🧠:[<think>Alright, so I need to find the equation of a plane that passes through point A and is perpendicular to the vector BC. Let me start by recalling what information I have here. First, the coordinates of the points are given: A is (-8, 0, 7), B is (-3, 2, 4), and C is (-1, 4, 5). The problem says the plane is perpendicular to vector BC, so I need to find vector BC first. Vector BC would be from point B to point C, right? So to find that vector, I subtract the coordinates of B from C. Let me do that component-wise. For the x-component: -1 (which is C's x-coordinate) minus (-3) (which is B's x-coordinate). That's -1 + 3 = 2. For the y-component: 4 (C's y) minus 2 (B's y) = 2. For the z-component: 5 (C's z) minus 4 (B's z) = 1. So vector BC is (2, 2, 1). Got that. Now, the plane is perpendicular to this vector BC. Wait, if the plane is perpendicular to a vector, that vector is the normal vector of the plane. Right? Because the normal vector is perpendicular to the plane. So that means vector BC is the normal vector of the plane we need. So the normal vector n is (2, 2, 1). The plane passes through point A(-8, 0, 7). To write the equation of the plane, I can use the point-normal form of the plane equation. The formula is:n • (r - r₀) = 0Where n is the normal vector, r is any point on the plane, and r₀ is the given point through which the plane passes. Alternatively, this can be written in component form as:a(x - x₀) + b(y - y₀) + c(z - z₀) = 0Where (a, b, c) is the normal vector and (x₀, y₀, z₀) is the point A.Plugging in the values, here's what I get:2(x - (-8)) + 2(y - 0) + 1(z - 7) = 0Simplify that:2(x + 8) + 2y + (z - 7) = 0Let me expand each term:2*x + 2*8 + 2*y + z - 7 = 0Calculating the constants:2*8 = 16, so:2x + 16 + 2y + z - 7 = 0Combine like terms (16 -7 = 9):2x + 2y + z + 9 = 0Hmm, wait, but usually the plane equation is written as ax + by + cz + d = 0. So in this case, the equation would be 2x + 2y + z + 9 = 0. Let me check my steps again to make sure I didn't make a mistake.Starting from the normal vector BC (2,2,1), point A(-8,0,7). The equation is 2(x +8) + 2(y -0) +1(z -7)=0. Then expanding:2x +16 +2y + z -7=0. Then combine constants: 16 -7=9. So yes, 2x +2y +z +9=0. That seems right.But maybe I can write it in standard form. Alternatively, sometimes people write the constant term on the other side. So 2x +2y +z = -9. But both forms are acceptable as long as it's clear. The problem doesn't specify a particular form, so I think either is okay, but probably the 2x +2y +z +9=0 is more standard.Wait, let me cross-verify. Let's think if this plane actually contains point A. Let's plug in A(-8,0,7) into the equation:Left-hand side: 2*(-8) +2*0 +7 +9 = (-16) +0 +7 +9 = (-16) +16=0. Perfect, that works. Also, the normal vector is (2,2,1). So the plane equation is correct. But just to make sure, let's check another way. Suppose I use the general plane equation: ax + by + cz + d =0. The normal vector is (a,b,c). So here, a=2, b=2, c=1. Then plug in point A(-8,0,7):2*(-8) +2*0 +1*7 +d =0 => -16 +0 +7 +d =0 => -9 +d=0 => d=9. Therefore, the equation is 2x +2y +z +9=0. Yep, same result. So that's correct.Alternatively, if I wanted to write it as 2x +2y +z = -9, that's equivalent. But since the problem says "write the equation", and doesn't specify the form, both are acceptable. But in most cases, the standard form is ax + by + cz + d =0, so 2x +2y +z +9=0.But let me check once more. Sometimes, people prefer the constant term to be positive, but in this case, it's already positive. Wait, 2x +2y +z +9=0 is correct. Alternatively, if we moved the 9 to the other side, it would be 2x +2y +z = -9. But both are correct, just different forms.Is there any other check I can perform? Maybe verify that vector BC is indeed perpendicular to the plane. Since the normal vector is BC, then the plane is perpendicular to BC. So that's already given. Another check: take another point on the plane? Wait, we only have point A. But if I can think of another point on the plane, but since I don't have more points, maybe not. Alternatively, maybe see if vector BA is on the plane? Wait, no, BA is a vector from B to A. But since the plane is passing through A and not necessarily B, BA is not necessarily on the plane.Wait, but the plane is perpendicular to vector BC, so any vector on the plane should be perpendicular to BC. Let me pick an arbitrary point on the plane, say (x,y,z). Then the vector from A to this point is (x +8, y -0, z -7). This vector should be perpendicular to BC (2,2,1). So their dot product should be zero.So (x +8, y, z -7) • (2,2,1) =0Which is 2(x +8) + 2y +1(z -7)=0, which is the same equation as before. So that's consistent. Therefore, the equation is correct.Therefore, I think the equation of the plane is 2x +2y +z +9=0.Final AnswerThe equation of the plane is boxed{2x + 2y + z + 9 = 0}.</think>To find the equation of the plane passing through point ( A(-8, 0, 7) ) and perpendicular to vector ( overrightarrow{BC} ):1. Find vector ( overrightarrow{BC} ): - Coordinates of ( B(-3, 2, 4) ) and ( C(-1, 4, 5) ). - ( overrightarrow{BC} = C - B = (-1 - (-3), 4 - 2, 5 - 4) = (2, 2, 1) ).2. Use the point-normal form of the plane equation: - The normal vector ( mathbf{n} ) is ( (2, 2, 1) ). - The point ( A(-8, 0, 7) ) lies on the plane. - The equation of the plane is given by: [ 2(x + 8) + 2(y - 0) + 1(z - 7) = 0 ]3. Simplify the equation: - Expand and combine like terms: [ 2x + 16 + 2y + z - 7 = 0 ] - Simplify the constants: [ 2x + 2y + z + 9 = 0 ]4. Verification: - Check that point ( A ) satisfies the equation: [ 2(-8) + 2(0) + 7 + 9 = -16 + 0 + 7 + 9 = 0 ]Thus, the equation of the plane is (boxed{2x + 2y + z + 9 = 0}).,<think>Okay, so I need to find the equation of a plane that passes through point A and is perpendicular to the vector BC. Let me recall how to do this. Hmm, right, the general equation of a plane is given by a(x - x₀) + b(y - y₀) + c(z - z₀) = 0, where (a, b, c) is the normal vector to the plane and (x₀, y₀, z₀) is a point on the plane. In this problem, point A is (-8, 0, 7), so that will be our (x₀, y₀, z₀). The normal vector is supposed to be vector BC. So first, I need to find vector BC by subtracting the coordinates of point B from point C. Let me confirm: vector BC is C minus B. Let me compute that. Point B is (-3, 2, 4) and point C is (-1, 4, 5). So subtracting B's coordinates from C's, we get:x-coordinate: -1 - (-3) = -1 + 3 = 2y-coordinate: 4 - 2 = 2z-coordinate: 5 - 4 = 1So vector BC is (2, 2, 1). Therefore, the normal vector (a, b, c) is (2, 2, 1). Now plug these into the plane equation. Using point A(-8, 0, 7):2(x - (-8)) + 2(y - 0) + 1(z - 7) = 0Simplify each term:First term: 2(x + 8)Second term: 2yThird term: 1(z - 7) which is z - 7So expanding that:2x + 16 + 2y + z - 7 = 0Combine the constants: 16 - 7 = 9So the equation becomes:2x + 2y + z + 9 = 0Wait, let me check the arithmetic again. 2(x +8) is 2x +16. Then 2y, and z -7. So when you add all those together: 2x +16 +2y + z -7. Combining 16 and -7 gives 9. So yes, 2x + 2y + z +9 =0. But let me verify once more. If vector BC is (2,2,1) and point A is (-8,0,7), then plugging into the plane equation:2*(x - (-8)) + 2*(y -0) +1*(z -7) =0Which simplifies to 2x +16 +2y +z -7 =016-7 is 9, so 2x +2y +z +9 =0. Yeah, that seems right.Alternatively, another way to check is to ensure that point A satisfies the equation. Let's plug in A(-8,0,7):2*(-8) +2*0 +7 +9 = ?2*(-8) is -16, plus 0 is -16, plus 7 is -9, plus 9 is 0. Perfect, it works. Also, the normal vector is BC, so the plane should be perpendicular to BC. To confirm, if we take another point on the plane, but actually since we derived it correctly, it should be okay. Maybe check if points B or C lie on the plane? Wait, but the plane is passing through A and perpendicular to BC, so points B and C might not necessarily lie on the plane. Let me check point B(-3,2,4):2*(-3) +2*2 +4 +9 = -6 +4 +4 +9 = (-6 +4) + (4 +9) = (-2) +13 =11 ≠0. So point B is not on the plane, which is expected. Similarly, point C(-1,4,5):2*(-1) +2*4 +5 +9 = -2 +8 +5 +9 = ( -2 +8 ) + (5 +9 ) = 6 +14=20 ≠0. So they are not on the plane, which is fine. Alternatively, maybe I can check if vector BC is indeed perpendicular to the plane. For that, the normal vector is BC, so any vector lying on the plane should be perpendicular to BC. Let's take a vector from point A to another arbitrary point on the plane. Let's pick a point on the plane, say, let me find another point. Let me choose x=0, y=0, then z would be from 2x +2y + z +9=0 => z = -9. So point (0,0,-9) is on the plane. Then vector from A(-8,0,7) to (0,0,-9) is (8,0,-16). Then the dot product with BC vector (2,2,1) should be zero if they are perpendicular. Let's compute:8*2 +0*2 + (-16)*1 =16 +0 -16=0. Yes, so that works. Another check: take another point. Let me pick y=1, x=0. Then 2*0 +2*1 +z +9=0 => z= -11. So point (0,1,-11). Vector from A is (0 - (-8),1 -0, -11 -7)=(8,1,-18). Dot product with BC: 8*2 +1*2 + (-18)*1=16 +2 -18=0. Perfect. So, the equation 2x +2y +z +9=0 is correct. Therefore, the answer should be 2x +2y +z +9=0. But since the problem might expect it written in a different form, let me check. The standard form is ax + by + cz + d =0, which is what we have here. Alternatively, sometimes written as 2x +2y +z = -9, but both are equivalent. The problem says "write the equation of the plane", so either form is acceptable. Since the coefficients are integers and there's no common factor, this is the simplest form. I think that's it. I can't see any mistakes here.Final AnswerThe equation of the plane is boxed{2x + 2y + z + 9 = 0}.</think>To find the equation of the plane passing through point ( A(-8, 0, 7) ) and perpendicular to the vector ( overrightarrow{BC} ):1. Determine the vector ( overrightarrow{BC} ): - Coordinates of points ( B(-3, 2, 4) ) and ( C(-1, 4, 5) ). - Calculate ( overrightarrow{BC} = C - B ): [ overrightarrow{BC} = (-1 - (-3), 4 - 2, 5 - 4) = (2, 2, 1) ]2. Use the point-normal form of the plane equation: - The normal vector is ( overrightarrow{BC} = (2, 2, 1) ). - The point ( A(-8, 0, 7) ) lies on the plane. - Substitute into the plane equation ( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 ): [ 2(x + 8) + 2(y - 0) + 1(z - 7) = 0 ]3. Simplify the equation: - Expand and combine like terms: [ 2x + 16 + 2y + z - 7 = 0 ] - Combine constants: [ 2x + 2y + z + 9 = 0 ]4. Verification: - Check that point ( A ) satisfies the equation: [ 2(-8) + 2(0) + 7 + 9 = -16 + 0 + 7 + 9 = 0 ] - Verify the normal vector with another point on the plane, confirming the dot product is zero.Thus, the equation of the plane is (boxed{2x + 2y + z + 9 = 0}).]

🔑:## SolutionLet's find the vector overrightarrow{B C}:overrightarrow{B C}={-1-(-3) ; 4-2 ; 5-4}={2 ; 2 ; 1}Since the vector overrightarrow{B C} is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:begin{aligned}& 2 cdot(x-(-8))+2 cdot(y-0)+(z-7)=0 & 2 x+16+2 y+z-7=0 & 2 x+2 y+z+9=0end{aligned}## Problem Kuznetsov Analytic Geometry 9-4

🧠:[<think>Okay, so I have this algebra problem here: factorize m cubed minus 4m. Let me think. Alright, factorization. Hmm. So, I need to write m³ - 4m as a product of simpler expressions. Let's start by recalling what factorization means. It means breaking down an expression into a product of its factors, right? So, like, factors are expressions that multiply together to give the original expression.First, I should check if there's a greatest common factor (GCF) that I can factor out from both terms. The terms here are m³ and -4m. Let's look at the coefficients first. The coefficients are 1 (from m³) and -4. The GCF of 1 and 4 is 1, so there's no common numerical factor. Now, let's look at the variables. The first term is m³ and the second term is m. So, both have an m. The GCF of m³ and m is m, since m is the lowest power. So, I can factor out an m from both terms.Let me try that. If I factor out an m, then inside the parentheses, I divide each term by m. So, m³ divided by m is m², and -4m divided by m is -4. So, that gives me m(m² - 4). Okay, so the expression becomes m times (m squared minus 4). That's better, but maybe it can be factored further. Let me check the remaining part: m² - 4.Wait, m² - 4 is a difference of squares, right? Because m squared is a square (of m) and 4 is a square (of 2). And the difference of squares formula is a² - b² = (a - b)(a + b). So, applying that here, a is m and b is 2. Therefore, m² - 4 can be factored as (m - 2)(m + 2). So, substituting back into the previous expression, the entire factorization would be m times (m - 2) times (m + 2).Let me verify that. Let's multiply it out step by step. First, multiply (m - 2)(m + 2). That's m² + 2m - 2m - 4, which simplifies to m² - 4. Then, multiply that by m: m*(m² - 4) = m³ - 4m. Yes, that's the original expression. So, the factorization is correct.But wait, let me check again. So, first factor out the GCF, which was m, then factor the difference of squares. Yep, that seems right. So, the factors are m, (m - 2), and (m + 2). So, putting it all together, the factorization of m³ - 4m is m(m - 2)(m + 2).I think that's it. Let me just make sure there's no other way to factor it further. The factors are all linear or constants now, so I don't think there's any more factoring possible. So, the final answer should be m(m - 2)(m + 2).Final AnswerThe factorization of ( m^{3} - 4m ) is boxed{m(m-2)(m+2)}.</think>To factorize ( m^3 - 4m ), we start by factoring out the greatest common factor (GCF), which is ( m ):[m^3 - 4m = m(m^2 - 4)]Next, we recognize that ( m^2 - 4 ) is a difference of squares, which can be factored as ( (m - 2)(m + 2) ):[m(m^2 - 4) = m(m - 2)(m + 2)]Verifying by expanding the factors confirms the original expression. Therefore, the factorization of ( m^3 - 4m ) is:[boxed{m(m-2)(m+2)}],<think>Okay, so I need to factor the expression ( m^{3} - 4m ). Let me think about how to approach this. Factoring usually involves looking for common factors first, right? Let me check if there's a common factor in both terms.The expression is ( m^{3} - 4m ). Let's see, both terms have an 'm' in them. The first term is ( m^3 ), which is ( m times m times m ), and the second term is ( 4m ), which is ( 4 times m ). So, yes, they both have an 'm' as a common factor. I can factor that out first.So, factoring out an 'm', we get:( m(m^{2} - 4) ).Okay, that's the first step. Now, looking at the remaining expression inside the parentheses, which is ( m^{2} - 4 ). Hmm, I think this is a difference of squares. Let me recall, a difference of squares is something like ( a^{2} - b^{2} ), which factors into ( (a - b)(a + b) ).Is ( m^{2} - 4 ) a difference of squares? Well, ( m^{2} ) is a square of 'm', and 4 is a square of 2, since ( 2^2 = 4 ). So yes, ( m^{2} - 4 = m^{2} - 2^{2} ). Therefore, it should factor into ( (m - 2)(m + 2) ).Putting it all together, after factoring out the 'm', and then factoring the difference of squares, the entire expression becomes:( m(m - 2)(m + 2) ).Let me double-check to make sure I didn't make a mistake. Let's expand my answer to see if it gets back to the original expression.First, multiply ( (m - 2)(m + 2) ). That should give ( m^{2} + 2m - 2m - 4 ). The middle terms cancel out, leaving ( m^{2} - 4 ). Then, multiplying by 'm' gives ( m^{3} - 4m ), which matches the original expression. So that seems correct.Wait, just to be thorough, let me check each step again.1. Original expression: ( m^3 - 4m ).2. Factor out the greatest common factor, which is 'm': ( m(m^2 - 4) ).3. Recognize that ( m^2 - 4 ) is a difference of squares: ( (m)^2 - (2)^2 ).4. Factor as ( (m - 2)(m + 2) ).5. Combine all factors: ( m(m - 2)(m + 2) ).Expanding this:- First, ( (m - 2)(m + 2) = m^2 - 4 ).- Then, ( m(m^2 - 4) = m^3 - 4m ).Yep, that's the original expression. So the factorization is correct.I wonder if there's another way to factor it? Let's see. Maybe factor by grouping? But in this case, since there are only two terms, factoring out the GCF first is the right approach. If there were four terms, maybe grouping would work, but here with two terms, GCF and then difference of squares makes sense.Alternatively, could we factor it as a cubic expression in some other form? Well, cubics can sometimes be factored by sum or difference of cubes, but this isn't a sum or difference of cubes. The terms here are ( m^3 ) and ( 4m ). The first is a cube, but the second is linear in m. So probably the method I used is the correct one.Another check: if I plug in a value for m, say m = 1, into the original expression and the factored form, they should give the same result.Original expression when m = 1: ( 1^3 - 4(1) = 1 - 4 = -3 ).Factored form when m = 1: ( 1(1 - 2)(1 + 2) = 1(-1)(3) = -3 ). Same result.Try m = 2: Original expression: ( 8 - 8 = 0 ). Factored form: ( 2(0)(4) = 0 ). Also matches.How about m = 3: Original: ( 27 - 12 = 15 ). Factored: ( 3(1)(5) = 15 ). Correct again.Alright, so all my checks indicate that the factorization is correct. Therefore, the answer should be ( m(m - 2)(m + 2) ).Final AnswerThe factorization of ( m^{3} - 4m ) is boxed{m(m - 2)(m + 2)}.</think>To factor the expression ( m^3 - 4m ):1. Factor out the greatest common factor (GCF): Both terms ( m^3 ) and ( -4m ) have a common factor of ( m ). [ m^3 - 4m = m(m^2 - 4) ]2. Factor the difference of squares: The remaining expression ( m^2 - 4 ) is a difference of squares, since ( m^2 - 4 = m^2 - 2^2 ). A difference of squares factors into ( (a - b)(a + b) ). [ m^2 - 4 = (m - 2)(m + 2) ]3. Combine the factors: Putting it all together, we get: [ m(m^2 - 4) = m(m - 2)(m + 2) ]Verification by expanding the factors confirms the original expression:- ( (m - 2)(m + 2) = m^2 - 4 )- ( m(m^2 - 4) = m^3 - 4m )Thus, the factorization of ( m^3 - 4m ) is boxed{m(m - 2)(m + 2)}.]

🔑:To factorize the given expression m^{3}-4m, we follow these steps:1. First, we identify a common factor in both terms. Here, the common factor is m.2. We factor out the common factor: m^{3}-4m = m(m^{2}-4).3. Next, we recognize that the expression inside the parentheses is a difference of squares, which can be factored further: m^{2}-4 = (m+2)(m-2).4. Substituting this back into our expression, we get: m(m^{2}-4) = m(m+2)(m-2).Therefore, the factorized form of the given expression is boxed{m(m+2)(m-2)}.

🧠:[<think>Okay, let me try to work through this problem step by step. First, let's parse the problem statement again to make sure I understand all parts correctly.We have triangle ABC with known information:1. sin A = √5 / 52. b = 2a cos AThen, part (Ⅰ) asks: If ac = 5, find the area of ABC.Part (Ⅱ) asks: If angle B is acute, find the value of sin C.Starting with part (Ⅰ). Let's note down all given information and see how to relate them.First, in triangle ABC, sides are opposite angles as usual: side a opposite angle A, side b opposite angle B, and side c opposite angle C.Given that sin A = √5 / 5. Since we are dealing with triangles, maybe we can find other trigonometric ratios related to angle A. Since sin A is given, we can find cos A using the Pythagorean identity.Recall that sin²θ + cos²θ = 1. So, cos²A = 1 - sin²A = 1 - (√5 / 5)² = 1 - (5/25) = 1 - 1/5 = 4/5. Therefore, cos A = ±2√5 / 5. However, in a triangle, all angles are between 0 and π radians (0 and 180 degrees), so cosine can be positive or negative depending on whether the angle is acute or obtuse. Since angle A has sin A = √5 /5, which is positive, and since sin is positive in both the first and second quadrants, angle A could be acute or obtuse. However, since cos A is given as part of the relation b = 2a cos A, and sides in a triangle are positive lengths, then 2a cos A must be positive. Since a is a positive length, this implies cos A must be positive. Therefore, angle A is acute, so cos A = 2√5 / 5.So, cos A = 2√5 / 5. Then, from the given relation, b = 2a cos A. Substituting cos A here:b = 2a * (2√5 /5) = (4√5 /5) a.So, side b is (4√5 /5) times side a.Also, we are given that ac = 5. So, the product of sides a and c is 5.Our goal is to find the area of triangle ABC. The area of a triangle can be found using several formulas. One common formula is (1/2)ab sin C, but since we might not know angle C here, perhaps another formula would be better. Alternatively, using sides and angles, maybe we can use the formula:Area = (1/2) * b * c * sin AWait, let's see. If we have sides b and c, and the angle between them, which would be angle A, then the area is (1/2)*b*c*sin A. But we need to check if angle A is between sides b and c.Wait, in triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. Therefore, angle A is between sides b and c. Therefore, yes, the area is (1/2)*b*c*sin A. That seems promising.But to use this formula, we need to know b and c. However, we know that ac = 5, so c = 5/a. Also, from earlier, we have b = (4√5 /5) a. Therefore, if we can express b and c in terms of a, then plug into the area formula, and see if the a cancels out.Let's try that.Express b and c in terms of a:b = (4√5 /5) ac = 5 / aTherefore, the area would be (1/2) * b * c * sin A = (1/2) * (4√5 /5 a) * (5 / a) * (√5 /5)Let's compute this step by step.First, multiply (4√5 /5 a) and (5 / a):(4√5 /5 a) * (5 / a) = 4√5 /5 * 5 /1 * a /a = 4√5 /1 * 1 = 4√5Then multiply that by sin A which is √5 /5:4√5 * √5 /5 = 4 * (√5 * √5) /5 = 4 * 5 /5 = 4Then multiply by (1/2):(1/2) * 4 = 2So, the area is 2. Wait, that seems straightforward. Let me check that again.Alternatively, maybe there's a different approach. Let's verify using another formula.Another formula for area is (1/2) * a * b * sin C, but we might not know angle C here. Alternatively, using Heron's formula, but that requires knowing all three sides, which we don't have yet. Alternatively, use the formula involving the sine of angle A and sides a and c.Wait, the formula (1/2)*a*c*sin B. Hmm, but we don't know angle B. Wait, perhaps using the Law of Sines or Law of Cosines.Wait, let's recall the Law of Sines: a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius.But perhaps we can relate the sides using Law of Sines.Given that, we can express b and c in terms of a. Since we have a, b, and c related through ac =5, and also b = (4√5 /5) a.So, from ac =5, c = 5 / a.From Law of Sines: a / sin A = b / sin B = c / sin C.But since we have expressions for b and c in terms of a, maybe we can find sin B and sin C in terms of a, but I'm not sure if that's necessary here.Alternatively, maybe use the Law of Cosines to find another relation.But given that we already computed the area as 2 using the formula (1/2)*b*c*sin A, and that worked out nicely with all variables cancelling, perhaps that's the correct answer. Let me check again.Given:Area = (1/2)*b*c*sin AWe know:b = (4√5 /5) ac = 5 / asin A = √5 /5So substituting:(1/2) * (4√5 /5 a) * (5 / a) * (√5 /5)First, let's handle the a terms: (4√5 /5 a) * (5 / a) = 4√5 /5 * 5 /1 * a /a = 4√5 * 1 = 4√5Then multiply by sin A (√5 /5):4√5 * √5 /5 = 4 * 5 /5 = 4Multiply by 1/2: 4 * 1/2 = 2Yes, that's correct. So the area is 2. So part (Ⅰ) answer is 2. Let me just verify once more if there's another approach.Alternatively, since we know sides a, b, c in terms of a, perhaps using Heron's formula. Let's see:First, express all sides in terms of a:a = ab = (4√5 /5) ac = 5 / aThen, the semi-perimeter s = (a + b + c)/2 = [a + (4√5 /5 a) + (5 /a)] /2But this seems complicated because we have a and 1/a terms. It would be messy to compute the area via Heron's formula here. So perhaps the first method is better.Alternatively, another area formula is (1/2)*a*b*sin C, but we don't know angle C here. Alternatively, (1/2)*a*c*sin B, but again, we don't know angle B.But given that angle A is between sides b and c, the formula (1/2)*b*c*sin A is the correct one, and it worked out nicely. So I think that's solid.Therefore, part (Ⅰ) answer is 2.Moving on to part (Ⅱ): If angle B is acute, find the value of sin C.Okay, so here we need to find sin C given that angle B is acute. Let's recall that in a triangle, the sum of angles is π radians. So A + B + C = π. Therefore, knowing angles A and B would allow us to find angle C. But we only have information about angle B being acute. Let's see how we can relate this.First, we still know sin A = √5 /5 and cos A = 2√5 /5, so angle A is known in terms of its sine and cosine. Also, we have the relation b = 2a cos A, which gives us a relation between sides a and b. So perhaps using Law of Sines or Cosines.Let me try to use Law of Sines. From Law of Sines:a / sin A = b / sin B = c / sin CWe know that b = 2a cos A. Let's substitute that into the Law of Sines:From a / sin A = b / sin B => a / sin A = (2a cos A) / sin BWe can cancel a from both sides:1 / sin A = 2 cos A / sin B => sin B = 2 sin A cos ABut 2 sin A cos A is sin 2A. Therefore, sin B = sin 2A.So sin B = sin 2A. Then, this implies that either B = 2A or B = π - 2A. However, since angle B is acute, and angles in a triangle are less than π, we can consider the possible cases.First, if B = 2A, then we can check if that leads to an acute angle B. If B = π - 2A, we need to check if that angle is acute.But let's compute angle A first. Since sin A = √5 /5 ≈ 0.447, which is approximately 26.565 degrees (since sin 26.565 ≈ 0.447). Then 2A would be approximately 53.13 degrees, which is acute. On the other hand, π - 2A would be 180 - 53.13 = 126.87 degrees, which is obtuse. But since angle B is given to be acute, we must have B = 2A.Therefore, angle B = 2A. Then angle C = π - A - B = π - A - 2A = π - 3A.Therefore, angle C = π - 3A. Then, sin C = sin(π - 3A) = sin 3A. So, sin C = sin 3A. So, if we can compute sin 3A, we can find sin C.Alternatively, we can use trigonometric identities to express sin 3A in terms of sin A and cos A. Let's recall that sin 3θ = 3 sin θ - 4 sin³θ.Given that sin A = √5 /5, and cos A = 2√5 /5. Let's compute sin 3A:sin 3A = 3 sin A - 4 sin³ASubstituting sin A = √5 /5:= 3*(√5 /5) - 4*( (√5 /5)^3 )Compute each term:First term: 3*(√5 /5) = (3√5)/5Second term: 4*( (√5)^3 / 5^3 ) = 4*(5√5 / 125) = 4*(√5 /25) = (4√5)/25Therefore, sin 3A = (3√5)/5 - (4√5)/25 = (15√5 - 4√5)/25 = (11√5)/25Therefore, sin C = sin 3A = 11√5 /25.But wait, let me verify if angle C = π - 3A. Since angle B = 2A, which is acute, then 2A < π/2, so A < π/4. Wait, but angle A has sin A = √5 /5 ≈ 0.447, which is sin inverse of approximately 26.565 degrees. So angle A is about 26.565 degrees, which is less than 45 degrees (π/4 radians), so 2A is approximately 53.13 degrees, which is acute. Then angle C = π - 3A ≈ 180 - 79.695 ≈ 100.305 degrees, which is obtuse. Therefore, sin C is sin(100.305°) ≈ sin(π - 79.695°) = sin(79.695°) ≈ 0.984. But according to our calculation, sin 3A = 11√5 /25 ≈ 11*2.236 /25 ≈ 24.596 /25 ≈ 0.983, which matches.But let's check if there is another approach.Alternatively, using the Law of Cosines. If angle B is acute, then cos B > 0.We can also try to express angle C in terms of other angles or sides.Alternatively, since we have relations between sides a, b, c.We have b = 2a cos A, so using Law of Cosines on angle B:Wait, angle B is acute. Let's recall that in a triangle, the square of the side opposite an acute angle is less than the sum of the squares of the other two sides. So, for angle B to be acute, a² + c² > b².But maybe that can be another way to approach the problem.But perhaps the first method via Law of Sines is more straightforward.Wait, but in the earlier step, we found that sin B = sin 2A, and since angle B is acute, then B = 2A. Therefore, angle C = π - 3A, so sin C = sin 3A. Then, using the identity sin 3A = 3 sin A - 4 sin³ A, which gives 11√5 /25. That seems correct.Let me verify with another identity. Alternatively, express sin 3A as sin(2A + A) = sin 2A cos A + cos 2A sin A.Compute sin 2A and cos 2A first.sin 2A = 2 sin A cos A = 2*(√5/5)*(2√5/5) = 2*(5*2)/25 = 20/25 = 4/5.cos 2A = 1 - 2 sin²A = 1 - 2*(5/25) = 1 - 2*(1/5) = 1 - 2/5 = 3/5.Therefore, sin 3A = sin(2A + A) = sin 2A cos A + cos 2A sin A= (4/5)*(2√5/5) + (3/5)*(√5/5)= (8√5)/25 + (3√5)/25= (11√5)/25Which matches the earlier result.Therefore, sin C = 11√5 /25.Therefore, the answer to part (Ⅱ) is 11√5 /25.But wait, let's confirm if angle C is indeed π - 3A. Since angle B = 2A, then angles sum to π: A + B + C = A + 2A + C = 3A + C = π. Therefore, C = π - 3A. So yes, that's correct. Therefore, sin C = sin(3A), and as shown above, sin(3A) = 11√5 /25.Therefore, the answer is 11√5 /25.Before finalizing, let's check if there are any alternative interpretations or possible mistakes.Wait, when we derived sin B = sin 2A, we concluded that B = 2A because angle B is acute. But sin B = sin 2A could also mean that B = π - 2A. However, if B = π - 2A, then since angle A is acute (as cos A is positive), π - 2A would be greater than π - 2*(π/2) = π - π = 0, but actually, if angle A is acute (less than π/2), then 2A < π, so π - 2A would be greater than 0, but whether it's acute or not depends on the value.Given that angle B is acute, so B < π/2. Let's see if π - 2A < π/2. That would require π - 2A < π/2 => π/2 < 2A => A > π/4. But we know that sin A = √5 /5 ≈ 0.447, so angle A is approximately arcsin(0.447) ≈ 26.565 degrees, which is less than 45 degrees (π/4 radians). Therefore, π - 2A ≈ 180 - 53.13 ≈ 126.87 degrees, which is obtuse. Therefore, in that case, angle B cannot be π - 2A, since that would be obtuse, but angle B is given to be acute. Therefore, only possible solution is B = 2A.Therefore, our conclusion that angle C = π - 3A is correct, and sin C = sin 3A = 11√5 /25.Therefore, part (Ⅱ) answer is 11√5 /25.Final AnswerⅠ. The area of triangle (ABC) is boxed{2}.Ⅱ. The value of (sin C) is boxed{dfrac{11sqrt{5}}{25}}.</think>Given triangle (ABC) with (sin A = frac{sqrt{5}}{5}) and (b = 2a cos A):Ⅰ. If (ac = 5), find the area of (ABC).1. Calculate (cos A) using the Pythagorean identity: [ cos^2 A = 1 - sin^2 A = 1 - left(frac{sqrt{5}}{5}right)^2 = 1 - frac{5}{25} = frac{20}{25} = frac{4}{5} ] Thus, (cos A = frac{2sqrt{5}}{5}).2. Substitute (cos A) into (b = 2a cos A): [ b = 2a cdot frac{2sqrt{5}}{5} = frac{4sqrt{5}}{5}a ]3. Given (ac = 5), express (c) as (c = frac{5}{a}).4. Use the area formula (text{Area} = frac{1}{2}bc sin A): [ text{Area} = frac{1}{2} left(frac{4sqrt{5}}{5}aright) left(frac{5}{a}right) left(frac{sqrt{5}}{5}right) ] Simplify the expression: [ text{Area} = frac{1}{2} cdot 4sqrt{5} cdot frac{sqrt{5}}{5} = frac{1}{2} cdot 4 cdot frac{5}{5} = frac{1}{2} cdot 4 = 2 ]Ⅱ. If angle (B) is acute, find the value of (sin C).1. Using the Law of Sines: [ frac{a}{sin A} = frac{b}{sin B} implies sin B = 2 sin A cos A = sin 2A ] Since (B) is acute, (B = 2A).2. Calculate angle (C): [ C = pi - A - B = pi - A - 2A = pi - 3A ] Thus, (sin C = sin(3A)).3. Use the triple angle identity for sine: [ sin 3A = 3 sin A - 4 sin^3 A ] Substitute (sin A = frac{sqrt{5}}{5}): [ sin 3A = 3 left(frac{sqrt{5}}{5}right) - 4 left(frac{sqrt{5}}{5}right)^3 = frac{3sqrt{5}}{5} - 4 left(frac{5sqrt{5}}{125}right) = frac{3sqrt{5}}{5} - frac{4sqrt{5}}{25} = frac{11sqrt{5}}{25} ] Final AnswerⅠ. The area of triangle (ABC) is (boxed{2}).Ⅱ. The value of (sin C) is (boxed{dfrac{11sqrt{5}}{25}}).,<think>Okay, let's tackle this problem step by step. We have triangle ABC with given conditions: sin A = √5 / 5 and b = 2a cos A. Part (Ⅰ) asks for the area when ac = 5, and part (Ⅱ) wants sin C if angle B is acute. Let's start with part (Ⅰ).First, let me recall some trigonometric relationships in triangles. The Law of Sines and Law of Cosines might be useful here. Since we know sin A, maybe we can find other trigonometric functions related to angle A. Given sin A = √5 / 5, we can find cos A using the Pythagorean identity. Remember that sin² A + cos² A = 1. So cos² A = 1 - sin² A = 1 - (5/25) = 1 - 1/5 = 4/5. Therefore, cos A = 2/√5 or -2/√5. However, since A is an angle in a triangle, it must be between 0 and 180 degrees, so cos A could be positive or negative. But in a triangle, angles are between 0 and 180, so cosine is positive for acute angles (0-90) and negative for obtuse angles (90-180). But we might need to consider the given condition b = 2a cos A. Let's see.If cos A were negative, then b would be negative, which is impossible because side lengths are positive. Therefore, cos A must be positive, so angle A is acute. Therefore, cos A = 2/√5.So, cos A = 2/√5. Then, from the given condition, b = 2a cos A. Plugging in cos A, we get b = 2a*(2/√5) = 4a / √5. So b = (4a)/√5.Now, we need to relate sides a, b, c. Also, we know that ac = 5. So, c = 5/a.Let me write down the Law of Sines for triangle ABC. It states that a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius.Given sin A = √5 / 5, so a / (√5 / 5) = 5a / √5 = √5 a. Therefore, the Law of Sines ratio is √5 a. So, b = √5 a * sin B and c = √5 a * sin C.But we already have expressions for b and c in terms of a. From above, b = 4a / √5 and c = 5/a.Therefore, from Law of Sines:b = √5 a * sin B => 4a / √5 = √5 a * sin B. Divide both sides by a: 4 / √5 = √5 sin B. Multiply both sides by √5: 4 = 5 sin B => sin B = 4/5. So sin B = 4/5.Similarly, for c: c = √5 a * sin C => 5/a = √5 a * sin C. Multiply both sides by a: 5 = √5 a² sin C => sin C = 5 / (√5 a²) = 5 / (√5 a²) = √5 / a².But we need to find the area of triangle ABC. The area can be calculated using several formulas. One common formula is (1/2)ab sin C, or (1/2)bc sin A, etc. Let's see which sides and angles we know.We have a, b, c related, and some angles. Let me see if we can use the formula (1/2)ab sin C. But we need sin C. Alternatively, since we have sides a, b, c and angles related, maybe using Heron's formula? But Heron's formula requires knowing all three sides. Alternatively, maybe the formula (1/2)bc sin A? Let's check that.If we use (1/2) * b * c * sin A, we know sin A, and we have expressions for b and c in terms of a. Let's substitute:Area = (1/2) * b * c * sin A = (1/2) * (4a / √5) * (5 / a) * (√5 / 5). Let's compute this step by step.First, (4a / √5) * (5 / a) = (4a * 5) / (√5 * a) = 20 / √5 = 4√5. Then multiply by sin A (√5 / 5): 4√5 * (√5 /5 ) = 4 * 5 / 5 = 4. Then multiply by 1/2: 4 * 1/2 = 2. So area is 2. Wait, that seems straightforward. Let me check if that's correct.Wait, let's verify each step:First, b = 4a / √5, c = 5/a. Then b * c = (4a / √5) * (5 / a) = (4 * 5) / √5 = 20 / √5 = 4√5. Then multiply by sin A (√5 /5): 4√5 * √5 /5 = 4*(5)/5 = 4. Then multiply by 1/2: 4 * 1/2 = 2. Yes, that gives area as 2. So the area is 2. That seems straightforward. Let me see if there's another way to confirm this.Alternatively, using Law of Sines. We found sin B = 4/5. Since angle B is acute (since sin B = 4/5, cos B would be 3/5). Then angle C = 180 - A - B. Let's compute angles A, B, C.First, angle A: sin A = √5 /5, so A = arcsin(√5/5). Similarly, angle B: arcsin(4/5). Then angle C = 180 - angle A - angle B. Then we can compute sin C using sine addition formula. But that might be more complicated. Alternatively, since we have sides, maybe using Law of Cosines.Alternatively, since we know sides a, b, c, we can compute the area. Wait, but we have expressions for b and c in terms of a, and ac =5. So, if c =5/a, then we can write all sides in terms of a. Let's see:Given a, b = 4a / √5, c = 5/a. Then, using Heron's formula:First, compute the semi-perimeter s = (a + b + c)/2 = (a + 4a/√5 + 5/a)/2. That seems messy, but maybe we can proceed. However, this might be more complicated. Alternatively, since we found the area via (1/2)bc sin A, which gave 2, and that seems consistent, maybe that's correct.Alternatively, check if the area via another formula gives the same result. For example, using (1/2)ab sin C. We found sin C = √5 / a². So, (1/2)*a*b*sin C = (1/2)*a*(4a/√5)*(√5 / a²) = (1/2)*(4a / √5)*(√5 / a) = (1/2)*4*(a / √5)*(√5 /a ) = (1/2)*4 = 2. Same result. So that confirms it. Therefore, the area is 2.So part (Ⅰ) answer is 2.Now, part (Ⅱ): If angle B is acute, find the value of sin C.Wait, in part (Ⅰ), we found sin B = 4/5. But if angle B is acute, then cos B is positive, which would be 3/5 (since sin² B + cos² B =1 => cos B = 3/5). However, in part (Ⅰ), angle B might not necessarily be acute? Wait, in part (Ⅰ), we had sin B =4/5. But angle B could be acute or obtuse, since sine is positive in both first and second quadrants. However, in part (Ⅱ), we are told angle B is acute, so cos B = 3/5. So, we need to use this information to find sin C.But let me recall in part (Ⅰ), angle B's cosine was positive, but in part (Ⅱ), angle B is specified to be acute, so cos B is positive. So perhaps in part (Ⅱ), we have additional constraints.But wait, part (Ⅱ) is a separate question. It might be using the same triangle, but with an added condition that angle B is acute. Wait, but in the original problem, we have triangle ABC with sin A = √5 /5 and b = 2a cos A. Then part (Ⅰ) adds the condition ac =5, and part (Ⅱ) adds angle B is acute. So perhaps part (Ⅱ) is a different scenario, not necessarily with ac=5. Wait, the problem says "In triangle ABC, we know that sin A= √5 /5 and b=2a cos A. (Ⅰ) If ac=5, find the area... (Ⅱ) If angle B is acute, find sin C." So, it's the same triangle, but with different additional conditions for each part. So in part (Ⅱ), angle B is acute, but ac is not necessarily 5. So we have to find sin C under the condition that angle B is acute.So, let's re-examine the given information for part (Ⅱ):Given triangle ABC with sin A = √5 /5, and b = 2a cos A. Additionally, angle B is acute. Need to find sin C.First, from sin A = √5 /5, we found earlier that cos A = 2/√5 (since angle A must be acute because b = 2a cos A, which would be positive). So angle A is acute, cos A = 2/√5. Therefore, side b = 2a*(2/√5) = 4a / √5, as before.Now, using the Law of Sines: a / sin A = b / sin B = c / sin C.From this, since a / sin A = √5 a / (√5 /5) = 5a / √5 = √5 a. Wait, sin A = √5 /5, so a / (√5 /5) = 5a / √5 = √5 a. Therefore, the ratio is √5 a. Therefore, b = √5 a sin B, c = √5 a sin C.But we already have b =4a / √5, so √5 a sin B = 4a / √5. Divide both sides by a: √5 sin B = 4 / √5 => sin B = 4 /5. So sin B =4/5, same as in part (Ⅰ). But in part (Ⅱ), angle B is acute, so cos B =3/5. In part (Ⅰ), angle B could be acute or obtuse, but since in part (Ⅰ) we didn't have that condition, but in part (Ⅱ) we do. So in part (Ⅰ), even if angle B could be obtuse, but given that we found the area without needing to know whether B is acute or not, perhaps the area remains the same. But for part (Ⅱ), we need to ensure that angle B is acute, so cos B =3/5.Now, to find sin C. Let's recall that in a triangle, angles sum to 180 degrees. So angle C = 180 - A - B. Therefore, sin C = sin (A + B). Because sin(180 - x) = sin x, so sin C = sin (A + B). Then, using sine addition formula: sin(A + B) = sin A cos B + cos A sin B.We know sin A = √5 /5, cos A =2/√5, sin B =4/5, cos B =3/5 (since angle B is acute). Therefore:sin C = sin A cos B + cos A sin B = (√5 /5)(3/5) + (2/√5)(4/5)Compute each term:First term: (√5 /5)(3/5) = 3√5 /25Second term: (2/√5)(4/5) = 8 / (5√5) = 8√5 /25Therefore, sin C = 3√5 /25 +8√5 /25 =11√5 /25So sin C =11√5 /25. Let me check the calculation again.sin C = sin(A + B) = sin A cos B + cos A sin BPlug in the values:sin A = √5/5, cos B =3/5: (√5/5)(3/5)= 3√5/25cos A=2/√5, sin B=4/5: (2/√5)(4/5)=8/(5√5)=8√5/25Adding: 3√5/25 +8√5/25=11√5/25. Yes, that's correct.Alternatively, could we use the Law of Cosines? Let me see.We know sides a, b, c in terms of a. Wait, we have b=4a/√5, and from Law of Sines, c=√5 a sin C. But we need to find sin C.Alternatively, using the Law of Cosines on angle B: since angle B is acute, cos B =3/5. Then:cos B = (a² + c² - b²)/(2ac). Let's plug in the values.cos B=3/5=(a² +c² -b²)/(2ac). We have b=4a/√5, so b²=16a²/5.Thus:3/5=(a² +c² -16a²/5)/(2ac)Multiply numerator and denominator:Multiply numerator: a² +c² -16a²/5= (5a² +5c² -16a²)/5= (-11a² +5c²)/5Thus, 3/5= (-11a² +5c²)/(5*2ac)= (-11a² +5c²)/(10ac)Multiply both sides by 10ac:3/5 *10ac=6ac= -11a² +5c²Rearrange:5c² -11a² -6ac=0But this is a quadratic in c: 5c² -6a c -11a²=0. Let's solve for c in terms of a.Using quadratic formula:c = [6a ±√(36a² +220a²)]/(2*5)= [6a ±√(256a²)]/10= [6a ±16a]/10Thus:c=(6a +16a)/10=22a/10=11a/5 or c=(6a -16a)/10=(-10a)/10=-a. Since c is positive, we discard the negative solution. Thus c=11a/5.Therefore, c=11a/5. Then, from Law of Sines: c=√5 a sin C => 11a/5=√5 a sin C => sin C=11/(5√5)=11√5/25. Same result as before. So that confirms sin C=11√5/25.Therefore, the answer for part (Ⅱ) is 11√5 /25.So, summarizing:Ⅰ. Area is 2.Ⅱ. sin C=11√5/25.Final AnswerⅠ. boxed{2}Ⅱ. boxed{dfrac{11sqrt{5}}{25}}</think>Given triangle (ABC) with (sin A = frac{sqrt{5}}{5}) and (b = 2a cos A):(Ⅰ) If (ac = 5), find the area of (ABC).1. Calculate (cos A) using the Pythagorean identity: [ cos^2 A = 1 - sin^2 A = 1 - left(frac{sqrt{5}}{5}right)^2 = frac{4}{5} implies cos A = frac{2}{sqrt{5}} ]2. Given (b = 2a cos A), substitute (cos A): [ b = 2a cdot frac{2}{sqrt{5}} = frac{4a}{sqrt{5}} ]3. Given (ac = 5), we have (c = frac{5}{a}).4. Use the formula for the area of a triangle: [ text{Area} = frac{1}{2} cdot b cdot c cdot sin A ]5. Substitute the values: [ text{Area} = frac{1}{2} cdot frac{4a}{sqrt{5}} cdot frac{5}{a} cdot frac{sqrt{5}}{5} = frac{1}{2} cdot 4sqrt{5} cdot frac{sqrt{5}}{5} = frac{1}{2} cdot 4 = 2 ](Ⅱ) If angle (B) is acute, find the value of (sin C).1. From the Law of Sines, (sin B = frac{4}{5}). Since angle (B) is acute, (cos B = frac{3}{5}).2. Use the sine addition formula for (sin C = sin(180^circ - A - B) = sin(A + B)): [ sin C = sin(A + B) = sin A cos B + cos A sin B ]3. Substitute the known values: [ sin C = left(frac{sqrt{5}}{5} cdot frac{3}{5}right) + left(frac{2}{sqrt{5}} cdot frac{4}{5}right) ]4. Simplify each term: [ sin C = frac{3sqrt{5}}{25} + frac{8sqrt{5}}{25} = frac{11sqrt{5}}{25} ] Final AnswerⅠ. boxed{2}Ⅱ. boxed{dfrac{11sqrt{5}}{25}}]

🔑:(Ⅰ) Based on the given condition b=2acos A, we can apply the Sine Rule frac{a}{sin A} = frac{b}{sin B}, which gives us sin B = 2 sin A cos A. Since cos A = frac{b}{2a} > 0 and sin A= frac{sqrt{5}}{5}, we can find cos A using the Pythagorean identity cos^2 A = 1 - sin^2 A:cos A = sqrt{1 - sin^2 A} = sqrt{1 - left( frac{sqrt{5}}{5} right)^2} = frac{2sqrt{5}}{5}.Now we can determine sin B:sin B = 2 times frac{sqrt{5}}{5} times frac{2sqrt{5}}{5} = frac{4}{5}.The area S of triangle ABC can now be computed using the formula for the area of a triangle with sides a and c and the sine of the included angle B:S = frac{1}{2}acsin B = frac{1}{2} times 5 times frac{4}{5} = boxed{2}.(Ⅱ) Since we know sin B = frac{4}{5} and angle B is acute, we can find cos B again using the Pythagorean identity:cos B = sqrt{1 - sin^2 B} = sqrt{1 - left( frac{4}{5} right)^2} = frac{3}{5}.Now we can compute sin C by using the sum of angles in a triangle (C=pi - A - B) and the sine sum formula:sin C = sin(pi - A - B) = sin(A+B) = sin A cos B + cos A sin B = frac{sqrt{5}}{5} times frac{3}{5} + frac{2sqrt{5}}{5} times frac{4}{5} = boxed{frac{11sqrt{5}}{25}}.

🧠:[<think>Okay, let's see. I need to figure out which of these operations A, B, C, D is correct. Hmm, let me go through each one step by step. Starting with option A: 1/a + 1/b equals 1/(a + b). Wait, I remember that when adding fractions, you need a common denominator. So, 1/a + 1/b should be (b + a)/ab, right? Because you multiply numerator and denominator to get the same denominator. So 1/a is b/(ab) and 1/b is a/(ab), so adding them gives (a + b)/ab. But the answer here is 1/(a + b). That doesn't seem right. So A is incorrect. Moving on to option B: y/(x - y) - y/(y - x) equals 0. Let me work this out. First, notice that (y - x) is the same as -(x - y). So y/(x - y) minus y/(-(x - y)) would be y/(x - y) + y/(x - y) because subtracting a negative is adding. So that becomes (y + y)/(x - y) = 2y/(x - y). But the problem says it equals 0. Hmm, so unless y is zero, which isn't specified here, this doesn't hold. So B is not correct either. Wait, maybe I made a mistake here. Let me check again. Original expression: y/(x - y) - y/(y - x). Let's rewrite the second term's denominator as -(x - y). So the expression becomes y/(x - y) - y/(- (x - y)) = y/(x - y) + y/(x - y) = 2y/(x - y). So unless y is zero, which isn't given, this can't be zero. So B is incorrect. Okay, moving to option C: 1 divided by (b/a) multiplied by (a/b) equals a²/b². Let's parse this. 1 ÷ (b/a) is the same as 1 multiplied by (a/b), so that becomes (a/b). Then multiplying by (a/b) would be (a/b) * (a/b) = a²/b². So that seems correct. Let me verify step by step. 1 divided by (b/a) is indeed a/b because dividing by a fraction is multiplying by its reciprocal. Then multiplying by (a/b) gives (a/b)*(a/b) = a²/b². So yes, option C is correct. But wait, let me check the order of operations. The expression is 1 ÷ (b/a) × (a/b). Since division and multiplication are left associative, you do them from left to right. So first 1 ÷ (b/a) = a/b, then multiplied by (a/b) gives a²/b². So yes, that's correct. So C is correct. But let me check D as well, just to be thorough. Option D: (x² + y²)/(x + y) equals x + y. Let's test this with numbers. Let x = 1 and y = 1. Then numerator is 1 + 1 = 2, denominator is 2, so 2/2 = 1. But x + y is 2. So 1 ≠ 2. So D is not correct. Alternatively, algebraically, x² + y² doesn't factor into (x + y)^2. Because (x + y)^2 is x² + 2xy + y². So unless 2xy is zero, which isn't the case in general, (x² + y²)/(x + y) is not x + y. So D is incorrect. Therefore, the correct answer is C. Let me just verify once more each step to make sure I didn't make a mistake.A: Definitely wrong, common denominator needed.B: Ends up being 2y/(x - y), not zero.C: Correct steps, order of operations followed properly.D: Test case shows it's wrong.So yes, C is correct.</think>To determine which operation is correct, we analyze each option:A: (frac{1}{a} + frac{1}{b} = frac{1}{a+b}) Adding fractions requires a common denominator: [frac{1}{a} + frac{1}{b} = frac{b + a}{ab} neq frac{1}{a+b}] A is incorrect.B: (frac{y}{x-y} - frac{y}{y-x} = 0) Notice (y - x = -(x - y)), so: [frac{y}{x-y} - frac{y}{-(x-y)} = frac{y}{x-y} + frac{y}{x-y} = frac{2y}{x-y} neq 0] B is incorrect.C: (1 div frac{b}{a} times frac{a}{b} = frac{a^2}{b^2}) Following order of operations: [1 div frac{b}{a} = frac{a}{b}, quad text{then} quad frac{a}{b} times frac{a}{b} = frac{a^2}{b^2}] C is correct.D: (frac{x^2 + y^2}{x + y} = x + y) Testing (x = 1) and (y = 1): [frac{1^2 + 1^2}{1 + 1} = frac{2}{2} = 1 neq 2 = 1 + 1] D is incorrect.Final Answer: boxed{C},<think>Okay, let's see. I have this problem here asking which of the operations A through D is correct. I need to check each one step by step. Hmm, where should I start? Maybe with option A.Option A says 1/a + 1/b equals 1/(a + b). Wait, that doesn't sound right. When adding fractions, you need a common denominator. The correct way is to add them by converting both fractions to have the denominator ab. So, 1/a is b/(ab) and 1/b is a/(ab). Adding them together gives (b + a)/ab, which is (a + b)/ab. But the answer given here is 1/(a + b), which is different. So A must be incorrect. Let me write that down: A is wrong because you can't just add denominators like that.Moving on to option B: y/(x - y) - y/(y - x) equals 0. Hmm, let's see. Let me look at the denominators. The first term is x - y, and the second is y - x. Notice that y - x is equal to -(x - y). So the second denominator is the negative of the first. Let me rewrite the second fraction. If I have -y/(y - x), but the denominator is y - x, which is -(x - y), so the second term is y/(-(x - y)) which is -y/(x - y). So the expression becomes y/(x - y) - (-y/(x - y)) which is y/(x - y) + y/(x - y). That equals 2y/(x - y). But the option says it's equal to 0, which is not correct unless 2y/(x - y) is 0. That would require y = 0, but there's no restriction given here. So unless y is zero, which isn't specified, this isn't zero. Therefore, option B is incorrect. Wait, but let me check again. Maybe I messed up the signs. Let's do it step by step.Original expression: y/(x - y) - y/(y - x). Let me factor out a negative from the second denominator: y/(x - y) - y/(-(x - y)) = y/(x - y) + y/(x - y) = 2y/(x - y). Yeah, that's what I got before. So definitely not zero. So B is wrong.Option C: 1 ÷ (b/a) × (a/b) equals a²/b². Let's parse the order of operations here. Division and multiplication have the same precedence, so we do them left to right. So first, 1 ÷ (b/a) is the same as 1 multiplied by (a/b), so that becomes a/b. Then multiply that by (a/b). So (a/b) × (a/b) is a²/b². Which matches option C. Let me confirm. 1 divided by (b/a) is indeed a/b, then multiplied by a/b gives a squared over b squared. So yes, C seems correct. But let me check once more. Maybe there's a trick with the order? If it were 1 ÷ (b/a × a/b), then it would be 1 ÷ 1, which is 1, but according to the original expression, it's 1 ÷ (b/a) × (a/b). So left to right. So first division, then multiplication. So yeah, a/b × a/b is a²/b². So C is correct. So maybe C is the answer? Wait, let me check the other options just to be sure.Option D: (x² + y²)/(x + y) equals x + y. Hmm, is that true? Let's test with numbers. Suppose x = 1 and y = 1. Then the left side is (1 + 1)/(1 + 1) = 2/2 = 1. The right side is 1 + 1 = 2. So 1 ≠ 2. Therefore, D is not correct in general. Alternatively, algebraically, x² + y² doesn't factor into (x + y)(something). Unless it's (x + y)^2 which is x² + 2xy + y². Since there's no 2xy term, it doesn't factor like that. Therefore, (x² + y²)/(x + y) cannot be simplified to x + y. So D is definitely wrong.So summarizing:A: Incorrect because adding reciprocals doesn't work that way.B: Incorrect because it simplifies to 2y/(x - y), not zero.C: Correct, simplifies to a²/b².D: Incorrect as shown by counterexample.Therefore, the correct answer is option C.Final Answerboxed{C}</think>Let's analyze each option step by step:Option A: (frac{1}{a} + frac{1}{b} = frac{1}{a + b})- To add (frac{1}{a}) and (frac{1}{b}), we need a common denominator (ab). Converting the fractions, we get (frac{b}{ab} + frac{a}{ab} = frac{a + b}{ab}). This is not equal to (frac{1}{a + b}). Therefore, A is incorrect.Option B: (frac{y}{x - y} - frac{y}{y - x} = 0)- Notice that (y - x = -(x - y)). Rewriting the second term, we get (frac{y}{x - y} - frac{y}{-(x - y)} = frac{y}{x - y} + frac{y}{x - y} = frac{2y}{x - y}). This is not equal to 0 unless (y = 0), which is not specified. Therefore, B is incorrect.Option C: (1 div frac{b}{a} times frac{a}{b} = frac{a^2}{b^2})- Following left-to-right order: (1 div frac{b}{a} = 1 times frac{a}{b} = frac{a}{b}). Then, (frac{a}{b} times frac{a}{b} = frac{a^2}{b^2}). This matches the given result. Therefore, C is correct.Option D: (frac{x^2 + y^2}{x + y} = x + y)- Testing with (x = 1) and (y = 1), the left side is (frac{1 + 1}{1 + 1} = 1) and the right side is (2). They are not equal. Algebraically, (x^2 + y^2) does not factor into ((x + y)^2). Therefore, D is incorrect.The correct answer is (boxed{C}).]

🔑:To evaluate each option step-by-step:Option A:We start with the left side of the equation and attempt to find a common denominator:frac{1}{a} + frac{1}{b} = frac{b}{ab} + frac{a}{ab} = frac{a+b}{ab}This shows that frac{1}{a} + frac{1}{b} neq frac{1}{a+b}, disproving option A.Option B:For this option, we simplify the expression given:begin{align*}frac{y}{x-y} - frac{y}{y-x} &= frac{y}{x-y} + frac{y}{x-y} quad text{(since y-x = -(x-y))} &= frac{2y}{x-y}end{align*}This shows that frac{y}{x-y} - frac{y}{y-x} neq 0, disproving option B.Option C:We simplify the expression step by step:begin{align*}1 div frac{b}{a} times frac{a}{b} &= 1 times frac{a}{b} times frac{a}{b} &= frac{a^2}{b^2}end{align*}This matches the operation given in option C, proving it correct.Option D:The expression frac{x^2+y^2}{x+y} cannot be simplified to x+y without additional conditions (e.g., x=y), which are not given. Therefore, option D does not match the question.Based on the step-by-step evaluation, the correct operation is:boxed{text{C}}

🧠:[<think>Okay, let me try to figure out this probability problem. So the question is: Among the numbers 1, 2, 3, 4, 5, if three numbers are randomly selected, what is the probability that at least one of the remaining two numbers is even? Hmm. Let me parse this.First, the total number of elements is five: 1, 2, 3, 4, 5. We are selecting three numbers at random. Then, after selecting three, we look at the remaining two numbers. The problem asks for the probability that at least one of these remaining two numbers is even.So, essentially, when we pick three numbers, what's the chance that in the two left behind, there's at least one even number. Alternatively, this could be phrased as: what's the probability that not both remaining numbers are odd. Since "at least one even" is the complement of "both odd."Maybe it's easier to compute the probability of the complement event and subtract it from 1. Let me think. So, the complement of "at least one even" is "both remaining numbers are odd." So if I can find the probability that both remaining numbers are odd, then subtract that from 1, I will get the desired probability.Let's verify that. The total possible outcomes when selecting three numbers are combinations of 5 taken 3 at a time. The number of such combinations is C(5,3) which is 10. So total possible ways is 10.Now, the complement event: both remaining numbers are odd. So, how many ways can this happen? The original set has three odd numbers: 1, 3, 5 and two even numbers: 2, 4. If both remaining numbers are odd, that means all three selected numbers must include both even numbers. Wait, let me think again.Wait, if we are leaving two numbers, and they are both odd, then the three selected numbers must include all the even numbers? Because there are only two even numbers: 2 and 4. So if we leave two numbers, both odd, then the remaining two numbers must be odd. So that means the three selected numbers must include both even numbers and one odd number. Because there are three odd numbers in total (1,3,5). So if we leave two odd numbers, then in the selected three numbers, we have both even numbers (2 and 4) and one odd number. So the number of such combinations would be the number of ways to choose both even numbers and one odd number.But there are two even numbers. So to include both of them in the selected three, we have to choose both evens (only one way, since there are exactly two even numbers) and one odd. There are three odd numbers, so C(3,1) = 3. So the number of such combinations is 3. Therefore, the probability that both remaining numbers are odd is 3/10. Therefore, the probability that at least one remaining number is even is 1 - 3/10 = 7/10.Wait, is that correct? Let me double-check. Alternatively, maybe I can approach it directly. The total number of possible remaining two numbers is C(5,2) = 10, which is the same as selecting three numbers. So the problem is equivalent to looking at all possible pairs of remaining numbers and seeing how many of them have at least one even.Alternatively, since the remaining two numbers are determined by the three selected numbers, it's equivalent. So another approach: how many pairs of remaining numbers have at least one even. The total number of pairs is 10. The pairs that have at least one even would be total pairs minus the pairs that have both odds. The number of pairs with both odds is C(3,2) = 3 (since there are three odd numbers). So the number of pairs with at least one even is 10 - 3 = 7. Therefore, the probability is 7/10. So this confirms the previous result.Alternatively, let's think step by step. The numbers are 1,2,3,4,5. Three are selected, two are left. The two left can be any of the combinations. The question is about the two left. So we can model this as: when choosing three numbers, what's the probability that in the two not chosen, there's at least one even.Alternatively, think of the two numbers not chosen. The possible pairs not chosen are all possible pairs from the five numbers. So there are C(5,2) = 10 possible pairs. Out of these 10, how many have at least one even. The even numbers are 2 and 4. So pairs that include 2 or 4 or both.Alternatively, the pairs that have at least one even are the total pairs minus the pairs that have neither 2 nor 4. The pairs that have neither 2 nor 4 are the pairs consisting of 1,3,5. So C(3,2)=3. So pairs with at least one even are 10-3=7. Therefore, 7/10. So same answer.Therefore, the probability is 7/10.But let me check if there's another way to approach it. For instance, considering the selection of three numbers. The two numbers not selected must have at least one even. So the three numbers selected can contain at most one even number. Because if the selected numbers contain both even numbers, then the remaining two numbers would have zero even numbers (since there are only two even numbers total). Conversely, if the selected numbers contain one or zero even numbers, then the remaining numbers would have one or two even numbers, respectively. Wait, no. Wait: there are two even numbers in total. So if we select three numbers:Case 1: selected numbers include both even numbers (2 and 4). Then, the remaining two numbers are all odd (since there's no even numbers left). So in this case, the remaining two numbers are both odd.Case 2: selected numbers include one even number. Then, the remaining numbers include one even and one odd.Case 3: selected numbers include zero even numbers. Then, the remaining numbers include both even numbers (2 and 4) and one odd? Wait, no. If we selected three numbers with zero even numbers, that means all three selected numbers are odd (since only 1,3,5 are odd). Then, the remaining two numbers are 2 and 4, which are both even. So in this case, the remaining two numbers are both even.Wait, so in case 3, remaining numbers are both even, which would satisfy "at least one even." In case 2, remaining numbers are one even and one odd, which also satisfies "at least one even." Only in case 1, remaining numbers are both odd, which does not satisfy "at least one even."Therefore, the only time when the remaining two numbers have no evens is when the three selected numbers include both even numbers. So the number of such selections is the number of ways to choose both evens (2 and 4) and one odd. The number of ways to choose both evens is 1 (since there's exactly two even numbers), and then choose one odd from three, which is C(3,1) = 3. Therefore, 3 combinations. The total number of combinations is C(5,3) = 10. Therefore, the number of selections where remaining numbers have at least one even is 10 - 3 = 7. So probability is 7/10.Therefore, the answer is 7/10. Hmm. So I think that's consistent.Alternatively, to approach it through probability steps. When selecting three numbers, the first number selected is irrelevant. The key is whether 2 and 4 are both selected or not. The probability that both 2 and 4 are selected in the three numbers. If they are, then remaining numbers are both odd. Otherwise, at least one of 2 or 4 remains, so at least one even.So the probability that both 2 and 4 are selected. Let's compute that. How many ways to choose three numbers including both 2 and 4. As above, that's C(3,1) = 3 (choosing 2,4 and one more number from 1,3,5). Total ways C(5,3)=10. So probability 3/10. Therefore, the probability that at least one even remains is 1 - 3/10 = 7/10.Yes, that's the same as before. So I think this is solid.Another way: Suppose we fix the two remaining numbers. The probability that both are odd. The two remaining numbers are a random pair from the five. The probability that both are odd is C(3,2)/C(5,2) = 3/10. Hence, the probability that at least one is even is 1 - 3/10 = 7/10.Same answer.Therefore, all approaches converge to 7/10. So the answer is 7/10, which is 0.7. So as a numerical value, that's 0.7. But the problem says "express the result as a numerical value." Wait, 7/10 is 0.7. But maybe they want it as a fraction? Wait, the problem says "numerical value," so maybe 0.7 is acceptable, but in the box, usually fractions are preferred unless specified otherwise. Wait, but let me check the original problem again.The problem says: "the probability that at least one of the remaining two numbers is even is ______. (Express the result as a numerical value)".So "numerical value" could mean decimal or fraction. Since 7/10 is 0.7. But in mathematical contexts, fractions are often considered numerical values. However, in some cases, decimal is preferred. The problem may want it in decimal form. Let me check.But in the original problem statement, since it's a probability, both 7/10 and 0.7 are correct. However, in the instructions, they said "put your final answer within boxed{}". Typically, for probability questions, fractions are preferred unless a decimal is specifically requested. Since the problem doesn't specify, but says "numerical value", which can include fractions. So maybe safest to put both. Wait, but in the problem statement, numbers are 1,2,3,4,5. So the answer is 7/10. If we write that as a numerical value, is 0.7 acceptable? Maybe. But in the context of the problem, since all numbers are integers, and the answer is a fraction, 7/10 is exact. 0.7 is also exact. So either is fine. But maybe the user wants a fraction. Wait, the original problem had numbers written as words: "five numbers 1, 2, 3, 4, 5". So maybe follow the same style.But in the Chinese curriculum, fractions are often preferred for probabilities unless decimals are necessary. So maybe write boxed{dfrac{7}{10}}.But wait, in the problem statement, it says "Express the result as a numerical value". Hmm. In Chinese math problems, "numerical value" can sometimes mean decimal, but sometimes just a number. However, 7/10 is a numerical value. So maybe either is acceptable, but given that 7/10 is exact, it's better to present that. Alternatively, maybe the problem expects 0.7. Wait, let me check.If the user is Chinese, in Chinese math problems, probabilities can be expressed as fractions or decimals. However, in many cases, fractions are preferred unless specified otherwise. For example, when the answer is a simple fraction like 1/2 or 3/4, it's usually written as a fraction. So 7/10 is a simple fraction, so I think that's acceptable.Alternatively, if I can check with another approach, just to make sure. Let's list all possible combinations.The five numbers: 1,2,3,4,5. We need to select three. The remaining two will be the ones not selected. Let's list all possible combinations of three numbers and see how many leave at least one even.Total combinations C(5,3)=10. Let's list them:1. 1,2,3 – remaining: 4,5 (4 is even)2. 1,2,4 – remaining: 3,5 (both odd)3. 1,2,5 – remaining: 3,4 (4 is even)4. 1,3,4 – remaining: 2,5 (2 is even)5. 1,3,5 – remaining: 2,4 (both even)6. 1,4,5 – remaining: 2,3 (2 is even)7. 2,3,4 – remaining: 1,5 (both odd)8. 2,3,5 – remaining: 1,4 (4 is even)9. 2,4,5 – remaining: 1,3 (both odd)10. 3,4,5 – remaining: 1,2 (2 is even)Now, let's check which remaining pairs have at least one even number.1. 4,5 – 4 is even: yes2. 3,5 – both odd: no3. 3,4 – 4 is even: yes4. 2,5 – 2 is even: yes5. 2,4 – both even: yes6. 2,3 – 2 is even: yes7. 1,5 – both odd: no8. 1,4 – 4 is even: yes9. 1,3 – both odd: no10. 1,2 – 2 is even: yesSo out of 10 cases, the remaining pairs have at least one even in cases 1,3,4,5,6,8,10. That's 7 cases. So 7/10. Therefore, the probability is indeed 7/10.Therefore, the answer is 7/10, which as a numerical value is 0.7, but since 7/10 is exact, I think boxed{dfrac{7}{10}} is appropriate.Final Answerboxed{dfrac{7}{10}}</think>To determine the probability that at least one of the remaining two numbers is even when three numbers are randomly selected from the set {1, 2, 3, 4, 5}, we can use the complement rule. First, we calculate the total number of ways to select three numbers from five, which is given by the combination ( binom{5}{3} = 10 ).Next, we consider the complement event where both remaining numbers are odd. This happens if the three selected numbers include both even numbers (2 and 4) and one odd number. The number of ways to choose both even numbers and one odd number is ( binom{2}{2} times binom{3}{1} = 1 times 3 = 3 ).Thus, the probability that both remaining numbers are odd is ( frac{3}{10} ). The probability that at least one of the remaining numbers is even is the complement of this probability:[1 - frac{3}{10} = frac{7}{10}]To confirm, we list all possible combinations of three numbers and check the remaining pairs. Out of 10 combinations, 7 result in remaining pairs with at least one even number. This confirms the probability is indeed ( frac{7}{10} ).[boxed{dfrac{7}{10}}],<think>Okay, so I need to find the probability that when I randomly select three numbers out of 1, 2, 3, 4, 5, then at least one of the remaining two numbers is even. Hmm, let's break this down step by step.First, let me make sure I understand the problem correctly. We have the numbers 1 through 5. We select three numbers, and then look at the two that are left. The question is asking: what's the probability that at least one of those two remaining numbers is even. So, instead of looking at the three we picked, we're focusing on the two that are left. That makes sense.Probability problems can sometimes be tricky, especially with "at least one" conditions. I remember that often it's easier to calculate the probability of the complementary event and then subtract it from 1. The complementary event here would be that neither of the remaining two numbers is even, which means both are odd. So maybe I can compute the probability that both remaining numbers are odd and subtract that from 1 to get the desired probability.Let me confirm that approach. If I find the probability that both remaining numbers are odd, then 1 minus that probability should give me the probability that at least one is even. That seems right because "at least one even" and "no evens" (i.e., all odds) are complementary events. So, yes, that's a solid plan.Alright, so first, let's figure out the total number of ways to select three numbers from five. Since the order doesn't matter in selection, this is a combination problem. The formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items to choose.So, total number of possible ways to choose three numbers from five is C(5, 3). Let me calculate that:C(5, 3) = 5! / (3! * (5 - 3)!) = (5 * 4 * 3! ) / (3! * 2!) = (5 * 4) / 2! = 20 / 2 = 10.So there are 10 possible combinations of three numbers. Each of these combinations is equally likely if the selection is random.Now, moving on to the complementary event: both remaining numbers are odd. Let's figure out how many such combinations there are. If both remaining numbers are odd, that means the three numbers we picked must include all the even numbers. Wait, let's see. The original set has numbers 1, 2, 3, 4, 5. The even numbers here are 2 and 4, right? So there are two even numbers and three odd numbers (1, 3, 5). So if we want the remaining two numbers to both be odd, that would mean that the two numbers left are both odd. Therefore, the three numbers we picked must include the two even numbers and one odd number. Is that correct?Let me verify. Suppose we have two even numbers: 2 and 4. If we want the remaining two numbers to be both odd, then we must have removed all even numbers from the original set. But since there are two even numbers, removing all even numbers would require picking both of them. However, we are picking three numbers. So, to leave two numbers, both of which are odd, we need to have not picked those two odd numbers. Wait, this is getting confusing. Let me think again.Original set: 1, 2, 3, 4, 5.If we pick three numbers, the remaining two numbers are left. We want at least one of those two remaining numbers to be even. Alternatively, the complement is that both remaining numbers are odd. So, how can both remaining numbers be odd? That would mean that the two numbers not picked are both odd. Therefore, the three numbers that are picked must include all the even numbers and some odd numbers.Wait, there are two even numbers (2, 4) and three odd numbers (1, 3, 5). So, to leave two odd numbers, we need to have not picked two of the three odd numbers. Therefore, the three numbers we picked must include both even numbers and one odd number. Because if we pick both even numbers (2 and 4) and one odd number, then the remaining two numbers would be the other two odd numbers, which are both odd. That's the scenario we want to avoid for the original probability, but since we're calculating the complementary probability, that's exactly the case we need to count.So, the number of ways to pick three numbers such that both even numbers are included and one odd number is included. The number of such combinations would be C(2, 2) * C(3, 1). Because we need to choose both even numbers (which can only be done in 1 way, since there are two even numbers and we choose both) and then one odd number out of the three available.Calculating that:C(2, 2) = 1 (since there's only one way to choose both even numbers).C(3, 1) = 3 (choosing one odd number from three).Therefore, the number of favorable combinations for the complementary event is 1 * 3 = 3.Therefore, the probability that both remaining numbers are odd is 3 / 10.Hence, the probability that at least one of the remaining two numbers is even is 1 - 3/10 = 7/10.Wait, let me double-check that. So total combinations are 10. The number of combinations where both remaining numbers are odd is 3. So the probability of the complement is 3/10, so the desired probability is 7/10. That seems right.Alternatively, maybe I can approach this problem directly instead of using the complement. Let's try that to confirm.We need the number of ways that at least one of the remaining two numbers is even. So, in the remaining two numbers, there can be one even and one odd, or two evens. Wait, but the original set only has two even numbers. So in the remaining two numbers, the possible even counts are 0, 1, or 2.But since we are selecting three numbers, the remaining two can have a maximum of two even numbers. Since there are only two even numbers in total, the remaining two can have 0, 1, or 2 evens.But the problem states "at least one even", which would include the cases where there is 1 even or 2 evens in the remaining two numbers.So, let's compute the number of combinations where the remaining two numbers have 1 even and 1 odd, plus the combinations where they have 2 evens.First, the case with two evens: To have both even numbers remaining, that would mean we didn't pick either of the even numbers. But we need to pick three numbers. However, there are only three odd numbers. So if we leave both evens, that means we picked three odd numbers. The original set has three odd numbers: 1, 3, 5. So the number of ways to pick all three odd numbers is C(3, 3) = 1. Therefore, the remaining two numbers in this case would be 2 and 4, both even. So that's one combination where the remaining two numbers are both even.Second, the case with one even and one odd remaining: That would mean we picked one even and two odd numbers. Because the remaining two numbers would be one even and one odd. So, how many such combinations are there?We need to compute the number of ways to pick one even and two odd numbers. Since there are two even numbers and three odd numbers:Number of ways to pick one even: C(2, 1) = 2.Number of ways to pick two odds: C(3, 2) = 3.So total combinations: 2 * 3 = 6.Therefore, the number of combinations where the remaining two numbers have one even and one odd is 6.But wait, hold on. If we pick one even and two odd numbers, then the remaining numbers would be the other one even and the one remaining odd. Wait, no. Let's see. If we pick one even and two odds, the remaining numbers are the other even and the other odd. Because there are two evens total. So if we pick one even, the remaining is one even. And we picked two odds, so the remaining is one odd. So remaining two numbers would be one even and one odd. Therefore, yes, that case gives remaining numbers as 1 even and 1 odd.Therefore, the total number of favorable combinations (remaining at least one even) is the combinations where remaining numbers have one even (6 combinations) plus the combinations where remaining numbers have two evens (1 combination). So 6 + 1 = 7.Therefore, the probability is 7/10. Which matches the previous result. So that's reassuring.So whether I calculate the complement and subtract from 1, or calculate the favorable cases directly, I get 7/10. So the answer should be 7/10. Therefore, the probability is 0.7, or 70%.Let me just verify once again with another approach to be thorough.Another way is to list all possible combinations of three numbers and see how many leave at least one even number.Total combinations are 10. Let's list them:1. 1, 2, 3: Remaining numbers are 4, 5. 4 is even. So at least one even.2. 1, 2, 4: Remaining numbers are 3, 5. Both odd. So not favorable.3. 1, 2, 5: Remaining numbers are 3, 4. 4 is even.4. 1, 3, 4: Remaining numbers are 2, 5. 2 is even.5. 1, 3, 5: Remaining numbers are 2, 4. Both even.6. 1, 4, 5: Remaining numbers are 2, 3. 2 is even.7. 2, 3, 4: Remaining numbers are 1, 5. Both odd.8. 2, 3, 5: Remaining numbers are 1, 4. 4 is even.9. 2, 4, 5: Remaining numbers are 1, 3. Both odd.10. 3, 4, 5: Remaining numbers are 1, 2. 2 is even.Now, let's count the cases where the remaining numbers have at least one even:1. Yes2. No3. Yes4. Yes5. Yes (both even)6. Yes7. No8. Yes9. No10. YesSo how many "Yes" cases are there? Let's count:1. Yes3. Yes4. Yes5. Yes6. Yes8. Yes10. YesThat's 7 out of 10. So 7/10. Which confirms the result. So that's solid.Therefore, the answer is 7/10, which is 0.7. So expressed as a numerical value, that would be 0.7. But depending on the problem's requirement, it might need a fraction or a decimal. However, the problem says "numerical value," which could mean either, but since fractions are exact, maybe we should present it as 7/10. Wait, the original problem says "express the result as a numerical value." Hmm. Wait, "numerical value" could mean decimal or fraction. But in the context of probability, both are acceptable. But given that 7/10 is exact, maybe we should present that. However, the problem might want a decimal. Let me check the problem statement again.The problem says: "the probability that at least one of the remaining two numbers is even is ______. (Express the result as a numerical value)"So "numerical value" is a bit ambiguous. But in math problems, especially in probability, fractions are often preferred unless specified otherwise. But since the problem is in Chinese (assuming the original is Chinese), maybe they expect a decimal. Wait, no, the problem is presented in English here. Hmm. Wait, but the user hasn't specified. However, given the way the problem is written, and the fact that fractions are exact, whereas 0.7 is also exact here, both are acceptable. But in the context of the problem, maybe the answer is expected as a fraction. Alternatively, perhaps the system expects a decimal. However, to be safe, since 7/10 is 0.7, maybe present both? Wait, but the answer must be in a box, so I need to check the instructions again.Wait, the user says: "put your final answer within boxed{}". So in the problem statement, they just need the numerical value. So in the box, we can write either 0.7 or 7/10. But in mathematical contexts, fractions are often preferred. Let me check the original problem again. Wait, the original problem is presented in English, with numbers 1, 2, 3, 4, 5. So I think 7/10 is acceptable. However, in case, let me check with the listing approach.Wait, when I listed all combinations, 7 out of 10 cases were favorable. So 7/10. So 7/10 is the exact probability. Therefore, the answer is 7/10, which is 0.7. But as a fraction, 7/10 is more precise. So probably 7/10. Let me see the options if it were a multiple choice, but since it's not, we have to write the numerical value. However, in the problem statement, it's written as "numerical value", which can be a fraction or a decimal. Since both are numerical. But in the Chinese curriculum, sometimes they prefer fractions, but in this case, given that the user hasn't specified, but given that 7/10 is a simpler form, I think that's acceptable. However, the problem says "the probability... is ______. (Express the result as a numerical value)". So as a numerical value, 7/10 or 0.7 are both acceptable. Since the problem didn't specify decimal or fraction, either is fine, but in most probability problems, fractions are preferred unless the answer is a simple decimal like 0.5. However, 0.7 is also simple. Hmm. The original calculation gives 7/10. So to be precise, I should put 7/10. Alternatively, if the answer is to be put in a box, the user might expect boxed{dfrac{7}{10}} or boxed{0.7}. Given that the problem is mathematical, and in mathematical writing, fractions are often preferred for exactness, I would go with 7/10.But to be thorough, let me check once more. In the list of combinations, 7 out of 10 are favorable. Therefore, the probability is 7/10, which is 0.7. Either is correct, but since the problem says "numerical value," perhaps 0.7 is acceptable. However, in many mathematical contexts, fractions are considered numerical values. For example, 7/10 is a rational number, which is a numerical value. So both 0.7 and 7/10 are correct. However, in the context of the problem, if it's in a country where decimal is preferred, or fraction, but given that it's written with LaTeX formatting, and the answer is to be boxed, fractions are more common in such formatting. Therefore, I think the answer is boxed{dfrac{7}{10}}.But to confirm, let's see if there's a standard way. In probability, fractions are often used unless the answer requires a decimal. Since 7/10 is exact and simple, it's preferable. So yes, 7/10 is the right answer.Final AnswerThe probability is boxed{dfrac{7}{10}}.</think>To find the probability that at least one of the remaining two numbers is even when three numbers are randomly selected from 1, 2, 3, 4, 5, we can use the complementary probability approach.1. Total number of ways to select three numbers from five: [ binom{5}{3} = 10 ]2. Complementary event (both remaining numbers are odd): - This happens if the three selected numbers include both even numbers (2 and 4) and one odd number. - Number of ways to choose both even numbers: (binom{2}{2} = 1) - Number of ways to choose one odd number from three: (binom{3}{1} = 3) - Total combinations for complementary event: (1 times 3 = 3)3. Probability of complementary event: [ frac{3}{10} ]4. Probability of at least one even number in the remaining two: [ 1 - frac{3}{10} = frac{7}{10} ]Verification by listing all combinations confirms that 7 out of 10 combinations result in at least one even number remaining.Thus, the probability that at least one of the remaining two numbers is even is boxed{dfrac{7}{10}}.]

🔑:To solve this problem, consider the five numbers 1, 2, 3, 4, 5. When three numbers are randomly selected, the total number of basic events is n= C_{5}^{3}=10, The complementary event of having at least one even number among the remaining two is that both remaining numbers are odd, Therefore, the probability that at least one of the remaining two numbers is even is: p=1- dfrac{C_{3}^{2}}{C_{5}^{2}}=0.7. Thus, the answer is boxed{0.7}. The total number of basic events is n= C_{5}^{3}=10, and the complementary event of having at least one even number among the remaining two is that both remaining numbers are odd. By using the formula for the probability of complementary events, we can calculate the probability that at least one of the remaining two numbers is even. This question tests the method of calculating probability and is a basic question. When solving it, it is important to carefully read the question and correctly apply the formula for the probability of complementary events.

🧠:[<think>Okay, let's see. The problem is about determining whether the octal number N = (12345677654321)₈ is divisible by 7, 9, both, or neither. The options are A, B, C, D. Hmm. I need to check divisibility by 7 and 9. Since the number is given in base 8, I should remember how to handle divisibility rules in different bases. First, maybe converting the entire number to decimal would work, but that number is pretty long. Let me count the digits. Starting from the right: 1,2,3,4,5,6,7,7,6,5,4,3,2,1. That's 13 digits in octal. Converting that directly to decimal would be a huge number. Maybe there's a smarter way.Alternatively, remember that in base 8, each digit represents a power of 8. So, the number is N = 1×8¹² + 2×8¹¹ + 3×8¹⁰ + 4×8⁹ + 5×8⁸ + 6×8⁷ + 7×8⁶ + 7×8⁵ + 6×8⁴ + 5×8³ + 4×8² + 3×8¹ + 2×8⁰. Wait, actually, the last digit is 1×8⁰, then next is 2×8¹, up to the first digit which is 1×8¹². Hmm, that's 13 digits. But calculating this entire number is impractical. So, maybe I can use modular arithmetic properties. For divisibility by 7 and 9, since 8 ≡ 1 mod 7 (since 8 - 7 = 1), and 8 ≡ -1 mod 9 (since 8 +1 = 9). So, maybe I can use these congruences to compute N mod 7 and N mod 9.Let's start with divisibility by 7. Since 8 ≡ 1 mod 7, then 8^k ≡ 1^k = 1 mod 7 for any integer k. Therefore, each digit multiplied by 8^k will be congruent to the digit itself mod 7. Therefore, the entire number N mod 7 is equal to the sum of its digits mod 7. So, if I sum all the digits in octal and take mod 7, that will give me N mod 7.Similarly, for divisibility by 9. Since 8 ≡ -1 mod 9, then 8^k ≡ (-1)^k mod 9. Therefore, the number N mod 9 is equal to the alternating sum of its digits. That is, starting from the right (the least significant digit), subtract and add digits alternately. Wait, but the direction might matter. Let me confirm.In base 10, for divisibility by 11, which is 10 ≡ -1 mod 11, you take the alternating sum. Similarly, here, since 8 ≡ -1 mod 9, we can use the same rule. So, starting from the right, each digit is multiplied by (-1)^k, where k is the position. But actually, the exponents of 8 will correspond to the digits. Let's be precise.The number N in base 8 is d₁₂d₁₁...d₀, where d₁₂ is the leftmost digit (1) and d₀ is the rightmost digit (1). Then, N = Σ_{k=0}^{12} d_k × 8^k. So, modulo 9, since 8 ≡ -1, N ≡ Σ_{k=0}^{12} d_k × (-1)^k. Therefore, it's an alternating sum where the rightmost digit (d₀) is multiplied by (-1)^0 = 1, then d₁ is multiplied by (-1)^1 = -1, d₂ by (-1)^2 = 1, etc. So, starting from the right, it's 1×1 - 2×1 + 3×1 - 4×1 + 5×1 - 6×1 + 7×1 - 7×1 + 6×1 - 5×1 + 4×1 - 3×1 + 2×1 - 1×1? Wait, no. Wait, actually, since the digits are from d₀ (rightmost) to d₁₂ (leftmost), but in our case, the digits of N are (from left to right) 1,2,3,4,5,6,7,7,6,5,4,3,2,1. Wait, actually, hold on, let's parse the number correctly. Wait, the number is (12345677654321)₈. Let's count the digits:Starting from the left: 1,2,3,4,5,6,7,7,6,5,4,3,2,1. That's 14 digits. Wait, earlier I thought 13 digits, but actually, 1 followed by 2,3,4,5,6,7,7,6,5,4,3,2,1. Let's count: 1 (1), 2 (2), 3 (3), 4 (4), 5 (5), 6 (6), 7 (7), 7 (8), 6 (9), 5 (10), 4 (11), 3 (12), 2 (13), 1 (14). So, it's a 14-digit number in octal. So, positions are from d₁₃ (leftmost) to d₀ (rightmost). Wait, but in standard notation, the rightmost digit is d₀. So, the number is:d₁₃ d₁₂ d₁₁ d₁₀ d₉ d₈ d₇ d₆ d₅ d₄ d₃ d₂ d₁ d₀1 2 3 4 5 6 7 7 6 5 4 3 2 1Therefore, when considering N mod 9, which is Σ_{k=0}^{13} d_k × (-1)^k. So, starting from the rightmost digit (d₀ =1), multiply by (-1)^0=1, then d₁=2 multiplied by (-1)^1=-1, d₂=3 multiplied by (-1)^2=1, etc., up to d₁₃=1 multiplied by (-1)^13=-1.Alternatively, since the exponents go from 0 to 13, each digit d_k is multiplied by (-1)^k. So, if we list the digits from right to left as positions 0 to 13, then the rightmost digit (1) is position 0, the next (2) is position 1, etc., up to the leftmost digit (1) at position 13. Therefore, the alternating sum would be:Starting from the right (position 0): 1 (position 0: +1), 2 (position 1: -2), 3 (position 2: +3), 4 (position 3: -4), 5 (position 4: +5), 6 (position 5: -6), 7 (position 6: +7), 7 (position 7: -7), 6 (position 8: +6), 5 (position 9: -5), 4 (position 10: +4), 3 (position 11: -3), 2 (position 12: +2), 1 (position 13: -1).So, the total sum is:1 - 2 + 3 - 4 + 5 - 6 + 7 -7 +6 -5 +4 -3 +2 -1.Let me compute that step by step:Start with 1.1 -2 = -1-1 +3 = 22 -4 = -2-2 +5 = 33 -6 = -3-3 +7 = 44 -7 = -3-3 +6 = 33 -5 = -2-2 +4 = 22 -3 = -1-1 +2 = 11 -1 = 0So, the alternating sum is 0. Therefore, N ≡ 0 mod 9, which means N is divisible by 9.Now, check divisibility by 7. As earlier, since 8 ≡1 mod7, each digit multiplied by 8^k is congruent to the digit itself mod7. Therefore, the sum of all digits multiplied by 1 mod7. So, the total sum of digits is:From left to right (digits are 1,2,3,4,5,6,7,7,6,5,4,3,2,1). Let's compute the sum:1+2+3+4+5+6+7+7+6+5+4+3+2+1.Let me add them step by step:Start with 1+2=33+3=66+4=1010+5=1515+6=2121+7=2828+7=3535+6=4141+5=4646+4=5050+3=5353+2=5555+1=56.Total sum is 56. Now, 56 mod7. Since 7×8=56, 56 is divisible by7. Therefore, N ≡56 mod7≡0 mod7. Therefore, N is divisible by7 as well.Wait, but according to the options: A. divisible by7 not9; B. divisible by9 not7; C. neither; D. both. So, according to my calculations, N is divisible by both7 and9, so answer D.But wait, let me double-check. Maybe I made a mistake in the alternating sum for 9.Wait, the calculation for the alternating sum: starting from the right, the digits are 1,2,3,4,5,6,7,7,6,5,4,3,2,1. So positions 0 to13:Position:0:1, 1:2,2:3,3:4,4:5,5:6,6:7,7:7,8:6,9:5,10:4,11:3,12:2,13:1.So, the coefficients are (-1)^k for k=0 to13. So:k=0: (-1)^0=1 --> 1×1=1k=1: (-1)^1=-1 -->2×(-1)=-2k=2: (-1)^2=1 -->3×1=3k=3: (-1)^3=-1 -->4×(-1)=-4k=4: (-1)^4=1 -->5×1=5k=5: (-1)^5=-1 -->6×(-1)=-6k=6: (-1)^6=1 -->7×1=7k=7: (-1)^7=-1 -->7×(-1)=-7k=8: (-1)^8=1 -->6×1=6k=9: (-1)^9=-1 -->5×(-1)=-5k=10: (-1)^10=1 -->4×1=4k=11: (-1)^11=-1 -->3×(-1)=-3k=12: (-1)^12=1 -->2×1=2k=13: (-1)^13=-1 -->1×(-1)=-1So adding all these up:1 -2 +3 -4 +5 -6 +7 -7 +6 -5 +4 -3 +2 -1Compute step by step:1 -2 = -1-1 +3 = 22 -4 = -2-2 +5 =33 -6 =-3-3 +7 =44 -7 =-3-3 +6 =33 -5 =-2-2 +4=22 -3=-1-1 +2=11 -1=0Yes, total is 0. So N is divisible by9.Sum for 7: sum of digits is 56, which is 7×8=56, so divisible by7. Therefore, N is divisible by both7 and9. So answer is D.But wait, let me check the sum of digits again. The digits are:First digit (leftmost):1Then 2,3,4,5,6,7,7,6,5,4,3,2,1.Wait, total digits:1,2,3,4,5,6,7,7,6,5,4,3,2,1. That's 14 digits. Let me recount:1 (1)2 (2)3 (3)4 (4)5 (5)6 (6)7 (7)7 (8)6 (9)5 (10)4 (11)3 (12)2 (13)1 (14)Yes, 14 digits. So sum is:1+2=33+3=66+4=1010+5=1515+6=2121+7=2828+7=3535+6=4141+5=4646+4=5050+3=5353+2=5555+1=56Yes, 56. 56 divided by7 is8. So divisible by7.Therefore, the answer is D.But wait, the options are A, B, C, D. The problem says "then among the following four statements, the correct one is". So, D says "N is divisible by both 7 and9", which is correct according to our calculations. Therefore, the answer should be D.But the original problem was presented in octal. Wait, octal digits go up to7, which is correct here. So all digits are valid.But let me cross-verify with another method. Maybe converting a small part of the number.Alternatively, maybe check if the conclusions are correct. Since both sums lead to divisibility by7 and9, then D is correct. So the answer is D.Final Answerboxed{D}</think>To determine whether the octal number ( N = (12345677654321)_8 ) is divisible by 7, 9, both, or neither, we use modular arithmetic.Divisibility by 7:Since ( 8 equiv 1 pmod{7} ), each digit multiplied by ( 8^k ) is congruent to the digit itself modulo 7. Therefore, the sum of the digits of ( N ) modulo 7 will determine divisibility by 7. The digits of ( N ) are:[ 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1 ]Summing these digits:[1 + 2 + 3 + 4 + 5 + 6 + 7 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 56]Since ( 56 div 7 = 8 ), ( 56 ) is divisible by 7. Hence, ( N ) is divisible by 7.Divisibility by 9:Since ( 8 equiv -1 pmod{9} ), the alternating sum of the digits (starting from the rightmost digit) modulo 9 will determine divisibility by 9. The digits from right to left (positions 0 to 13) are:[ 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1 ]Calculating the alternating sum:[1 - 2 + 3 - 4 + 5 - 6 + 7 - 7 + 6 - 5 + 4 - 3 + 2 - 1]Step-by-step calculation:[1 - 2 = -1 -1 + 3 = 2 2 - 4 = -2 -2 + 5 = 3 3 - 6 = -3 -3 + 7 = 4 4 - 7 = -3 -3 + 6 = 3 3 - 5 = -2 -2 + 4 = 2 2 - 3 = -1 -1 + 2 = 1 1 - 1 = 0]The alternating sum is 0, which is divisible by 9. Hence, ( N ) is divisible by 9.Since ( N ) is divisible by both 7 and 9, the correct answer is:[boxed{D}],<think>Okay, let's see. The problem is about determining whether the octal number N = (12345677654321)₈ is divisible by 7, 9, both, or neither. The options are A, B, C, D. I need to figure out which one is correct. First, I remember that checking divisibility in different number systems can sometimes be tricky. Since octal is base 8, maybe there are some tricks similar to how we check divisibility by 9 or 3 in base 10 by summing the digits. Let me think.In base 10, a number is divisible by 9 if the sum of its digits is divisible by 9. Similarly, for 3. Maybe in base 8, there's a similar rule for 7 or 9? Wait, 8 is 2³, so the factors are 2, 4, 8. But 7 and 9 are co-prime with 8. Hmm.Alternatively, maybe I can convert the octal number to decimal and then check divisibility by 7 and 9. But converting such a large number from octal to decimal might be time-consuming. Let me see how long the number is. The octal number is 12345677654321. Let's count the digits: 1,2,3,4,5,6,7,7,6,5,4,3,2,1. That's 13 digits. So in decimal, it's going to be a huge number. Converting directly might not be feasible.Alternatively, maybe there's a way to compute N mod 7 and N mod 9 without converting the entire number? Because if I can find the remainder when N is divided by 7 and 9, then I can determine divisibility.Yes, modular arithmetic can help here. For a number in base b, the value mod (b-1) is the sum of its digits mod (b-1). Since we're in base 8, then mod 7 (which is 8 -1), the sum of the digits mod 7 will give the remainder. Similarly, for mod 9, since 8 ≡ -1 mod 9, then each digit's contribution can be calculated as d_i * (-1)^i mod 9. Wait, maybe?Let me recall. For modulus m, if we can express the number in terms of powers of b, then N = a_0 + a_1*b + a_2*b² + ... + a_n*b^n. Then N mod m can be calculated by considering each term a_i*b^i mod m and summing them up.So for modulus 7 and modulus 9, we can compute N mod 7 and N mod 9 using the fact that b = 8. Let's do that.First, for modulus 7: Since 8 ≡ 1 mod 7, because 8 -7 =1. Therefore, 8^i ≡ 1^i = 1 mod 7 for any i. Therefore, N mod 7 is equal to the sum of its digits in base 8 mod 7. So if we sum all the digits of the octal number and take mod 7, that will give N mod 7.Similarly, for modulus 9: 8 ≡ -1 mod 9, since 8 +1 =9. Therefore, 8^i ≡ (-1)^i mod 9. Therefore, N mod 9 is equal to the alternating sum of the digits. That is, starting from the rightmost digit (the least significant digit), we alternate between adding and subtracting the digits. Wait, but the exponent depends on the position. Let me clarify.Each digit d_i (starting from i=0 at the right) is multiplied by 8^i. So 8^i ≡ (-1)^i mod 9. Therefore, the contribution of each digit d_i is d_i * (-1)^i mod 9. Therefore, N mod 9 is equal to the sum over i of d_i * (-1)^i mod 9. So if we start from the rightmost digit (the units place, which is 8^0), then the coefficients alternate between +1 and -1. So starting from the right, the first digit (position 0) is multiplied by (-1)^0 =1, position 1 is multiplied by (-1)^1 = -1, position 2 is multiplied by (-1)^2 =1, etc.But wait, the digits in the octal number are written from left to right, starting with the most significant digit. So the leftmost digit is the highest power of 8. Let's make sure about the positions.Suppose N is written as d_12 d_11 d_10 ... d_0 in octal, where d_12 is the leftmost digit (1), and d_0 is the rightmost digit (1). Then, N = d_12 *8^12 + d_11*8^11 + ... + d_0*8^0.So when we compute N mod 9, each term d_i *8^i mod 9 is d_i * (-1)^i mod 9. Therefore, the entire sum is (sum_{i=0}^{12} d_i * (-1)^i) mod 9. So starting from the rightmost digit (d_0), which is position 0, coefficient is (-1)^0 =1, then d_1 is position 1, coefficient -1, d_2 position 2, coefficient 1, etc., up to d_12 which is position 12, coefficient (-1)^12=1.But since the digits are written from left to right as d_12 d_11 ... d_0, when we list them out, we have to reverse the order for the coefficients. Wait, maybe not. Let's clarify.Let me write out the digits of N: (1)(2)(3)(4)(5)(6)(7)(7)(6)(5)(4)(3)(2)(1). Wait, that's 14 digits. Wait, original problem says N=(12345677654321)₈. Let's count: 1,2,3,4,5,6,7,7,6,5,4,3,2,1. That's 14 digits. So the positions are from 0 (rightmost) to 13 (leftmost). Wait, octal digits are 0-7, so each digit is 0-7, but the number starts with 1, so leading zeros are not present. So the number is 14 digits long. So positions are from 0 (rightmost) to 13 (leftmost). So when computing N mod 9, we need to take each digit, starting from the rightmost (position 0) with coefficient (-1)^0=1, then the next (position 1) with coefficient (-1)^1=-1, etc., up to position 13 with coefficient (-1)^13=-1.But the digits as given are left to right: d_13 d_12 ... d_0. So if I list the digits:Position 13: 1Position 12: 2Position 11:3Position10:4Position9:5Position8:6Position7:7Position6:7Position5:6Position4:5Position3:4Position2:3Position1:2Position0:1So each digit at position i (from 0 to 13) is d_i. Wait, actually, in standard notation, the rightmost digit is position 0, so if the number is written as (d_n ... d_0), then d_0 is the units place. Therefore, for N=(12345677654321)_8, the digits are:Leftmost digit: 1 (position 13)Then 2 (position 12)3 (position 11)4 (position10)5 (position9)6 (position8)7 (position7)7 (position6)6 (position5)5 (position4)4 (position3)3 (position2)2 (position1)1 (position0)Yes, so positions 13 to 0. Therefore, when computing N mod 9, each digit d_i (i from 0 to 13) is multiplied by (-1)^i. Wait, no. Wait, the exponent in 8^i is i, so for each digit d_i at position i (from right, starting at 0), it's multiplied by 8^i. So 8^i mod9 is (-1)^i. Therefore, each digit d_i contributes d_i*(-1)^i to the total modulo 9. Therefore, to compute N mod9, we need to compute the sum over all digits d_i*(-1)^i, then take mod9.But since the digits are given from left to right, which corresponds to positions 13 down to 0, we need to assign exponents accordingly. So the leftmost digit is position13, so it's multiplied by 8^13, which is (-1)^13= -1 mod9. The next digit is position12, multiplied by8^12= (-1)^12=1 mod9, and so on.Therefore, to compute the sum, starting from the leftmost digit (position13):Digit1 (position13): coefficient (-1)^13= -1Digit2 (position12): coefficient (-1)^12=1Digit3 (position11): coefficient (-1)^11= -1Digit4 (position10): coefficient (-1)^10=1Digit5 (position9): coefficient (-1)^9= -1Digit6 (position8): coefficient (-1)^8=1Digit7 (position7): coefficient (-1)^7= -1Digit8 (position6): coefficient (-1)^6=1Digit9 (position5): coefficient (-1)^5= -1Digit10 (position4): coefficient (-1)^4=1Digit11 (position3): coefficient (-1)^3= -1Digit12 (position2): coefficient (-1)^2=1Digit13 (position1): coefficient (-1)^1= -1Digit14 (position0): coefficient (-1)^0=1Therefore, each digit from left to right alternates starting with -1, then +1, -1, etc. So the coefficients for the digits from left to right are: -1, +1, -1, +1, -1, +1, -1, +1, -1, +1, -1, +1, -1, +1.Therefore, to compute N mod9, we can list the digits from left to right and multiply them by these coefficients and sum them up.Similarly, for N mod7, since 8 ≡1 mod7, so 8^i ≡1 mod7 for all i. Therefore, N mod7 is just the sum of all digits mod7.So first, let's compute N mod7 by summing all the digits.Digits of N from left to right:1,2,3,4,5,6,7,7,6,5,4,3,2,1.Sum them up:1+2=33+3=66+4=1010+5=1515+6=2121+7=2828+7=3535+6=4141+5=4646+4=5050+3=5353+2=5555+1=56.Total sum is 56.56 mod7: 56 divided by7 is 8, remainder 0. So 56 ≡0 mod7. Therefore, N is divisible by7. So options C and B are out (since C says not divisible by7, and B says not divisible by7 but by9). So the answer is either A or D.Now, check divisibility by9. For that, compute the alternating sum as per the coefficients.Digits from left to right with coefficients:Position13:1 * (-1) = -1Position12:2 * (+1) = +2Position11:3 * (-1) = -3Position10:4 * (+1) = +4Position9:5 * (-1) = -5Position8:6 * (+1) = +6Position7:7 * (-1) = -7Position6:7 * (+1) = +7Position5:6 * (-1) = -6Position4:5 * (+1) = +5Position3:4 * (-1) = -4Position2:3 * (+1) = +3Position1:2 * (-1) = -2Position0:1 * (+1) = +1Now, let's sum these up step by step:Start with -1 (from position13)-1 +2 =11 -3 = -2-2 +4=22 -5= -3-3 +6=33 -7= -4-4 +7=33 -6= -3-3 +5=22 -4= -2-2 +3=11 -2= -1-1 +1=0.Total sum is 0. Therefore, N ≡0 mod9. So N is divisible by9 as well.Therefore, N is divisible by both7 and9. Therefore, the correct answer is D.Wait, but let me verify again. The sum for mod9 was 0? Let me recheck the calculation:Starting with the digits left to right with coefficients:1*(-1) = -12*(+1)=+2 → total: -1+2=13*(-1)=-3 → total:1-3=-24*(+1)=+4 → total:-2+4=25*(-1)=-5 → total:2-5=-36*(+1)=+6 → total:-3+6=37*(-1)=-7 → total:3-7=-47*(+1)=+7 → total:-4+7=36*(-1)=-6 → total:3-6=-35*(+1)=+5 → total:-3+5=24*(-1)=-4 → total:2-4=-23*(+1)=+3 → total:-2+3=12*(-1)=-2 → total:1-2=-11*(+1)=+1 → total:-1+1=0Yes, that's correct. So the total is 0 mod9. Therefore, N is divisible by9. So since it's divisible by both7 and9, the correct answer is D.But let me double-check the mod7 calculation. The sum of digits was 56, which is 7*8, so indeed 56 mod7=0. Correct. So N is divisible by7 and9. So answer D.But wait, the options are A, B, C, D. Let me check again:A. divisible by7 but not by9B. divisible by9 but not by7C. not divisible by eitherD. divisible by bothYes, D is correct.But let me cross-verify with another method. For example, maybe compute N in decimal and check. However, converting such a large number is tedious, but perhaps for mod7 and mod9, the methods we used are reliable.Alternatively, let's check for mod9 using another approach. Since 9 is 3^2, and in base10, the alternating sum is similar. But in base8, as we saw, the rule is different.Alternatively, maybe convert N to base10 step by step mod7 and mod9. Wait, but since we already used modular arithmetic, and the result is 0 for both, that should be sufficient. Maybe the error is in the alternating sum. Let me check that again.Wait, the key idea is that 8 ≡ -1 mod9, so 8^i ≡ (-1)^i mod9. Therefore, N = sum_{i=0}^{13} d_i *8^i ≡ sum_{i=0}^{13} d_i*(-1)^i mod9. So starting from the rightmost digit (d_0) which is multiplied by (-1)^0=1, next digit d_1 is multiplied by (-1)^1=-1, etc. However, in our previous calculation, we considered the leftmost digit (position13) as multiplied by (-1)^13. Since the leftmost digit is d_13, which is multiplied by8^13 ≡ (-1)^13 mod9. Therefore, yes, our calculation is correct.Therefore, the conclusion is correct. N is divisible by both7 and9. Therefore, answer D.Final Answerboxed{D}</think>To determine whether the octal number ( N = (12345677654321)_8 ) is divisible by 7, 9, both, or neither, we use modular arithmetic.Divisibility by 7:Since ( 8 equiv 1 mod 7 ), each digit of ( N ) contributes to the sum modulo 7. Summing the digits:[1 + 2 + 3 + 4 + 5 + 6 + 7 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 56]Since ( 56 equiv 0 mod 7 ), ( N ) is divisible by 7.Divisibility by 9:Since ( 8 equiv -1 mod 9 ), each digit ( d_i ) at position ( i ) contributes ( d_i cdot (-1)^i mod 9 ). Calculating the alternating sum of the digits (starting from the rightmost digit with coefficient ( 1 )):[begin{align*}1 cdot (-1)^{13} &+ 2 cdot (-1)^{12} + 3 cdot (-1)^{11} + 4 cdot (-1)^{10} + 5 cdot (-1)^9 + 6 cdot (-1)^8 + 7 cdot (-1)^7 + 7 cdot (-1)^6 &+ 6 cdot (-1)^5 + 5 cdot (-1)^4 + 4 cdot (-1)^3 + 3 cdot (-1)^2 + 2 cdot (-1)^1 + 1 cdot (-1)^0end{align*}]Simplifying each term:[-1 + 2 - 3 + 4 - 5 + 6 - 7 + 7 - 6 + 5 - 4 + 3 - 2 + 1 = 0]Since the result is ( 0 mod 9 ), ( N ) is divisible by 9.Thus, ( N ) is divisible by both 7 and 9. The correct answer is (boxed{D}).]

🔑:39 D. Since 8 equiv 1(bmod 7), we have 8^{i} equiv 1(bmod 7), henceN=left(a_{2 n-1} cdot a_{2 n-2} cdots a_{1} a_{0}right)_{8}=sum_{i=0}^{2 n-1} a_{i} 8^{i} equiv sum_{i=0}^{2 n-1} a_{i}(bmod 7),i.e., N is divisible by 7 Leftrightarrow the sum of the digits in the octal representation of N is divisible by 7. The given N clearly satisfies this.Furthermore, since 8 equiv-1(bmod 9), we have 8^{i} equiv(-1)^{i}(bmod 9), hencebegin{aligned}N & =left(a_{2 n-1} a_{2 n-2} cdots a_{1} a_{0}right)_{8} equiv sum_{i=0}^{2 n-1} a_{i}(-1)^{i} & equivleft(a_{0}+a_{2}+cdots+a_{2 n-2}right)-left(a_{1}+a_{3}+cdots+a_{2 n-1}right)(bmod 9),end{aligned}i.e., N is divisible by 9 Leftrightarrow in the octal representation of N, the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 9. The given N clearly satisfies this condition.