac{1}{m^2} implies 2 - 2m = m implies 3m = 2 implies m = frac{2}{3}]Substituting ( m = frac{2}{3} ) back into the constraint:[2 left( frac{2}{3} right) + n = 2 implies n = frac{2}{3}]Thus, the minimum value of ( frac{2}{m} + frac{1}{n} ) is:[frac{2}{frac{2}{3}} + frac{1}{frac{2}{3}} = 3 + frac{3}{2} = frac{9}{2}]The minimum value is (boxed{dfrac{9}{2}}).,<think>Okay, let me try to figure this problem out. So, the question is about the graph of the function y = log_a(x + 3) - 1, which always passes through a fixed point A. Then, this point A lies on the line mx + ny + 2 = 0, where mn > 0. We need to find the minimum value of 2/m + 1/n. Hmm, okay, let's break this down step by step.First, the function given is y = log_a(x + 3) - 1. They mention that it always passes through a fixed point A. Now, a fixed point for a logarithmic function... Usually, logarithmic functions can have points that are independent of the base a, right? For example, if we can find x and y values that make the equation true regardless of the base a, that would be a fixed point. So, maybe we need to find such a point where the equation holds no matter what a is.Let me recall that log_a(b) = c means that a^c = b. So, if we can manipulate the equation y = log_a(x + 3) - 1 to get a^something = something else, but have that equation hold true for any a, then the exponents must be zero because a^0 = 1 for any a ≠ 1. That might be a strategy here.So, starting with the equation:y + 1 = log_a(x + 3)Converting this to exponential form:a^{y + 1} = x + 3Now, we want this equation to hold true for all a > 0, a ≠ 1. The only way this can happen is if both sides are equal to 1, because a^0 = 1 for any a. So, if we set the exponent y + 1 = 0, then the left side becomes a^0 = 1, and the right side x + 3 must also equal 1. Therefore:y + 1 = 0 ⇒ y = -1x + 3 = 1 ⇒ x = -2So, the fixed point A must be (-2, -1). Let me check that. If we plug x = -2 and y = -1 into the original equation:y = log_a(-2 + 3) - 1 ⇒ y = log_a(1) - 1 ⇒ y = 0 - 1 ⇒ y = -1. Perfect, that works for any a. So, point A is (-2, -1).Okay, so point A is (-2, -1). Now, this point lies on the line mx + ny + 2 = 0. Let's substitute x = -2 and y = -1 into the line equation:m*(-2) + n*(-1) + 2 = 0 ⇒ -2m - n + 2 = 0Simplifying:-2m - n = -2 ⇒ 2m + n = 2So, the equation relating m and n is 2m + n = 2. And we know that mn > 0. Since mn > 0, this means that m and n must be both positive or both negative. But let's see if that's possible given 2m + n = 2.If m and n are both positive, then 2m + n = 2 would have solutions where m and n are positive. If they are both negative, then 2m + n would be negative, but 2 is positive, so that's impossible. Therefore, m and n must both be positive. So, mn > 0 implies m > 0 and n > 0.So, our constraints are 2m + n = 2, m > 0, n > 0. We need to find the minimum value of 2/m + 1/n.Alright, so this is an optimization problem with a linear constraint and a function to minimize. The variables m and n are positive, and they satisfy 2m + n = 2. We can use methods like substitution or Lagrange multipliers, but since it's a simple constraint, substitution might be easier.Let me solve for n from the constraint:n = 2 - 2mSince n > 0, then 2 - 2m > 0 ⇒ 2m < 2 ⇒ m < 1. Also, since m > 0, the domain for m is 0 < m < 1.Now, substitute n = 2 - 2m into the expression to minimize:2/m + 1/n = 2/m + 1/(2 - 2m)So, our function to minimize is f(m) = 2/m + 1/(2 - 2m) where 0 < m < 1.Now, to find the minimum of f(m), we can take the derivative with respect to m and set it to zero.First, let's compute f'(m):The derivative of 2/m is -2/m².The derivative of 1/(2 - 2m) is [0 - (-2)]/(2 - 2m)² = 2/(2 - 2m)².So, f'(m) = -2/m² + 2/(2 - 2m)².Set derivative equal to zero:-2/m² + 2/(2 - 2m)² = 0Multiply both sides by m²(2 - 2m)² to eliminate denominators:-2(2 - 2m)² + 2m² = 0Simplify:-2(4 - 8m + 4m²) + 2m² = 0Wait, let's compute (2 - 2m)² first:(2 - 2m)^2 = 4 - 8m + 4m²So, substituting back:-2*(4 - 8m + 4m²) + 2m² = 0Multiply out:-8 + 16m - 8m² + 2m² = 0Combine like terms:-8 + 16m -6m² = 0Multiply both sides by -1 to make the quadratic coefficient positive:6m² -16m +8 = 0Now, let's simplify this quadratic equation. Let's divide all terms by 2:3m² -8m +4 = 0Now, solving for m using quadratic formula:m = [8 ± sqrt(64 - 48)] / 6sqrt(64 -48) = sqrt(16) = 4So, m = [8 ±4]/6So, two solutions:m = (8 +4)/6 = 12/6 = 2m = (8 -4)/6 = 4/6 = 2/3But our domain for m is 0 < m < 1, so m = 2 is outside the domain. Therefore, the critical point is at m = 2/3.Now, check if this is a minimum.We can use the second derivative test or check intervals.Let's compute the second derivative f''(m):First derivative f'(m) = -2/m² + 2/(2 - 2m)^2Second derivative:d/dm [-2/m²] = 4/m³d/dm [2/(2 - 2m)^2] = 2*2*(2 - 2m)^{-3}*2 = 8/(2 - 2m)^3Wait, let me recast that properly.The derivative of 2/(2 - 2m)^2 is:Let u = 2 - 2m, so d/dm [2/u²] = 2*(-2)/u³ * (-2) = ?Wait, perhaps better to use chain rule:d/dm [2*(2 - 2m)^{-2}] = 2*(-2)*(2 - 2m)^{-3}*(-2) = 2*(-2)*(-2)*(2 - 2m)^{-3} = 8*(2 - 2m)^{-3}Wait, actually:Wait, derivative of (2 - 2m)^{-2} is (-2)(-2)(2 - 2m)^{-3} = 4(2 - 2m)^{-3}Wait, perhaps step by step:Let me compute derivative of 2/(2 - 2m)^2.Let f(m) = 2*(2 - 2m)^{-2}Then f'(m) = 2*(-2)*(2 - 2m)^{-3}*(-2) = 2*(-2)*(-2)*(2 - 2m)^{-3} = 8*(2 - 2m)^{-3}Wait, hold on. Let me see:The derivative of (2 - 2m)^k is k*(2 - 2m)^{k -1}* derivative of (2 - 2m). So, k = -2, so derivative is (-2)*(2 - 2m)^{-3}*(-2) = 4*(2 - 2m)^{-3}So, the derivative of 2*(2 - 2m)^{-2} is 2*4*(2 - 2m)^{-3} = 8*(2 - 2m)^{-3}Wait, no:Wait, first, the function is 2*(2 - 2m)^{-2}Derivative: 2 * (-2)*(2 - 2m)^{-3} * (-2) = 2*(-2)*(-2)*(2 - 2m)^{-3} = 8*(2 - 2m)^{-3}Yes, because chain rule: derivative of (2 - 2m) is -2, so multiplied by -2 from the exponent, then another -2 from the derivative of the inside. So, 2 * (-2) * (-2) = 8.Similarly, derivative of -2/m² is 4/m³.Therefore, f''(m) = 4/m³ + 8/(2 - 2m)^3At m = 2/3, let's compute f''(2/3):First term: 4/( (2/3)^3 ) = 4 / (8/27) ) = 4 * (27/8) = 108/8 = 13.5Second term: 8/(2 - 2*(2/3))^3 = 8/(2 - 4/3)^3 = 8/(2/3)^3 = 8/(8/27) ) = 8 * (27/8) = 27So, f''(2/3) = 13.5 + 27 = 40.5 > 0. So, since the second derivative is positive, this critical point is a local minimum. Since the function tends to infinity as m approaches 0 or 1, this local minimum is indeed the global minimum.Therefore, the minimum occurs at m = 2/3. Then, n = 2 - 2m = 2 - 2*(2/3) = 2 - 4/3 = 2/3.Wait, n = 2 - 2m. So, m = 2/3, so n = 2 - 4/3 = 2/3. Therefore, m = 2/3, n = 2/3.Then, plugging into 2/m + 1/n:2/(2/3) + 1/(2/3) = (2 * 3/2) + (3/2) = 3 + 1.5 = 4.5 = 9/2.Hmm, so the minimum value is 9/2. Let me verify this again.Wait, 2/m + 1/n with m = 2/3, n = 2/3:2/(2/3) = 2 * 3/2 = 31/(2/3) = 3/2So, total is 3 + 3/2 = 9/2, which is 4.5. That seems correct.Alternatively, could we use the method of Lagrange multipliers here? Let me check.We need to minimize f(m, n) = 2/m + 1/n subject to the constraint g(m, n) = 2m + n - 2 = 0, with m > 0, n > 0.The Lagrangian is L = 2/m + 1/n - λ(2m + n - 2)Taking partial derivatives:∂L/∂m = -2/m² - 2λ = 0∂L/∂n = -1/n² - λ = 0∂L/∂λ = -(2m + n - 2) = 0From ∂L/∂m: -2/m² - 2λ = 0 ⇒ λ = -1/m²From ∂L/∂n: -1/n² - λ = 0 ⇒ λ = -1/n²Therefore, -1/m² = -1/n² ⇒ 1/m² = 1/n² ⇒ m = n or m = -n. But since m and n are positive, m = n.But from the constraint 2m + n = 2, and m = n, so 2m + m = 2 ⇒ 3m = 2 ⇒ m = 2/3, then n = 2/3. So same result.Hence, the minimum is 9/2. So, that seems consistent.Alternatively, using the Cauchy-Schwarz inequality or AM-GM inequality.Let me try using AM-GM. We have 2/m + 1/n, with 2m + n = 2.Let me consider the variables 2/m and 1/n. To apply AM-GM, maybe we need to relate the coefficients. Alternatively, use substitution.Let me let u = m, v = n. Then, 2u + v = 2. Need to minimize 2/u + 1/v.Alternatively, write variables in terms of t.Let me set m = t, so n = 2 - 2t. Then, the expression becomes 2/t + 1/(2 - 2t). But maybe not helpful.Alternatively, use substitution x = 1/m and y = 1/n. Then, m = 1/x, n = 1/y. The constraint becomes 2*(1/x) + (1/y) = 2. And the expression to minimize is 2x + y. So, the problem becomes minimize 2x + y subject to 2/x + 1/y = 2, with x > 0, y > 0.Hmm, that's another form. Not sure if that helps, but maybe using Cauchy-Schwarz.Alternatively, let's use Cauchy-Schwarz on the original problem. The expression 2/m + 1/n can be thought of as a dot product of vectors (sqrt(2), 1) and (sqrt(2)/m, 1/n). But maybe that's overcomplicating.Alternatively, since we have 2m + n = 2, we can write this as (2m) + n = 2. Let’s denote a = 2m, b = n, so a + b = 2, and a, b > 0. Then, the expression to minimize becomes (2/m) + (1/n) = (2/(a/2)) + (1/b) = 4/a + 1/b. Now, we need to minimize 4/a + 1/b with a + b = 2, a > 0, b > 0.This might be easier. Let's set up the function f(a) = 4/a + 1/(2 - a), with 0 < a < 2.Take derivative f’(a) = -4/a² + 1/(2 - a)^2Set to zero:-4/a² + 1/(2 - a)^2 = 0So, 1/(2 - a)^2 = 4/a²Take square roots (since both sides positive):1/(2 - a) = 2/aCross-multiplying:a = 2(2 - a)a = 4 - 2a3a = 4a = 4/3Then, b = 2 - a = 2 - 4/3 = 2/3So, a = 4/3, b = 2/3. Therefore, original m = a/2 = (4/3)/2 = 2/3, and n = b = 2/3. So same result.Therefore, the minimum of 4/a +1/b is 4/(4/3) +1/(2/3) = 3 + 3/2 = 9/2. Same answer. So, this method also works.Alternatively, using the method of Lagrange multipliers, as above, gives the same result. Therefore, regardless of the method, the minimum value is 9/2.Therefore, the answer should be boxed{dfrac{9}{2}}.Final Answerboxed{dfrac{9}{2}}</think>The graph of the function ( y = log_a(x + 3) - 1 ) always passes through a fixed point ( A ). To find this point, we convert the logarithmic equation to its exponential form:[y + 1 = log_a(x + 3) implies a^{y + 1} = x + 3]For this equation to hold for all ( a > 0 ) and ( a neq 1 ), both sides must be 1. Thus, we set:[y + 1 = 0 implies y = -1 quad text{and} quad x + 3 = 1 implies x = -2]The fixed point ( A ) is ((-2, -1)).This point lies on the line ( mx + ny + 2 = 0 ). Substituting ((-2, -1)) into the line equation:[-2m - n + 2 = 0 implies 2m + n = 2]We need to minimize ( frac{2}{m} + frac{1}{n} ) under the constraint ( 2m + n = 2 ) with ( mn > 0 ). Solving for ( n ) in terms of ( m ):[n = 2 - 2m]Substituting ( n ) into the expression to minimize:[frac{2}{m} + frac{1}{2 - 2m}]Taking the derivative and setting it to zero, we find the critical point ( m = frac{2}{3} ). Substituting ( m = frac{2}{3} ) back into the constraint gives ( n = frac{2}{3} ). Evaluating the expression:[frac{2}{m} + frac{1}{n} = frac{2}{frac{2}{3}} + frac{1}{frac{2}{3}} = 3 + frac{3}{2} = frac{9}{2}]Thus, the minimum value is (boxed{dfrac{9}{2}}).]

🔑:AnalysisThis question examines the properties of logarithmic functions and the knowledge of the inequality of arithmetic and geometric means, using the idea of substitution, which is a key content in college entrance examinations.Based on the properties of logarithmic functions, we first determine the coordinates of A, substitute them into the equation of the line to find the relationship between m and n, and then solve it using substitution combined with the inequality of arithmetic and geometric means.SolutionSince when x=-2, y=log_a1-1=-1,Therefore, the graph of the function y=log_a(x+3)-1 (a > 0, a neq 1) always passes through the fixed point (-2,-1), that is, A(-2,-1).Since point A lies on the line mx+ny+2=0,Therefore, -2m-n+2=0, which means 2m+n=2.Since mn > 0,Therefore, m > 0, n > 0, frac{2}{m}+ frac{1}{n}= frac{1}{2}(2m+n)(frac{2}{m}+ frac{1}{n})= frac{1}{2}(5+ frac{2n}{m}+ frac{2m}{n})geqslant frac{1}{2}(5+4)= frac{9}{2}Therefore, the minimum value of frac{2}{m}+ frac{1}{n} is boxed{frac{9}{2}}.Hence, the answer is boxed{frac{9}{2}}.

🧠:[<think>Okay, let's try to figure out this problem. So, Xiao Wang plays a game five times, and his average score is 87. The question is asking what's the lowest score he could have gotten in any single game. Hmm. Alright, so if the average is 87 over five games, that means the total points he has is 87 multiplied by 5, right? Let me calculate that first.87 times 5. Let's see, 80 times 5 is 400, and 7 times 5 is 35, so 400 + 35 = 435. So, his total points over the five games is 435. Now, we need to find the minimum possible score he could have in one of those games. To find the minimum score in one game, we need to maximize the scores in the other four games, right? Because if the other four are as high as possible, then the fifth one will be as low as possible. Since the maximum score per game is 100, right? The problem says the full score is 100 points. So, each game can be at most 100.So, if he scored 100 in four of the games, then the total points from those four games would be 100 times 4, which is 400. Then, the fifth game would be 435 total minus 400, which is 35. So, 35 points. Therefore, his lowest possible score in any single game is 35. Is that right? Let me check again.Total points needed: 435. If four games are maxed at 100 each: 4*100=400. 435-400=35. So yes, 35. That seems straightforward. But wait, let me make sure there isn't a trick here. The problem says "the average score is the game's result". Hmm, maybe I need to verify if there's any different interpretation.But average score is the result, so they take the average of the five games. So, total is 5*87=435. To find the minimum possible score in any one game, we maximize the others. Since each game can't exceed 100, so setting four games to 100 gives the maximum total for four games, leaving the fifth as 35. So, unless there's a restriction I'm missing, like maybe scores have to be integers? Wait, the problem doesn't specify that scores have to be whole numbers. Hmm. But in most games, scores are whole numbers, but maybe here it's allowed to have fractions?Wait, but 100 is the maximum, so if scores can be any real number up to 100, then even if we have four games at 100, the fifth is 35. But if they have to be integers, then 35 is still possible. So, maybe the answer is 35. But let me check again. Let's think of another approach. Suppose we have variables for the five games: a, b, c, d, e. Each between 0 and 100, inclusive. The sum a + b + c + d + e = 435. We need to find the minimal possible value among a, b, c, d, e.To minimize one variable, we need to maximize the others. Since each of the others can be at most 100, so maximum total for four variables is 400, hence the fifth is 35. Therefore, the minimum possible score is 35. So, answer is 35. Wait, but let me check if 35 is possible. For example, if four games are 100, 100, 100, 100, and the fifth is 35. Then, total is 400 +35=435, average is 87. So, yes, that works. So, 35 is achievable. Therefore, the answer should be 35. So, the score in any single game cannot be lower than 35. So, 35 is the minimum. Therefore, the answer is boxed{35}. But wait, let me just think again if there's another scenario where maybe two games are lower, but to get a lower minimum. But if you spread the "remaining" points more, but we need the minimal single score. So, actually, to get the minimal possible score, you need to set the other four as high as possible. If you set three to 100 and one to something else, then the minimum might be lower? Wait, no, because if you set four to 100, then the fifth is 35, which is the lowest possible. If you set three to 100, and one to, say, 99, then the fifth would be 435 - (3*100 +99)=435-399=36, which is higher than 35. So, actually, by maximizing four of them, you get the minimum possible fifth. So, 35 is indeed the minimum. Therefore, the answer is 35. Another way: suppose all five games are 87. Then, total is 435. But if you want to lower one game, you have to compensate by raising others. But since others can be raised only up to 100, so the maximum you can compensate is 100 -87=13 per game. So, for each of the four games, you can add 13 to them. So, total compensation is 4*13=52. Therefore, the maximum you can lower the fifth game is 52. So, 87 -52=35. That's another way to see. So, 35. Yeah, that makes sense. Therefore, the answer is 35. So, I think that's solid.Final Answerboxed{35}</think>To determine the lowest score Xiao Wang could have received in any single game, we start by calculating the total points he scored over five games. Given his average score is 87, the total points is:[87 times 5 = 435]To find the minimum score in one game, we maximize the scores in the other four games. Since the maximum score per game is 100, the total points for the four highest games is:[100 times 4 = 400]Subtracting this from the total points gives the minimum score in the fifth game:[435 - 400 = 35]Thus, the lowest score Xiao Wang could have received in any single game is (boxed{35}).,<think>Okay, let's see. The problem is about Xiao Wang playing a game where the full score is 100 points each time. He can play five times, and his average score is 87. We need to find the minimum score he must have in any single game. Hmm, so the question is asking, given that he has an average of 87 over five games, what's the lowest possible score he could have gotten in any one of those games. Right?First, let's recall that average score is the total points divided by the number of games. So if his average is 87 over five games, his total points must be 87 multiplied by 5. Let me calculate that. 87 times 5... 80 times 5 is 400, and 7 times 5 is 35, so 400 + 35 = 435. So his total score across all five games is 435 points.Now, the problem is to find the minimum score he could have gotten in any single game. To find the lowest possible score in one game, we need to maximize the scores in the other four games because that would leave the remaining points for the fifth game as low as possible. Since the maximum score per game is 100, we can assume he scored 100 in four of the games. Let me check that logic. If he scores the maximum possible in four games, then the fifth game's score would be the total (435) minus the sum of those four maximums. Since each maximum is 100, four games would be 400. So 435 - 400 = 35. Therefore, the minimum he could have scored in one game is 35.Wait, but let me verify if that's correct. Let's add up four 100s and one 35: 100 + 100 + 100 + 100 + 35 = 435. Yep, that's correct. The average would be 435 / 5 = 87. So that works. Therefore, the lowest possible score he could have in any single game is 35. Is there a possibility of getting a lower score if the other games aren't all 100? Let's see. Suppose he scored 99 in four games. Then the total for those four would be 99*4 = 396. Then the fifth game would need to be 435 - 396 = 39, which is higher than 35. So that's actually a higher minimum. So, indeed, the lower the other scores are, the higher the minimum score would have to be. Therefore, to minimize the fifth score, we need to maximize the other four. Hence, the answer is 35.But hold on, the question says "his score in any single game cannot be lower than ( ) points." So, this is the minimal possible score he must have at least once? Wait, no. It's saying that in any single game, his score cannot be lower than a certain value. So, in other words, what's the minimal lower bound such that all his games are at least that number. Wait, maybe I misinterpreted the question. Let me read it again."Xiao Wang's average score is 87 points, so his score in any single game cannot be lower than ( ) points."Hmm, maybe the question is asking for the minimum possible lower bound, such that in all games, his scores are at least that number, given the average of 87. So, to find the minimal value x such that all five scores are ≥ x, and the average is 87. So, in that case, the minimal x would be the minimal possible score in the set of five scores that gives the total 435. So, in order to find the minimal x, we need to distribute the total 435 in such a way that the minimal score is as low as possible. Which is exactly what I did before: maximizing the other four, so the fifth is 35. Therefore, x is 35. So, in other words, there exists a scenario where he has one game with 35 and four with 100, so his average is 87, and thus his score cannot be lower than 35 because there's a possibility he scored 35. Wait, but the wording is a bit confusing.Wait, actually, the problem is saying "his score in any single game cannot be lower than ( ) points." So, it's not about the possible minimum he could have, but rather the guaranteed minimum. That is, given that his average is 87, what is the minimal score that he must have scored in every game. But that seems contradictory, because if you have an average, individual scores can vary. So maybe the problem is phrased as, "what is the minimum score he must have obtained in at least one game," but that's not how it's worded. Alternatively, maybe it's saying that given the average, it's impossible for all his scores to be above a certain value. Wait, the question is a bit ambiguous.Wait, let's parse the original problem again:"In a game with a full score of 100 points, each person can play five times, and the average score is the game's result. Xiao Wang's average score is 87 points, so his score in any single game cannot be lower than ( ) points."So, perhaps the question is: Given that Xiao Wang has an average of 87 over five games, each with a maximum of 100, what is the minimum possible score he must have in at least one of the games. In other words, what is the lowest possible score he could have gotten in any one game, with the others as high as possible, which would be 100. So, that gives 35, as we calculated before.Alternatively, if the question is asking for the minimal lower bound such that all his games are at least that number, then we need to find the minimal x where all five scores are ≥ x and average is 87. So, that would be x = minimal such that 5x ≤ 435, so x ≤ 87. But 435 divided by 5 is 87, so if all scores are equal, they are 87. Therefore, if all scores are at least x, then x can't be higher than 87. But that seems contradictory. Wait, maybe the problem is phrased differently.Wait, perhaps the problem is actually asking for the minimal score that he necessarily must have in every game. For example, even if he tried to score as low as possible in one game, he couldn't go below a certain score because the average requires him to have a minimum total. So, the answer would be 35, because he could have a 35, but he can't have lower than that. Hence, in any single game, his score cannot be lower than 35. Because if he had a score lower than 35, even with maximums in the other four, the total would be less than 435. Wait, let's check that.Suppose he scored 34 in one game, and 100 in the other four. Then total would be 34 + 4*100 = 434, which is less than 435. So that would give an average of 434 /5 = 86.8, which is less than 87. Therefore, to achieve an average of 87, he cannot have any score lower than 35. Because if he had a 34, even with four 100s, the total would be insufficient. Therefore, 35 is the minimal possible score in any game, so his scores cannot be lower than 35. Therefore, the answer is 35.So the answer is boxed{35}Final Answerboxed{35}</think>To determine the lowest score Xiao Wang could have received in any single game given his average score of 87 over five games, we start by calculating his total score. The total score is the average multiplied by the number of games:[87 times 5 = 435]Next, we need to find the minimum score in one game. To do this, we assume he scored the maximum of 100 points in four of the games. The total score for these four games would be:[100 times 4 = 400]Subtracting this from the total score gives the minimum score in the remaining game:[435 - 400 = 35]We verify that if Xiao Wang scored 35 in one game and 100 in the other four, the total would indeed be 435, which results in an average of 87. If he scored lower than 35 in any game, even with maximum scores in the other four games, the total would be insufficient to maintain the average of 87. Therefore, the minimum score he could have in any single game is:[boxed{35}]]

🔑:【Answer】35 points.【Analysis】Key point: Maximum and minimum value problem, average number problemThink Academy corresponding lecture: Grade 3 Fall Class Average Number Problem (Lecture 5)To achieve the lowest score in one attempt, the scores in the other attempts should be the highest, i.e., 100 points. The maximum number of times you can score 100 is 4, at which point the score in the 5th attempt is the lowest, and the lowest score is: 87 times 5 - 100 times 4 = 35 (points).