🧠:[<think>Okay, so I have this problem here about bus routes in a city. There are 57 bus routes, and some conditions are given. Let me try to parse through this step by step. First, the problem states that from any stop, you can get to any other stop without transferring. Hmm, that probably means the entire network is connected as a single system. So even if two stops are on different routes, there's a way to get from one to the other by riding just one bus. That makes sense because otherwise, you'd need to transfer, right? But condition one says you don't have to transfer. So maybe every pair of stops is on at least one common route? Wait, but no, if you can get from any stop to any other without transferring, that would mean that all stops are on the same route, which can't be true because there are 57 routes. Wait, maybe I'm misinterpreting. Let me read that again."From any stop, you can get to any other stop without transferring." So, if I'm at stop A and want to go to stop B, there exists a single route that connects them. So, every pair of stops is connected by at least one route. That would imply that the union of all routes covers all possible pairs of stops. But that seems too strong because each route has multiple stops. Wait, but the problem says each route has at least three stops. Hmm. So maybe the entire system is designed such that the stops form a structure where any two stops are on exactly one route. Wait, that sounds like a projective plane or a block design. Wait, the second condition says: "for each pair of routes, there is one, and only one, stop where you can transfer from one of these routes to the other." So any two routes intersect at exactly one stop. That's similar to a projective plane where any two lines intersect at exactly one point. And the first condition says that any two stops are connected by a route, which would mean that any two points lie on exactly one line. So this is analogous to a finite projective plane. In finite projective planes, the number of points is n^2 + n + 1 for some integer n, and each line (route) contains n + 1 points (stops). Also, each point is on n + 1 lines. The number of lines is equal to the number of points. But here, in our problem, there are 57 routes. So if this is analogous to a projective plane, we would have 57 routes, each corresponding to a line, and the number of stops would be 57 as well? Wait, but in a projective plane of order n, the number of lines is n^2 + n + 1. If we have 57 routes, then n^2 + n + 1 = 57. Let me check if 57 can be expressed in that form. Solving n^2 + n + 1 = 57: n^2 + n - 56 = 0. Using the quadratic formula: n = [-1 ± sqrt(1 + 224)] / 2 = [-1 ± sqrt(225)] / 2 = [-1 ± 15]/2. So positive solution is (14)/2 = 7. So n = 7. Then the number of stops (points) would be n^2 + n + 1 = 57. Wait, so there are 57 stops and 57 routes. Each route would have n + 1 = 8 stops. But the problem says each route has no fewer than three stops. 8 is more than three, so that satisfies the third condition. But let me verify if this structure satisfies all the given conditions. 1) Any two stops are on exactly one route. Therefore, you can get from any stop to any other without transferring because they are on a common route. So that's condition one satisfied.2) Any two routes intersect at exactly one stop. That's condition two. 3) Each route has n + 1 = 8 stops, which is ≥ 3. Therefore, if this is a projective plane of order 7, then all conditions are satisfied. So the answer would be that each route has 8 stops. But wait, let me check again. If n = 7, then the number of stops is 7^2 + 7 + 1 = 57, which matches the number of routes. Each route (line) has 8 stops (points). Each stop (point) is on 8 routes (lines). Each pair of routes (lines) intersects at exactly one stop (point). Each pair of stops (points) is on exactly one route (line). So yes, this seems to fit. Therefore, the number of stops per route is 8. But let me think again. The problem states there are 57 bus routes. In the projective plane analogy, the number of lines (routes) is equal to the number of points (stops). So here, since there are 57 routes, there would be 57 stops. Each route has 8 stops. Each stop is part of 8 routes. But the problem doesn't mention the total number of stops, only the number of routes. However, the answer is about how many stops each route has, which would be 8. Wait, but just to be thorough, is there any other possible structure that could satisfy these conditions? For example, a Balanced Incomplete Block Design (BIBD). In a BIBD, parameters are usually given as (v, k, λ), where v is the number of elements (stops), b is the number of blocks (routes), each block has k elements, each element is in r blocks, each pair of elements is in λ blocks. In our case, condition 1 says that any two stops are on exactly one route, so λ = 1. Condition 2 says that any two routes intersect at exactly one stop, which is another property of a symmetric BIBD, where b = v, and the number of blocks is equal to the number of elements. In a symmetric BIBD with λ = 1, the parameters satisfy certain equations. Indeed, for a symmetric BIBD, we have that:b = v,and each block has k elements,each element is in r blocks,and the relations:bk = vr,andλ(v - 1) = r(k - 1).Since λ = 1, we have:v - 1 = r(k - 1).But since it's symmetric, b = v, so the first equation becomes:bk = vr → vk = vr → k = r.Therefore, substituting into the second equation:v - 1 = k(k - 1).So v - 1 = k^2 - k.Thus,k^2 - k - (v - 1) = 0.But in our problem, we are given that the number of routes (which is b) is 57. Since it's a symmetric BIBD, b = v, so v = 57. Therefore, we can solve for k:k^2 - k - (57 - 1) = 0 → k^2 - k - 56 = 0.Solving quadratic equation: k = [1 ± sqrt(1 + 224)] / 2 = [1 ± 15]/2. So k = (1 + 15)/2 = 8, or k = (1 -15)/2 = -7. Since k must be positive, k = 8. Therefore, each route has 8 stops. Therefore, whether we think of it as a projective plane of order 7 (since 7^2 + 7 + 1 = 57) or as a symmetric BIBD with parameters (57, 8, 1), we arrive at the conclusion that each route has 8 stops. Thus, the answer should be 8. I need to confirm once more if each route having 8 stops meets all the problem's conditions. 1) Any two stops are on exactly one route: yes, because λ = 1 in the BIBD.2) Any two routes share exactly one stop: yes, since in a symmetric BIBD with λ = 1, the intersection of any two blocks (routes) is λ = 1.3) Each route has at least three stops: 8 is more than three. Therefore, all conditions are satisfied. So the answer is 8.Final Answerboxed{8}</think>Given the problem of determining the number of stops for each of the 57 bus routes in a city with the following conditions:1. From any stop, you can get to any other stop without transferring.2. For each pair of routes, there is exactly one stop where you can transfer from one route to the other.3. Each route has no fewer than three stops.We recognize that this structure is analogous to a finite projective plane or a symmetric Balanced Incomplete Block Design (BIBD). 1. Condition 1 implies that any two stops lie on exactly one route, which is characteristic of a projective plane or a BIBD with (lambda = 1).2. Condition 2 implies that any two routes intersect at exactly one stop, which is also a property of a projective plane or a symmetric BIBD.3. Condition 3 requires each route to have at least three stops.For a symmetric BIBD with parameters ((v, k, lambda)) where (v) is the number of stops, (b) is the number of routes (57), (k) is the number of stops per route, and (lambda = 1):- Since (b = v), we have (v = 57).- The equations for a symmetric BIBD are (bk = vr) and (lambda(v - 1) = r(k - 1)). Given (b = v), we get (k = r) and (v - 1 = k(k - 1)).Solving for (k):[k^2 - k - (57 - 1) = 0 implies k^2 - k - 56 = 0]Using the quadratic formula:[k = frac{1 pm sqrt{1 + 224}}{2} = frac{1 pm 15}{2}]The positive solution is (k = 8).Thus, each route has 8 stops, satisfying all conditions. The final answer is (boxed{8}).,<think>Okay, let's try to tackle this problem. So, we have 57 bus routes in a city with some specific conditions. The goal is to figure out how many stops each route has, and they all have the same number of stops. Let me break down the given information first.First condition: From any stop, you can get to any other stop without transferring. Hmm, that sounds like the entire network is connected. So, the graph of stops is connected, right? But wait, actually, it's saying that you can get from any stop to any other without transferring. Transferring would mean changing buses, right? So, if you can go from any stop to any other without transferring, that means that every pair of stops is connected by at least one bus route. Wait, but that can't be, because there are 57 bus routes. If every pair of stops was on a single bus route, then you would need a huge number of routes. Maybe I misinterpret that.Wait, no. Let me read it again. "From any stop, you can get to any other stop without transferring." So, that suggests that the entire system is connected as a single transit network, but the connectivity is such that you don't need to transfer. So, maybe every pair of stops is on at least one common route? But if that were the case, then each pair of stops is connected by a route. But that would mean that the number of routes is at least the number of pairs of stops divided by the number of stops per route. But given that there are 57 routes, each with at least three stops, it's not immediately obvious.Wait, maybe it's a connected graph where the routes are the edges, and the stops are the vertices? But no, because each route is a bus route, which would correspond to a set of stops (vertices) connected in sequence. But in graph theory terms, a route would be more like a path or a cycle. But the problem states that from any stop, you can get to any other without transferring, so the entire graph is connected, but each route is a subset of stops. Hmm.Alternatively, maybe the routes are designed such that the union of all routes covers all stops, and the intersection of any two routes is exactly one stop (second condition). Wait, the second condition says that for each pair of routes, there is exactly one stop where you can transfer between them. That is, any two routes intersect at exactly one stop. So, each pair of routes shares exactly one common stop.So, if we model each route as a set of stops, then any two sets (routes) intersect in exactly one element (stop). Also, the first condition says that the network is connected, so the entire system is connected, meaning that you can go from any stop to any other via some sequence of routes, but since you can do it without transferring, that actually implies that all stops are on the same route? Wait, no, that can't be, because there are 57 routes. So, that must mean that any two stops are on some common route. Wait, but if that's the case, then every pair of stops is contained in at least one route. But given that each route has at least three stops, then the number of routes would be at least the number of pairs of stops divided by the number of pairs of stops per route. Let's think.Suppose there are S stops in total. Then the number of pairs of stops is C(S, 2). Each route has k stops, so the number of pairs of stops per route is C(k, 2). If every pair of stops is covered by exactly one route (since any two routes share exactly one stop, so two different routes cannot share more than one stop, but maybe also each pair is covered by exactly one route?), wait, but the first condition says you can get from any stop to any other without transferring, which would imply that for any two stops, there exists a route that contains both. Therefore, every pair of stops is in exactly one route. Because if they were in two routes, then those two routes would share two stops, which would violate condition two, which says that any two routes share exactly one stop.Therefore, this seems like a type of block design. Specifically, a Steiner system S(2, k, S), where each pair of elements is contained in exactly one block (route), and each block (route) contains k elements (stops). The parameters of a Steiner system S(t, k, v), where t=2, k is the block size, and v is the number of elements. In this case, the stops are the elements, the routes are the blocks, each block has size k, each pair of elements is in exactly one block, and each pair of blocks (routes) intersect in exactly one element (stop). Wait, no. Wait, the Steiner system S(2, k, v) ensures that each pair is in exactly one block, but the condition here is also that each pair of blocks intersect in exactly one element. Is that necessarily true?Wait, in a Steiner system, the intersection of two blocks can vary. For example, in a projective plane, which is a Steiner system S(2, k, v) where each pair of blocks intersects in exactly one point, which would correspond to our problem. So, in that case, a projective plane is a Steiner system where each pair of points is in exactly one line (block), and each pair of lines intersects in exactly one point. Therefore, this problem is exactly describing a finite projective plane. The number of routes would correspond to the number of lines, each line (route) has k stops, each pair of lines intersects in exactly one stop, and each pair of stops is on exactly one line. Therefore, the problem is essentially asking for the order of a projective plane with 57 lines.In a finite projective plane of order n, the number of lines is n^2 + n + 1. Each line has n + 1 points, and each point is on n + 1 lines. The number of points is also n^2 + n + 1. So, if we have 57 routes, which correspond to lines, then n^2 + n + 1 = 57. Let's solve for n:n^2 + n + 1 = 57n^2 + n - 56 = 0Using the quadratic formula:n = [-1 ± sqrt(1 + 224)] / 2sqrt(225) = 15, so n = (-1 + 15)/2 = 14/2 = 7, or n = (-1 -15)/2 = -8, which is discarded.Therefore, n = 7. So, a projective plane of order 7 would have 7^2 + 7 + 1 = 57 lines, and each line has 7 + 1 = 8 stops. So, each route has 8 stops. Moreover, each stop is on 7 + 1 = 8 routes. Since the problem states that each route has no fewer than three stops, which is satisfied because 8 ≥ 3.Therefore, the answer should be 8. Each route has 8 stops.But let me double-check. Let me verify that the number of stops in total would be n^2 + n + 1 = 57 as well? Wait, no. Wait, in a projective plane of order n, the number of points is n^2 + n + 1, and the number of lines is the same. So, in our case, if there are 57 routes (lines), then the number of stops (points) is also 57. But each route has n + 1 stops, which is 8. Therefore, each route has 8 stops, and there are 57 stops in total? Wait, that can't be, because each route has 8 stops, and there are 57 routes. But if there are 57 stops, then each stop is on how many routes? In a projective plane, each point is on n + 1 lines, which would be 8 routes. So, each stop is part of 8 routes, and each route has 8 stops. The total number of stop-route memberships is 57 routes * 8 stops = 456, which should also equal 57 stops * 8 routes per stop = 456. That checks out.But wait, the problem says "each route has no fewer than three stops," which is satisfied because 8 ≥ 3. So, all conditions are satisfied. Therefore, each route has 8 stops.But let me just confirm that the initial problem doesn't mention the number of stops, only the number of routes. So, in a projective plane, the number of routes (lines) is equal to the number of stops (points). Therefore, if there are 57 routes, there are 57 stops. Each route has 8 stops, and each stop is on 8 routes.But let me verify another way. If we have 57 routes, each pair of routes intersects at one stop. The number of pairs of routes is C(57, 2) = 57*56/2 = 1596. Each stop is the intersection of how many pairs of routes? If each stop is on 8 routes, then the number of pairs of routes that intersect at that stop is C(8, 2) = 28. Therefore, the total number of intersecting pairs is 57 stops * 28 = 1596, which matches the total number of pairs of routes. So that works out.Alternatively, using the formula from design theory: in a projective plane of order n, the number of lines is v = n^2 + n + 1, each line has k = n + 1 points, each point is on r = n + 1 lines, the number of lines is equal to the number of points, and the number of lines through each point is r = n + 1. The number of pairwise intersections between lines is equal to the number of pairs of lines, which is C(v, 2), and each intersection point accounts for C(r, 2) pairs. Wait, actually, each intersection is a unique point, so each point is the intersection of C(r, 2) pairs of lines? Wait, no. Each point is on r lines, so the number of pairs of lines that intersect at that point is C(r, 2). Therefore, the total number of intersecting pairs is v * C(r, 2) = v * r*(r -1)/2. And this must equal the total number of pairs of lines, which is C(v, 2) = v*(v -1)/2. Therefore, setting them equal:v * r*(r -1)/2 = v*(v -1)/2Cancel v/2 from both sides:r*(r -1) = v -1But in a projective plane, v = n^2 + n + 1 and r = n +1. Therefore:(n +1)*n = (n^2 + n +1) -1Left side: n(n +1) = n^2 + nRight side: n^2 + n +1 -1 = n^2 + nTherefore, equality holds. So, this checks out.Therefore, given that the number of routes (lines) is 57, which equals n^2 + n +1, solving for n gives n=7, so each route has n +1=8 stops. Thus, the answer is 8.Final Answerboxed{8}</think>Given the problem of determining the number of stops each of the 57 bus routes has, we start by analyzing the conditions provided:1. From any stop, you can get to any other stop without transferring.2. For each pair of routes, there is exactly one stop where you can transfer from one route to the other.3. Each route has no fewer than three stops.These conditions suggest a structure similar to a finite projective plane. In a projective plane of order ( n ), the number of lines (routes) is ( n^2 + n + 1 ), and each line (route) has ( n + 1 ) points (stops). Additionally, each pair of lines intersects at exactly one point, and each pair of points lies on exactly one line.Given there are 57 routes, we solve for ( n ) in the equation ( n^2 + n + 1 = 57 ):[n^2 + n - 56 = 0]Using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[n = frac{-1 pm sqrt{1 + 224}}{2} = frac{-1 pm 15}{2}]This gives ( n = 7 ) (discarding the negative solution). Therefore, each route has ( n + 1 = 8 ) stops.Verifying, a projective plane of order 7 has 57 routes, each with 8 stops, and each stop is on 8 routes. This satisfies all conditions: the network is connected, each pair of routes intersects at exactly one stop, and each route has at least three stops.Thus, the number of stops each route has is (boxed{8}).]

🔑:Let there be n stops on some route a. Take a stop B that route a does not pass through. From B, there is a route to each of the n stops on route a, and there is exactly one such route, since two different routes cannot have two common stops. Each route passing through B intersects route a. Therefore, exactly n routes pass through B.Now it is clear that any route b that does not pass through stop B has exactly n stops.Indeed, as above, the number of stops on it is equal to the number of routes passing through B. Consider an arbitrary point A on route a, take a stop C on route A B different from A and B, and draw a route c through it, different from A B. Route c does not pass through B, so it has n stops. Stop A does not lie on c, so, as shown above, n routes (including a) pass through A. The same is true for each point on route a.Thus, a total of n(n-1) routes, different from a, pass through the points of route a, and there is also the route a itself. These are all the routes in the city.Solving the equation n(n-1)+1=57, we find that n=8.## Answer8 stops.

🧠:[<think>Okay, so I need to find the angle between these two planes given by the equations x + 2y - 2z - 7 = 0 and x + y - 35 = 0. Hmm, let me recall how to approach this. I remember that the angle between two planes is related to the angle between their normal vectors. Each plane equation can be written in the form Ax + By + Cz + D = 0, where the coefficients A, B, C correspond to the normal vector of the plane. So, for the first plane, x + 2y - 2z - 7 = 0, the normal vector should be (1, 2, -2). For the second plane, x + y - 35 = 0, there's no z term, so the normal vector would be (1, 1, 0). Wait, right. The normal vectors are just the coefficients of x, y, z in each equation. So normal vector n1 is (1, 2, -2) and n2 is (1, 1, 0). Now, to find the angle between the planes, I need to find the angle between these two normal vectors. The formula for the angle θ between two vectors is the dot product divided by the product of their magnitudes. So cosθ = (n1 · n2) / (|n1| |n2|). Then θ would be the arccos of that value. Let me compute the dot product first. n1 · n2 = (1)(1) + (2)(1) + (-2)(0) = 1 + 2 + 0 = 3. Now the magnitudes. |n1| is sqrt(1^2 + 2^2 + (-2)^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3. |n2| is sqrt(1^2 + 1^2 + 0^2) = sqrt(1 + 1 + 0) = sqrt(2). So cosθ = 3 / (3 * sqrt(2)) = 3 / (3sqrt(2)) = 1/sqrt(2). Therefore, θ = arccos(1/sqrt(2)). I know that arccos(1/sqrt(2)) is 45 degrees or π/4 radians. But wait, angles between planes are defined as the smallest angle between them. So even if the angle between the normals is obtuse, we take the acute angle. But in this case, since cosθ is positive, the angle is acute, so 45 degrees. Wait a second, let me check again. If the dot product is positive, the angle is acute, which is correct. So 45 degrees is the angle between the two planes. Let me just verify my calculations once more to be sure. Normal vectors: (1, 2, -2) and (1, 1, 0). Dot product: 1*1 + 2*1 + (-2)*0 = 1 + 2 + 0 = 3. Correct. |n1|: sqrt(1 + 4 + 4) = 3. Correct. |n2|: sqrt(1 + 1 + 0) = sqrt(2). Correct. So 3/(3*sqrt(2)) = 1/sqrt(2). Arccos(1/sqrt(2)) is indeed 45 degrees. So the angle between the two planes is 45 degrees. Alternatively, sometimes angles in 3D can be ambiguous, but since planes have two sides, the angle between them is defined as the smallest angle between their normals. So, yes, 45 degrees is correct. I think that's all. The key steps are identifying the normal vectors, computing their dot product and magnitudes, then taking the arccos of the ratio. Everything checks out.Final AnswerThe angle between the planes is boxed{dfrac{pi}{4}}.</think>To find the angle between the planes (x + 2y - 2z - 7 = 0) and (x + y - 35 = 0), we start by identifying their normal vectors. The normal vector for the first plane is (mathbf{n_1} = (1, 2, -2)) and for the second plane is (mathbf{n_2} = (1, 1, 0)).Next, we calculate the dot product of the normal vectors:[mathbf{n_1} cdot mathbf{n_2} = (1)(1) + (2)(1) + (-2)(0) = 1 + 2 + 0 = 3]We then find the magnitudes of the normal vectors:[|mathbf{n_1}| = sqrt{1^2 + 2^2 + (-2)^2} = sqrt{1 + 4 + 4} = sqrt{9} = 3][|mathbf{n_2}| = sqrt{1^2 + 1^2 + 0^2} = sqrt{1 + 1 + 0} = sqrt{2}]Using the dot product formula to find the cosine of the angle (theta) between the vectors:[costheta = frac{mathbf{n_1} cdot mathbf{n_2}}{|mathbf{n_1}| |mathbf{n_2}|} = frac{3}{3 sqrt{2}} = frac{1}{sqrt{2}}]Thus, the angle (theta) is:[theta = arccosleft(frac{1}{sqrt{2}}right) = frac{pi}{4}]The angle between the planes is (boxed{dfrac{pi}{4}}).,<think>Okay, so I need to find the angle between these two planes given by the equations x + 2y - 2z -7 = 0 and x + y -35 = 0. Hmm, I remember that the angle between two planes is related to their normal vectors. Let me recall how that works.First, I think each plane has a normal vector, which is perpendicular to the plane. For a plane equation ax + by + cz + d = 0, the normal vector is (a, b, c). So, I need to find the normal vectors of both planes and then find the angle between those two vectors. The angle between the planes should be the same as the angle between their normals, right? Or is it the supplement? Wait, maybe the angle between the planes is defined as the angle between their normals, but sometimes people might refer to the dihedral angle. I should check that.But according to what I remember, the angle θ between two planes is given by the angle between their normal vectors. However, if the angle between the normals is obtuse, then the angle between the planes would actually be acute, so we might need to take the acute angle in that case. Hmm, maybe I need to take the minimum between θ and 180 - θ. Let me think.But actually, the formula for the angle between two planes is the same as the angle between their normals. So, if the normals are n1 and n2, then the cosine of the angle is the dot product of n1 and n2 divided by the product of their magnitudes. So cosθ = (n1 ⋅ n2)/(|n1||n2|). Then θ is the angle between the two planes. That makes sense, right? Because the dihedral angle between two planes is the angle between their normals. Wait, maybe I should confirm this with a quick example.Suppose two planes are parallel, then their normals are scalar multiples, so the angle between them would be 0 or 180 degrees. But the angle between the planes is considered to be 0 in that case. So, perhaps we take the acute angle between the normals? Wait, no, if the normals are pointing in opposite directions, the angle between the planes is still 0 because they are parallel. So maybe we take the angle between the normals in [0, 180], and the angle between the planes is the acute or right angle between them, so if the angle between the normals is obtuse, we take 180 minus that. Wait, this is getting confusing. Let me check.Wait, actually, the angle between two planes is defined as the angle between them when you look at the smallest angle between them. So if the normals make an angle θ, then the angle between the planes is either θ or π - θ, whichever is smaller. Therefore, the angle between the planes is the acute angle between the two possible angles. Therefore, if the angle between the normals is acute, that's the angle between the planes; if it's obtuse, then the angle between the planes is π minus that. So, in formulas, the angle φ between the planes is given by φ = arccos(|n1 ⋅ n2| / (|n1||n2|)). Wait, so taking the absolute value of the cosine. That way, regardless of the orientation of the normals, you get the acute angle. That makes sense.Therefore, the formula for the angle between two planes is the arccosine of the absolute value of the dot product of the normals divided by the product of their magnitudes. So first, I need to find the normal vectors of both planes.Let me start by writing down the two equations:First plane: x + 2y - 2z -7 = 0. So the coefficients of x, y, z are 1, 2, -2. Therefore, the normal vector n1 is (1, 2, -2).Second plane: x + y -35 = 0. Wait, this equation is x + y -35 = 0. There is no z term. So, the coefficients are 1, 1, 0. Therefore, the normal vector n2 is (1, 1, 0).So, now I have n1 = (1, 2, -2) and n2 = (1, 1, 0). Now, I need to compute the dot product of n1 and n2.Dot product n1 ⋅ n2 = (1)(1) + (2)(1) + (-2)(0) = 1 + 2 + 0 = 3.Next, compute the magnitudes of n1 and n2.|n1| = sqrt(1^2 + 2^2 + (-2)^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3.|n2| = sqrt(1^2 + 1^2 + 0^2) = sqrt(1 + 1 + 0) = sqrt(2) ≈ 1.4142.So, the cosine of the angle between the normals is (n1 ⋅ n2)/(|n1||n2|) = 3 / (3 * sqrt(2)) = 1/sqrt(2). Then, taking the absolute value, which is still 1/sqrt(2).Therefore, the angle between the planes is arccos(1/sqrt(2)) which is 45 degrees, or π/4 radians.Wait, but let me make sure I didn't make a mistake. Let's verify each step.First, the normal vectors: n1 is (1, 2, -2) from the first plane equation. Correct. The second plane equation is x + y -35 = 0, which can be written as x + y + 0*z -35 = 0, so the normal vector is (1, 1, 0). Correct.Dot product: 1*1 + 2*1 + (-2)*0 = 1 + 2 + 0 = 3. Correct.|n1|: sqrt(1 + 4 + 4) = sqrt(9) = 3. Correct.|n2|: sqrt(1 + 1 + 0) = sqrt(2). Correct.Therefore, cosine theta is 3 / (3*sqrt(2)) = 1/sqrt(2). So theta is arccos(1/sqrt(2)) = 45 degrees. Since this is already acute, the angle between the planes is 45 degrees.Therefore, the answer should be 45 degrees or π/4 radians. But let me check if the formula is correct with taking absolute value. Wait, in the formula for the angle between planes, we take the absolute value of the dot product over the product of magnitudes. Let's confirm that.Yes, according to the definition, the angle between two planes is the angle between their normals, but if the normals form an obtuse angle, the planes actually form an acute angle. So, to get the smallest angle between the planes, you take the acute angle. Therefore, using the absolute value ensures that even if the dot product is negative, you take the absolute value to get the acute angle. In this case, the dot product is positive (3), so the angle is acute. Therefore, 45 degrees is correct.Therefore, the angle between the two planes is 45 degrees. So, converting that into radians if needed, but since the problem didn't specify, probably degrees is fine, or they might want the answer in radians. Let me check the original problem statement.Wait, in the example, the user just provided the problem and the answer format. The example is "Find the angle between the planes..." So, in the answer, we need to put it in boxed notation, probably in degrees. But maybe they want radians? However, in typical math problems, unless specified, angles between planes are often given in degrees. But since 45 degrees is π/4 radians, both are correct. Let me check which one is more appropriate.Wait, in calculus and linear algebra, sometimes angles are in radians, but in geometry problems, degrees are also common. The problem didn't specify. However, since the answer is a standard angle, 45 degrees is π/4. Let me check the calculation again to confirm.Yes, 1/sqrt(2) is approximately 0.7071, which is the cosine of 45 degrees. So yes, arccos(1/sqrt(2)) is π/4 radians or 45 degrees.Therefore, the answer is π/4 radians or 45 degrees. Since the problem is presented in an equation format without specifying, maybe we can check common conventions. In the context of planes in 3D geometry, both are acceptable, but in academic settings, radians are often preferred. However, since the problem is given with integer coefficients and likely expects a simple answer, 45 degrees is straightforward.But let me confirm once more. The formula for the angle between two planes is:cosθ = |(n1 ⋅ n2)| / (|n1||n2|)So, θ is the angle between 0 and 90 degrees, because we take the absolute value, right? So regardless of the orientation of the normals, the angle between the planes is acute or right. So, the angle is always reported as the smallest one between the two planes. Therefore, in this case, θ = arccos(1/sqrt(2)) = 45 degrees.Therefore, the answer is 45 degrees, which is π/4 radians. Depending on the required format, both are correct. The problem statement here is in Chinese, but since mathematical notation is universal, but given that the equations have integer coefficients and the answer is a common angle, probably acceptable to answer in degrees. However, in higher mathematics, radians are more standard. Wait, the example given by the user uses the same problem and the answer is likely expected in radians or degrees? Let me check the example.Wait, the user wrote "Example. Find the angle between the planes..." followed by the equations, but didn't provide an answer. Wait, no, actually, the user is giving me a problem to solve, similar to the example. Wait, wait, no. Wait, actually, the user says: "You will be given a problem. Please reason step by step, and put your final answer within boxed{}: Example. Find the angle between the planes [equations]".Wait, perhaps the user is giving an example of the problem format? The actual problem to solve is the example. Wait, maybe not. Let me check again.The user wrote:"You will be given a problem. Please reason step by step, and put your final answer within boxed{}:Example. Find the angle between the planesx+2 y-2 z-7=0, quad x+y-35=0"So, the example is the problem to solve. So, the user is providing an example of the problem that I need to solve. Therefore, the actual task is to find the angle between those two planes given by the equations above. Therefore, in the answer, I need to box the answer. So, the answer is 45 degrees, or π/4 radians. But as I said, maybe both are acceptable, but the problem might expect degrees. But in the absence of units, the answer is often given in radians. Wait, but in the problem statement, if I were to write the answer as 45°, then boxed{45^circ}, or boxed{dfrac{pi}{4}}.But in many calculus textbooks, angles between planes are expressed in radians unless specified. Wait, but I need to check. Let me see. For instance, in Stewart's calculus book, angles between planes are computed using the formula with the normals and then presented in radians. But since 45 degrees is a common angle, it might be presented in degrees. Hmm.Alternatively, the problem might just want the exact value in terms of inverse cosine, but since 1/sqrt(2) is a standard value, the angle is π/4. Therefore, to present the answer as π/4 radians. Alternatively, 45 degrees. But in the box, the user might expect radians. Let me think.Since in mathematical analysis, angles in radians are the standard measure, and π/4 is a more compact expression, I think it's safer to present the answer in radians. Therefore, the angle is π/4 radians.But let me confirm once again. The formula gives θ = arccos(1/sqrt(2)) which is exactly π/4. So, π/4 is 45 degrees. So both are correct, but radians are more standard in higher mathematics. So, I think the answer is π/4. Therefore, the boxed answer should be boxed{dfrac{pi}{4}}.Alternatively, if the problem expects degrees, then boxed{45^circ}. But since the problem is presented without units, maybe we can just put 45 in degrees? Wait, but in international math competitions, angles without units are typically assumed to be radians. But in the problem here, given that the equations are in integers, and the angle is a common one, maybe the answer is expected in degrees. Hmm. But to be safe, perhaps write both? Wait, no, the user probably expects one answer.Wait, in China, where the user might be from, math education typically uses degrees for angles in geometry problems unless specified otherwise. But I'm not sure. Alternatively, in 3D geometry, the angle between planes is often given in degrees. For example, in engineering, degrees are more common. But in pure math, radians. Since the problem is a pure math problem, maybe radians. However, given the problem is written with simple coefficients, and the angle is a common degree measurement, 45 degrees, I think the answer is 45 degrees. But to confirm, let me check similar problems online.Wait, actually, if I recall, in the formula, since we take the arccos of the absolute value of the dot product over the product of magnitudes, the result is always between 0 and π/2 radians (0 and 90 degrees). Therefore, the angle between two planes is always acute or right. Therefore, π/4 radians or 45 degrees. Therefore, either is correct. But since the problem didn't specify, and given that π/4 is the exact value, whereas 45 degrees is the degree equivalent, perhaps the answer is expected in radians as π/4.Therefore, I will go with π/4 radians. Therefore, the boxed answer is boxed{dfrac{pi}{4}}.Final AnswerThe angle between the planes is boxed{dfrac{pi}{4}}.</think>To find the angle between the planes given by the equations (x + 2y - 2z - 7 = 0) and (x + y - 35 = 0), we start by identifying the normal vectors of each plane. For the first plane (x + 2y - 2z - 7 = 0), the normal vector (mathbf{n_1}) is ((1, 2, -2)). For the second plane (x + y - 35 = 0), the normal vector (mathbf{n_2}) is ((1, 1, 0)).Next, we calculate the dot product of the normal vectors:[mathbf{n_1} cdot mathbf{n_2} = (1)(1) + (2)(1) + (-2)(0) = 1 + 2 + 0 = 3]We then find the magnitudes of the normal vectors:[|mathbf{n_1}| = sqrt{1^2 + 2^2 + (-2)^2} = sqrt{1 + 4 + 4} = sqrt{9} = 3][|mathbf{n_2}| = sqrt{1^2 + 1^2 + 0^2} = sqrt{1 + 1 + 0} = sqrt{2}]Using the formula for the cosine of the angle between two vectors:[cos theta = frac{|mathbf{n_1} cdot mathbf{n_2}|}{|mathbf{n_1}| |mathbf{n_2}|} = frac{3}{3 sqrt{2}} = frac{1}{sqrt{2}}]Taking the arccosine of this value gives the angle:[theta = arccosleft(frac{1}{sqrt{2}}right) = frac{pi}{4}]Thus, the angle between the planes is (boxed{dfrac{pi}{4}}).]

🔑:Solution. The dihedral angle between the planes is equal to the angle between their normal vectors vec{n}_{1}={1,2,-2} and vec{n}_{2}={1,1,0}. Therefore, the angle varphi between the planes is determined by the equalitycos varphi=frac{left(vec{n}_{1}, vec{n}_{2}right)}{left|vec{n}_{1}right| cdotleft|vec{n}_{2}right|}=frac{1 cdot 1+2 cdot 1-2 cdot 0}{sqrt{1^{2}+2^{2}+(-2)^{2}} sqrt{1^{2}+1^{2}}}=frac{1}{sqrt{2}}Thus, varphi=arccos (1 / sqrt{2})=pi / 4.Answer. The angle between the planes varphi=pi / 4.Problem Conditions. Find the angle between the planes.1. 3 x-y+3=0,2. x-y+3 z-5=0,3. 5 x-4 y+3 z-3=0,4. 5 x-3 y+2 z+5=0,5. 6 x+2 y-4 z+17=0,6. x-y+z sqrt{2}-5=0,7. y-3 z+5=0,8. 6 x+2 y-3 z+1=0,9. 2 x+y+2 z-5=0,10. 5 x-y+2 z+12=0, x-2 y+5 z-10=0.x+z-2=0.4 x-y-z+2=0.3 x+3 y-3 z-8=0.9 x+3 y-6 z-4=0.x+y-z sqrt{2}+7=0.y+2 z-3=0.x+6 y+2 z-10=0.12 x+16 y-15 z+2=0.3 x+2 y+z+10=0.Answers. 1. varphi=arccos frac{1}{2 sqrt{3}}. 2. varphi=arccos sqrt{frac{8}{11}}. 3. varphi=arccos frac{7}{10}. 4. varphi=frac{pi}{2}. 5. varphi=0. 6. varphi=frac{2 pi}{3}. 7. varphi=frac{pi}{4}. 8. varphi=frac{pi}{2}. 9. varphi=arccos frac{2}{15}. 10. varphi=arccos sqrt{frac{15}{28}}. 1.10. Canonical Equations of a LineProblem Statement. Write the canonical equations of the line defined as the intersection of two planes (by general equations)left{begin{array}{l}A_{1} x+B_{1} y+C_{1} z+D_{1}=0 A_{2} x+B_{2} y+C_{2} z+D_{2}=0end{array}right.Solution Plan.1. Check that the vectors vec{n}_{1}=left{A_{1}, B_{1}, C_{1}right} and vec{n}_{2}=left{A_{2}, B_{2}, C_{2}right} are non-collinear and, therefore, the planes intersect along some line.The canonical equations of a line with a direction vector vec{a}={l, m, n} passing through a given point M_{0}left(x_{0}, y_{0}, z_{0}right) are of the formfrac{x-x_{0}}{l}=frac{y-y_{0}}{m}=frac{z-z_{0}}{n}Therefore, to write the equations of the line, it is necessary to find its direction vector and some point on the line.2. Since the line belongs to both planes simultaneously, its direction vector vec{a} is orthogonal to the normal vectors of both planes, i.e., vec{a} perp vec{n}_{1}=left{A_{1}, B_{1}, C_{1}right} and vec{a} perp vec{n}_{2}=left{A_{2}, B_{2}, C_{2}right}. Therefore, the direction vector vec{a} is found using the formulavec{a}=left[vec{n}_{1}, vec{n}_{2}right]=left|begin{array}{ccc}vec{i} & vec{j} & vec{k} A_{1} & B_{1} & C_{1} A_{2} & B_{2} & C_{2}end{array}right|3. Now choose some point on the line. Since the direction vector of the line is not parallel to at least one of the coordinate planes, the line intersects this coordinate plane. Therefore, a point on the line can be taken as the point of intersection with this coordinate plane.4. Substitute the found direction vector and point into the line equations (1) and write the answer.

🧠:[<think>Alright, let me try to figure out how to solve this problem. So, I need to find the number of terms in the arithmetic sequence 6, 9, 12, …, 300. Hmm, arithmetic sequences... right, they have a common difference between consecutive terms. Let me recall some formulas related to arithmetic sequences.First, an arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. In this case, the first term is 6, and then each subsequent term increases by 3. Let me check: 6, 9, 12... yes, 9 - 6 = 3, 12 - 9 = 3, so the common difference (d) is 3.Now, the question is asking for the number of terms in this sequence up to 300. So, the last term given is 300. I need to find how many terms there are from 6 up to 300 with a common difference of 3.I remember that the formula for the nth term of an arithmetic sequence is a_n = a_1 + (n - 1)d, where:- a_n is the nth term,- a_1 is the first term,- d is the common difference,- n is the number of terms.In this problem, I need to find n when a_n = 300. Let me plug the values into the formula:300 = 6 + (n - 1)*3Okay, so let's solve for n. Subtract 6 from both sides:300 - 6 = (n - 1)*3294 = (n - 1)*3Now, divide both sides by 3:294 / 3 = n - 198 = n - 1Then, add 1 to both sides:98 + 1 = nn = 99Wait, so is the number of terms 99? Let me verify that. Let's check with smaller numbers to make sure I didn't make a mistake.For example, take the first few terms: 6, 9, 12. Let's see if the formula works here. The first term is 6, the third term should be 12. Let's plug n=3 into the formula:a_3 = 6 + (3 - 1)*3 = 6 + 2*3 = 6 + 6 = 12. That's correct.Similarly, if n=2, a_2 = 6 + (2 - 1)*3 = 6 + 3 = 9. That also checks out.So, the formula seems to work. Therefore, applying it to the last term 300 gives n=99. So, the number of terms should be 99. But let me confirm once again to avoid any off-by-one errors.Another way to think about this is: the sequence starts at 6 and each term increases by 3. So, the sequence can be represented as 6 + 3k, where k starts at 0. So, the nth term would correspond to k = n - 1. Wait, actually, if we let k be the number of steps from the first term, then the nth term is 6 + 3*(n-1). Which is the same as the formula we used before.So, setting that equal to 300:6 + 3*(n - 1) = 3003*(n - 1) = 294n - 1 = 98n = 99So, that's consistent. But maybe another way to check is to calculate how many steps of 3 are needed to get from 6 to 300.The total difference between the first term and the last term is 300 - 6 = 294. Since each step is 3, the number of steps would be 294 / 3 = 98. But since we started counting from the first term, which is term 1, we need to add 1 to include the first term. So, 98 + 1 = 99 terms. That also gives the same answer.Therefore, the number of terms is 99. Hmm, I think that makes sense. Let me just check with another example. Suppose we have a sequence 3, 6, 9. The difference is 3, first term 3, last term 9. The total difference is 9 - 3 = 6, divided by 3 is 2 steps, plus 1 equals 3 terms. Which is correct. So, same logic applies here.So, applying that same method to the original problem: (300 - 6)/3 + 1 = 294/3 + 1 = 98 + 1 = 99. Yep, that's the same answer. So, I think that's correct.Therefore, the number of terms in the arithmetic sequence is 99.Final AnswerThe number of terms is boxed{99}.</think>To find the number of terms in the arithmetic sequence 6, 9, 12, …, 300, we use the formula for the nth term of an arithmetic sequence:[ a_n = a_1 + (n - 1)d ]where:- ( a_n ) is the nth term,- ( a_1 ) is the first term,- ( d ) is the common difference,- ( n ) is the number of terms.Given:- The first term ( a_1 = 6 ),- The common difference ( d = 3 ),- The last term ( a_n = 300 ).We substitute these values into the formula:[ 300 = 6 + (n - 1) cdot 3 ]Solving for ( n ):1. Subtract 6 from both sides: [ 300 - 6 = (n - 1) cdot 3 ] [ 294 = (n - 1) cdot 3 ]2. Divide both sides by 3: [ frac{294}{3} = n - 1 ] [ 98 = n - 1 ]3. Add 1 to both sides: [ 98 + 1 = n ] [ n = 99 ]Thus, the number of terms in the arithmetic sequence is (boxed{99}).,<think>Okay, let's see. I need to find the number of terms in this arithmetic sequence: 6, 9, 12, …, 300. Hmm, arithmetic sequences have a constant difference between consecutive terms, right? So first, I should confirm that the difference is consistent here. Let me check the first few terms. From 6 to 9, that's an increase of 3. Then 9 to 12 is also an increase of 3. So the common difference, which is usually denoted as 'd', is 3. Got that down.Now, the problem is asking for the number of terms in this sequence. I remember there's a formula for the nth term of an arithmetic sequence. Let me recall. It should be something like a_n = a_1 + (n - 1)d, where a_n is the nth term, a_1 is the first term, d is the common difference, and n is the number of terms. Yeah, that sounds right. So if I can express the last term, which is 300, in terms of this formula, I can solve for n.Let me write down what I know. The first term a_1 is 6, the common difference d is 3, and the last term a_n is 300. Plugging these into the formula: 300 = 6 + (n - 1)*3. Now I need to solve for n. Let me do the algebra step by step. Subtract 6 from both sides: 300 - 6 = (n - 1)*3. That simplifies to 294 = (n - 1)*3. Then divide both sides by 3: 294 / 3 = n - 1. Let me compute that. 294 divided by 3... 3 goes into 294. 3*90 is 270, subtract that from 294, we get 24. 3 goes into 24 eight times. So 90 + 8 is 98. Therefore, 98 = n - 1. Then add 1 to both sides: n = 99. So the number of terms should be 99.Wait, let me verify that. Let me check if the 99th term is indeed 300. Using the formula again: a_99 = 6 + (99 - 1)*3. That's 6 + 98*3. 98*3 is 294, so 6 + 294 is 300. Yes, that checks out. So that seems right. But maybe I should think about another way to cross-verify.Another approach: the number of terms in an arithmetic sequence can also be thought of as how many steps of size d you take from the first term to reach the last term. Since each step increases the term by d, the total increase from a_1 to a_n is 300 - 6 = 294. Each step is 3, so the number of steps is 294 / 3 = 98. But since you start counting from the first term, you need to add 1 to include the starting term. Hence, the number of terms is 98 + 1 = 99. Yep, same answer. So that confirms it.But just to make sure I didn't make any arithmetic errors, let's do 294 divided by 3 again. 3*90 is 270, 294 - 270 is 24. 24 divided by 3 is 8. So 90 + 8 is 98. Then 98 steps, plus the initial term gives 99 terms. Yep, that's solid.Alternatively, maybe list out a few terms and see the pattern. The first term is 6 when n=1. Second term is 9 when n=2. Third term is 12 when n=3. So term n is 6 + 3(n-1). Let's set that equal to 300:6 + 3(n - 1) = 300Subtract 6: 3(n - 1) = 294Divide by 3: n - 1 = 98Add 1: n = 99Same result every time. I think that's pretty conclusive. Maybe another way to think about it: starting at 6, how many times do you add 3 to get to 300? So 6, then 6+3=9, 9+3=12, and so on. Each addition of 3 is a new term. The total number of additions needed is (300 - 6)/3 = 294/3 = 98. Since each addition corresponds to a term after the first one, the total number of terms is 98 + 1 = 99. Yep, that works too.Wait, but let me check with a smaller example to see if the formula works. Suppose the sequence is 3, 6, 9. How many terms? Obviously 3. Let's apply the formula. a_1=3, d=3, a_n=9. So 9 = 3 + (n-1)*3. Subtract 3: 6 = (n-1)*3. Divide by 3: 2 = n -1. So n=3. Correct. Another example: 5, 8. Two terms. Apply the formula. a_1=5, d=3, a_n=8. 8 = 5 + (n-1)*3. Subtract 5: 3 = (n-1)*3. Divide by 3: 1 = n -1. So n=2. Correct. So the formula works here.Therefore, applying the formula to the original problem should give the correct answer. So 99 terms. I think that's right. I don't see any mistakes in my reasoning. The key steps are identifying the first term, common difference, applying the nth term formula, and solving for n. Then verifying through multiple methods: step counting, formula application, and smaller examples. All lead to the same answer, so I can be confident it's correct.Final AnswerThe number of terms in the arithmetic sequence is boxed{99}.</think>To find the number of terms in the arithmetic sequence 6, 9, 12, ..., 300, we start by noting the first term (a_1 = 6), the common difference (d = 3), and the last term (a_n = 300).The formula for the (n)-th term of an arithmetic sequence is:[a_n = a_1 + (n - 1)d]Substituting the known values into the formula:[300 = 6 + (n - 1) cdot 3]Solving for (n):1. Subtract 6 from both sides: [ 294 = (n - 1) cdot 3 ]2. Divide both sides by 3: [ 98 = n - 1 ]3. Add 1 to both sides: [ n = 99 ]We verified the result using different methods: calculating the number of steps, checking smaller examples, and confirming the formula. All methods confirm the number of terms is 99.[boxed{99}]]

🔑:Given the problem, we can determine that the first term of the arithmetic sequence a_1=6, and the common difference d=9-6=3, with the last term a_n=300.Therefore, by using the formula for the nth term of an arithmetic sequence, we get 300=6+3(n-1).Solving for n, we find n=99. Thus, the number of terms in the sequence is boxed{99}.

🧠:[<think>Okay, so I need to simplify the radical √[2(6 - 2√3 - 2√5 + √15)]. Hmm, let's break this down step by step. First, let me look at the expression inside the square root. It's 2 multiplied by the quantity (6 - 2√3 - 2√5 + √15). Maybe I should start by distributing the 2 into the parentheses. Wait, but before I do that, maybe there's a smarter approach. Sometimes these kinds of expressions can be rewritten as a square of a sum or difference involving radicals, which would make simplifying easier. Let me think.So the original expression is √[2*(6 - 2√3 - 2√5 + √15)]. Let me compute the inner part first. Let's denote A = 6 - 2√3 - 2√5 + √15. Then the expression is √(2A). If I can express A in a form that is a perfect square, then multiplying by 2 might still allow for simplification. Alternatively, maybe 2A is a perfect square. Let's see.Let me try to write A as a product of two terms. Let's see: 6 - 2√3 - 2√5 + √15. Hmm. Let's group terms. Maybe group the constants and the radicals:(6) + (-2√3 -2√5) + (√15). Hmm, not sure. Alternatively, maybe factor something out. The terms -2√3 -2√5 can be factored as -2(√3 + √5). So then A = 6 - 2(√3 + √5) + √15. Maybe that helps?So A = 6 - 2(√3 + √5) + √15. Hmm. Let's see if we can express this as (a - b - c)^2 or something. Wait, but squaring usually gives cross terms. Let me think of possible terms. Suppose I consider expressions like (√3 + √5). Let's compute (√3 + √5)^2 = 3 + 2√15 + 5 = 8 + 2√15. That's not quite helpful here. Alternatively, (√3 - √5)^2 = 3 - 2√15 + 5 = 8 - 2√15. Hmm, also not matching.Alternatively, maybe consider (a√3 + b√5 + c)^2. Let's expand that. That would be a²*3 + b²*5 + c² + 2ab√15 + 2ac√3 + 2bc√5. Comparing this to the original expression A, which is 6 - 2√3 - 2√5 + √15. So maybe we can set up equations:The non-radical terms: 3a² + 5b² + c² = 6.The coefficient of √15: 2ab = 1 (since we have +√15 in A, which is 1*√15).The coefficient of √3: 2ac = -2.The coefficient of √5: 2bc = -2.So we have a system of equations:1) 3a² + 5b² + c² = 62) 2ab = 1 => ab = 1/23) 2ac = -2 => ac = -14) 2bc = -2 => bc = -1Let me try solving these equations. From equation 3: ac = -1. From equation 4: bc = -1. So ac = bc => a = b (since c ≠ 0). So a = b. Then from equation 2: ab = a² = 1/2 => a = ±1/√2. Let's take a = 1/√2 (we can check the negative later). Then b = 1/√2. From equation 3: ac = -1 => c = -1/a = -√2. Similarly, check equation 4: bc = (1/√2)(-√2) = -1, which matches.So now we have a = 1/√2, b = 1/√2, c = -√2. Let's check equation 1:3a² + 5b² + c² = 3*(1/2) + 5*(1/2) + (√2)^2 = (3/2 + 5/2) + 2 = (8/2) + 2 = 4 + 2 = 6. Perfect, that matches. So this works.Therefore, A = (a√3 + b√5 + c)^2 = ( (1/√2)√3 + (1/√2)√5 - √2 )². Let's simplify each term:(1/√2)√3 = √(3/2)(1/√2)√5 = √(5/2)So combining terms: √(3/2) + √(5/2) - √2.But let's write √(3/2) as (√6)/2 and √(5/2) as (√10)/2, and √2 is just √2. So:(√6/2 + √10/2 - √2) = ( √6 + √10 - 2√2 ) / 2.Wait, hold on. Let me compute that:√6/2 + √10/2 - √2 = (√6 + √10)/2 - √2. To combine these, write √2 as 2√2/2. So:(√6 + √10 - 2√2)/2. Therefore, A is equal to [ (√6 + √10 - 2√2)/2 ] squared.But let me confirm:Let me compute (√(3/2) + √(5/2) - √2)^2.First, expand the terms:[√(3/2) + √(5/2) - √2]^2 = [√(3/2) + √(5/2)]^2 - 2[√(3/2) + √(5/2)]√2 + (√2)^2.Compute each part:First part: [√(3/2) + √(5/2)]^2 = (3/2) + 2√(15/4) + (5/2) = (8/2) + 2*(√15)/2 = 4 + √15.Second part: -2[√(3/2) + √(5/2)]√2 = -2√2[√(3/2) + √(5/2)] = -2√2*(√3/√2 + √5/√2) = -2√2*( (√3 + √5)/√2 ) = -2*(√3 + √5).Third part: (√2)^2 = 2.Combine all parts: 4 + √15 - 2(√3 + √5) + 2 = (4 + 2) + √15 - 2√3 - 2√5 = 6 + √15 - 2√3 - 2√5. Which matches A! Perfect.Therefore, A = [√(3/2) + √(5/2) - √2]^2. Therefore, the original expression inside the square root is 2A = 2*[√(3/2) + √(5/2) - √2]^2.Therefore, √(2A) = √[2*(√(3/2) + √(5/2) - √2)^2] = √2 * |√(3/2) + √(5/2) - √2|. Since all terms under the square roots are positive, but we need to check if √(3/2) + √(5/2) - √2 is positive.Compute approximate values:√(3/2) ≈ 1.2247√(5/2) ≈ 1.5811√2 ≈ 1.4142So 1.2247 + 1.5811 = 2.8058; 2.8058 - 1.4142 ≈ 1.3916, which is positive. Therefore, the absolute value can be removed.Therefore, √(2A) = √2 * (√(3/2) + √(5/2) - √2). Let's simplify each term:√2 * √(3/2) = √( (2)*(3/2) ) = √3Similarly, √2 * √(5/2) = √5√2 * √2 = 2Therefore, the entire expression becomes √3 + √5 - 2.Wait, let's verify:√2*(√(3/2) + √(5/2) - √2) = √2*√(3/2) + √2*√(5/2) - √2*√2 = √(2*(3/2)) + √(2*(5/2)) - 2 = √3 + √5 - 2. Yes, that's correct.Therefore, the simplified radical is √3 + √5 - 2. Let me check if that squares back to the original expression inside.Compute (√3 + √5 - 2)^2 = (√3)^2 + (√5)^2 + (-2)^2 + 2*(√3*√5) + 2*(√3*(-2)) + 2*(√5*(-2)) = 3 + 5 + 4 + 2√15 - 4√3 - 4√5 = 12 + 2√15 - 4√3 - 4√5. Then multiply by 2 (since the original expression was √[2A] where A was 6 - 2√3 - 2√5 + √15). Wait, hold on, actually:Wait, the original expression is √[2A], where A = 6 - 2√3 - 2√5 + √15. Therefore, 2A = 12 - 4√3 - 4√5 + 2√15. And (√3 + √5 - 2)^2 = 12 + 2√15 - 4√3 - 4√5. Which matches 2A. Therefore, √(2A) = |√3 + √5 - 2|. Since √3 + √5 ≈ 1.732 + 2.236 ≈ 3.968, minus 2 is ≈1.968, which is positive, so absolute value is redundant. Hence, √(2A) = √3 + √5 - 2.Therefore, the simplified form is √3 + √5 - 2.But let me double-check. Let's compute both the original expression and the simplified expression numerically to confirm.Original expression inside the square root: 2*(6 - 2√3 - 2√5 + √15). Compute each term:6 ≈ 6-2√3 ≈ -2*1.732 ≈ -3.464-2√5 ≈ -2*2.236 ≈ -4.472√15 ≈ 3.872Adding them up: 6 - 3.464 -4.472 +3.872 ≈ 6 - 3.464 = 2.536; 2.536 -4.472 = -1.936; -1.936 +3.872 ≈ 1.936. Then multiply by 2: ≈3.872. Then take the square root: √3.872 ≈1.968.Now compute √3 + √5 -2 ≈1.732 +2.236 -2 ≈1.968. Same result. So the numerical check confirms.Therefore, the simplified radical is √3 + √5 - 2.Final AnswerThe simplified radical is boxed{sqrt{3} + sqrt{5} - 2}.</think>To simplify the radical (sqrt{2(6 - 2 sqrt{3} - 2 sqrt{5} + sqrt{15})}), we start by denoting the inner expression as (A = 6 - 2 sqrt{3} - 2 sqrt{5} + sqrt{15}). We aim to express (2A) as a perfect square.We hypothesize that (A) can be written as a square of a binomial involving radicals. After some algebraic manipulation and solving a system of equations, we find that:[A = left( sqrt{frac{3}{2}} + sqrt{frac{5}{2}} - sqrt{2} right)^2]Multiplying (A) by 2, we get:[2A = 2 left( sqrt{frac{3}{2}} + sqrt{frac{5}{2}} - sqrt{2} right)^2]Taking the square root of (2A), we have:[sqrt{2A} = sqrt{2} left| sqrt{frac{3}{2}} + sqrt{frac{5}{2}} - sqrt{2} right|]Since all terms are positive, we can remove the absolute value. Simplifying each term:[sqrt{2} left( sqrt{frac{3}{2}} + sqrt{frac{5}{2}} - sqrt{2} right) = sqrt{3} + sqrt{5} - 2]Thus, the simplified radical is:[boxed{sqrt{3} + sqrt{5} - 2}],<think>Okay, let's see. I need to simplify the radical √[2(6 - 2√3 - 2√5 + √15)]. Hmm, that looks a bit complicated. Maybe I can start by working inside the parentheses first and see if I can simplify that expression. Let me write down what's inside the square root:2 times (6 - 2√3 - 2√5 + √15). So first, maybe I should distribute the 2 into the parentheses. Wait, no, actually, the expression is √[2*(...)], so maybe I need to simplify the entire expression under the radical first. Let me check that again.The original problem is to simplify √[2(6 - 2√3 - 2√5 + √15)]. So the expression inside the square root is 2 multiplied by (6 - 2√3 - 2√5 + √15). Let me compute that first. Let's multiply 2 by each term inside the parentheses:2*6 = 122*(-2√3) = -4√32*(-2√5) = -4√52*(√15) = 2√15So putting it all together, the expression under the radical becomes:12 - 4√3 - 4√5 + 2√15.Hmm, now I need to take the square root of that. So the problem reduces to simplifying √[12 - 4√3 - 4√5 + 2√15]. That still looks complicated. Maybe there's a way to write this expression as a square of some binomial involving radicals. Typically, when simplifying radicals like this, if the expression is a perfect square, we can express it as (a + b√3 + c√5 + ...)^2, but this might get messy.Alternatively, perhaps group some terms. Let me see. Let's look at 12 - 4√3 - 4√5 + 2√15. Maybe group the terms without √15 and the terms with √15. So:(12 - 4√3 - 4√5) + 2√15.Alternatively, maybe factor some terms. The terms -4√3 and -4√5 can be factored as -4(√3 + √5). So:12 - 4(√3 + √5) + 2√15.Hmm, maybe that helps. Let me think. Let's consider if this can be expressed as (a - b√3 - c√5 + d√15)^2. Wait, but squaring such a term would result in cross terms. Alternatively, perhaps (√a - √b - √c)^2 or something like that. Let me try.Suppose we think of the expression 12 - 4(√3 + √5) + 2√15 as a square of (something). Let me try to assume it's a square of (x + y√3 + z√5 + w√15). But that might be too complicated. Maybe a better approach is to note that √15 is √3*√5, so perhaps the expression can be factored into terms involving (√3 - √5) or something similar.Alternatively, let me consider that 12 - 4√3 - 4√5 + 2√15. Let me check if this can be written as (a√3 + b√5 + c)^2. Let's try expanding (a√3 + b√5 + c)^2:= 3a² + 5b² + c² + 2ab√15 + 2ac√3 + 2bc√5.Compare that to the expression we have: 12 - 4√3 - 4√5 + 2√15. So we can set up equations by matching the coefficients:1. The constant term: 3a² + 5b² + c² = 122. The coefficient of √15: 2ab = 2 ⇒ ab = 13. The coefficient of √3: 2ac = -4 ⇒ ac = -24. The coefficient of √5: 2bc = -4 ⇒ bc = -2So we have the system:ab = 1,ac = -2,bc = -2.Let me try solving these equations. From ab = 1, we can express a = 1/b. Then substitute into ac = -2: (1/b)c = -2 ⇒ c = -2b.Then substitute c = -2b into bc = -2: b*(-2b) = -2 ⇒ -2b² = -2 ⇒ b² = 1 ⇒ b = 1 or b = -1.Case 1: b = 1Then a = 1/b = 1,c = -2b = -2*1 = -2.Now compute the constant term 3a² + 5b² + c² = 3(1) + 5(1) + (-2)^2 = 3 + 5 + 4 = 12. Perfect! That matches the constant term.So this gives us a = 1, b = 1, c = -2.Therefore, (a√3 + b√5 + c)^2 = (√3 + √5 - 2)^2.Let's check if expanding this gives the original expression:(√3 + √5 - 2)^2 = (√3)^2 + (√5)^2 + (-2)^2 + 2*(√3*√5) + 2*(√3*(-2)) + 2*(√5*(-2))= 3 + 5 + 4 + 2√15 - 4√3 - 4√5= 12 + 2√15 - 4√3 - 4√5Which is exactly the expression we have under the square root: 12 - 4√3 - 4√5 + 2√15. Wait, but in our calculation, it's 12 + 2√15 - 4√3 - 4√5, which is the same as 12 - 4√3 - 4√5 + 2√15. So yes, that's correct.Therefore, the expression under the square root is equal to (√3 + √5 - 2)^2. Therefore, taking the square root of that would give |√3 + √5 - 2|. However, since √3 ≈ 1.732 and √5 ≈ 2.236, so √3 + √5 ≈ 3.968, which is greater than 2, so √3 + √5 - 2 is positive. Therefore, the square root simplifies to √3 + √5 - 2.Wait, but hold on. The original problem was to simplify √[2*(6 - 2√3 - 2√5 + √15)]. Then we simplified the inside to √[12 -4√3 -4√5 +2√15], which turned out to be (√3 + √5 - 2)^2. So the square root of that is √3 + √5 - 2. Therefore, the simplified radical is √3 + √5 - 2.But let me double-check that. Let's compute (√3 + √5 - 2)^2:First, square each term:(√3)^2 = 3,(√5)^2 = 5,(-2)^2 = 4.Then cross terms:2*(√3*√5) = 2√15,2*(√3*(-2)) = -4√3,2*(√5*(-2)) = -4√5.So adding all together: 3 + 5 + 4 + 2√15 - 4√3 - 4√5 = 12 + 2√15 -4√3 -4√5. Which is exactly what we had. So yes, that's correct. So √[12 -4√3 -4√5 +2√15] = √3 + √5 - 2.Therefore, the original problem simplifies to √3 + √5 - 2. Let me check if that's the simplest form. Alternatively, maybe there's another way to write it, but I think that's as simplified as it gets. So the answer should be √3 + √5 - 2.But let me make sure there isn't a step I missed. Let's recap:Original expression: √[2*(6 - 2√3 -2√5 +√15)].Multiply 2 inside: 12 -4√3 -4√5 +2√15.Then realize that this is equal to (√3 + √5 -2)^2.Therefore, square root of that is √3 + √5 -2.Yes, that seems right. Let me check with approximate numbers to verify:Compute the original expression numerically:First, compute inside the radical:6 - 2√3 -2√5 +√15 ≈ 6 - 2*1.732 - 2*2.236 + 3.872 ≈ 6 - 3.464 -4.472 +3.872 ≈ 6 - 3.464 = 2.536; 2.536 -4.472 = -1.936; -1.936 +3.872 ≈ 1.936.Multiply by 2: 1.936*2 ≈ 3.872.Take square root: √3.872 ≈ 1.9676.Now compute √3 + √5 -2 ≈ 1.732 + 2.236 -2 ≈ 3.968 -2 ≈ 1.968. Which is approximately equal to 1.9676. So that checks out. Therefore, the answer is correct.Therefore, the simplified radical is √3 + √5 -2.Final Answerboxed{sqrt{3} + sqrt{5} - 2}</think>To simplify the radical (sqrt{2(6 - 2 sqrt{3} - 2 sqrt{5} + sqrt{15})}), we start by distributing the 2 inside the parentheses:[2 times (6 - 2 sqrt{3} - 2 sqrt{5} + sqrt{15}) = 12 - 4 sqrt{3} - 4 sqrt{5} + 2 sqrt{15}]Next, we need to simplify (sqrt{12 - 4 sqrt{3} - 4 sqrt{5} + 2 sqrt{15}}). We assume this expression can be written as the square of a binomial involving radicals. We consider the expression ((sqrt{3} + sqrt{5} - 2)^2):[(sqrt{3} + sqrt{5} - 2)^2 = (sqrt{3})^2 + (sqrt{5})^2 + (-2)^2 + 2(sqrt{3} cdot sqrt{5}) + 2(sqrt{3} cdot -2) + 2(sqrt{5} cdot -2)]Calculating each term:[= 3 + 5 + 4 + 2 sqrt{15} - 4 sqrt{3} - 4 sqrt{5}]Combining the terms:[= 12 + 2 sqrt{15} - 4 sqrt{3} - 4 sqrt{5}]This matches the expression under the radical (12 - 4 sqrt{3} - 4 sqrt{5} + 2 sqrt{15}). Therefore, we have:[sqrt{12 - 4 sqrt{3} - 4 sqrt{5} + 2 sqrt{15}} = sqrt{(sqrt{3} + sqrt{5} - 2)^2}]Since (sqrt{3} + sqrt{5} - 2) is positive, the square root simplifies to:[sqrt{3} + sqrt{5} - 2]Thus, the simplified radical is:[boxed{sqrt{3} + sqrt{5} - 2}]]

🔑:text { 2. } begin{aligned}text { Original expression } & =sqrt{12-4 sqrt{3}-4 sqrt{5}+2 sqrt{15}} & =sqrt{12-2 sqrt{3 times 4}-2 sqrt{4 times 5}+2 sqrt{3 times 5}} & =sqrt{(1 overline{3}-sqrt{4}+sqrt{5})^{2}} & =sqrt{3}+sqrt{5}-2 .end{aligned}Note: 12=(sqrt{3})^{2}+(sqrt{14})^{2}+(sqrt{5})^{2}.

🧠:[<think>Okay, so I need to figure out how many different ways five students can be divided into three groups for volunteer activities, with each group having at least one person, and students A and B can't be in the same group. Hmm, let me start by recalling some combinatorics principles.First, when dividing people into groups, if the groups are distinct (like different activities here), then we can use the principle of inclusion-exclusion or maybe Stirling numbers of the second kind adjusted for labeled groups. But since each group must have at least one person, the total number of ways without any restrictions would be the number of onto functions from the set of students to the set of groups. That is, for each student, we assign them to one of three groups, but subtract the cases where one or more groups are empty.But wait, the formula for that is 3^5 - 3*2^5 + 3*1^5, right? Let me check. Using inclusion-exclusion: total functions are 3^5. Subtract the ones where at least one group is empty: C(3,1)*2^5. Then add back in the ones where two groups are empty: C(3,2)*1^5. So yes, 3^5 - 3*2^5 + 3*1^5. Let me compute that:3^5 = 2432^5 = 32, so 3*32 = 961^5 = 1, so 3*1 = 3So total is 243 - 96 + 3 = 150. So 150 ways without any restrictions. But we have the restriction that A and B cannot be in the same group. So now, I need to subtract the number of allocations where A and B are in the same group from the total 150.Alternatively, maybe it's easier to compute the total number of allocations where A and B are in the same group and subtract that from 150. Let's try that approach.First, how many ways are there where A and B are in the same group?If A and B must be in the same group, let's treat them as a single entity. So instead of five students, we have four entities: AB, C, D, E. Now, we need to assign these four entities into three groups, each group non-empty. But wait, the groups are the activities, so each entity (including AB) has to be assigned to one of the three activities. But again, each activity must have at least one entity.So the number of ways is similar to before: 3^4 - 3*2^4 + 3*1^4. Let's compute that:3^4 = 812^4 = 16, so 3*16 = 481^4 = 1, so 3*1 = 3Total is 81 - 48 + 3 = 36. So there are 36 ways where A and B are in the same group.But wait, hold on. Wait, is that correct? Because when we treat AB as a single entity, we're assuming that AB are in one group, and the other students (C, D, E) can be in any group. But in this calculation, we are counting the number of onto functions from four entities to three groups, which is 36. But the original problem counts the allocations where each group has at least one person. So if we subtract the 36 from the total 150, then we get 150 - 36 = 114. Is that the answer?Wait, but I need to verify if this approach is correct. Let me think again.Alternatively, maybe we can compute directly the number of allocations where A and B are in different groups. Let's try that.First, assign student A to one of the three groups. Then assign student B to one of the remaining two groups (since they can't be in the same group). Then assign the other three students (C, D, E) to any of the three groups, but ensuring that all groups are non-empty.Wait, but this might get complicated because after assigning A and B to different groups, we have to make sure that the remaining students are assigned such that all three groups are non-empty. But perhaps some groups might already have A or B, so we need to handle cases where the third group is empty, but we need to ensure all groups are non-empty.This seems a bit more involved, but let's try.First, assign A to group 1, B to group 2. Then, we need to assign C, D, E to groups 1, 2, or 3, but ensuring that group 3 is not empty. Similarly, if A is in group 1 and B is in group 3, then group 2 must be assigned at least one person. So maybe this approach requires considering different cases.Alternatively, let's model this as inclusion-exclusion.Total number of allocations where A and B are in different groups, and all groups are non-empty.Total allocations with all groups non-empty: 150 (as before).Total allocations where A and B are in the same group: 36 (as before).Thus, the answer should be 150 - 36 = 114. So 114 ways.But let me verify with another approach.Suppose we first place A and B into different groups. Since the groups are labeled (different activities), the number of ways to place A and B into different groups is 3 choices for A's group, then 2 choices for B's group. So 3*2 = 6 ways.Now, we have three remaining students: C, D, E. We need to assign them to the three groups, but we must ensure that all three groups are non-empty. However, since A and B are already in different groups, the third group (the one without A or B) must get at least one of the remaining students.Wait, so let's think of it as after assigning A and B to two different groups, group 1 and group 2, then group 3 is empty. So we need to assign C, D, E such that group 3 gets at least one person, and groups 1 and 2 can have any number (including those already with A or B). Wait, but actually, since all groups must have at least one person, group 3 must have at least one person, and groups 1 and 2 already have one person each (A and B). So the remaining students (C, D, E) can go to any groups, but group 3 must have at least one person. So the number of assignments is total assignments minus assignments where group 3 gets none. Assigning C, D, E to three groups with group 3 allowed: each has 3 choices. But we need to subtract the cases where all three are assigned to groups 1 and 2. The number of ways to assign C, D, E to groups 1 and 2 is 2^3 = 8. So total is 3^3 - 2^3 = 27 - 8 = 19. Then multiply by the number of ways to assign A and B to different groups, which is 3*2 = 6. So total is 6*19 = 114. That matches the previous result. So 114.Therefore, the answer is 114. Let me check again.Alternatively, let's think of the process as:1. Assign A to a group: 3 choices.2. Assign B to a different group: 2 choices.3. Assign the remaining three students (C, D, E) to any of the three groups, but ensuring that the third group (the one not containing A or B) gets at least one person.So as above, assignments for C, D, E: 3^3 = 27. Subtract the cases where all three are assigned to A's group or B's group. Wait, no. Wait, the third group (the one without A and B) must have at least one person. So actually, if we don't assign anyone to the third group, that's invalid. So the number of valid assignments is 3^3 - 2^3 = 27 - 8 = 19. Then, since A could be in any group and B in any other, but since the groups are labeled, the number of ways to assign A and B to different groups is 3*2 = 6. So total is 6*19 = 114. Yep, same result.Alternatively, suppose we don't use this approach but instead use inclusion-exclusion. Let's confirm once more.Total number of allocations with all groups non-empty: 150.Number of allocations where A and B are together: 36. Therefore, the answer is 150 - 36 = 114. So both methods agree.But just to be thorough, let's verify the 36 number again. When A and B are in the same group, we can treat them as a single entity. So we have four entities: AB, C, D, E. The number of onto functions from four entities to three groups is 3! * S(4,3), where S(4,3) is the Stirling numbers of the second kind. Wait, but S(4,3) is 6. So 3! * 6 = 36. Alternatively, using inclusion-exclusion: 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36. Yes, that's correct.Therefore, 150 - 36 = 114. So the answer is 114.Alternatively, another way to compute the total number of allocations where A and B are not in the same group is:Total allocations with all groups non-empty: 150.Subtract allocations where A and B are in the same group: 36.So 150 - 36 = 114.Alternatively, let's try to compute it directly. Assign students to groups such that A and B are in different groups and all groups are non-empty.First, assign A to any group. Let's say there are three choices for A. Then, assign B to a different group, which has two choices. Now, the third group must have at least one person from the remaining three students (C, D, E). The remaining three students can be assigned to any of the three groups, but group 3 (the one with neither A nor B) must have at least one student.So the number of ways is 3 (for A's group) * 2 (for B's group) * [3^3 - 2^3] = 3*2*(27 - 8) = 6*19 = 114. Same as before.Therefore, the answer is 114. I think that's solid. Let me check if any step has a mistake.Wait, when we compute the assignments for C, D, E, it's 3^3 total, minus the 2^3 where all are assigned to the groups with A or B. But since the third group must have at least one person, this subtraction is correct. Then multiply by the number of ways to place A and B, which is 3*2. So yes, that's correct.Alternatively, if we consider the groups as labeled (which they are, because different activities), then the total number is indeed 3*2*19 = 114.Alternatively, another approach is using the principle of inclusion-exclusion for the forbidden assignments. But since the forbidden assignments are A and B being together, and the total is 150, subtract the forbidden ones (36) gives 114. That's straightforward.Alternatively, if we use multinomial coefficients. The number of ways to partition 5 students into three groups of sizes k1, k2, k3 where each ki >=1. Since the groups are labeled, the number is 3! * S(5,3), where S(5,3) is the Stirling numbers of the second kind. Wait, S(5,3) is 25, so 3! * 25 = 150, which matches our previous total. Then, the number with A and B together: Let's compute using Stirling numbers. If A and B are in the same group, then we can consider them as a single entity, so we have 4 entities, and we need to partition them into 3 groups, which is S(4,3) = 6. Then multiplied by 3! (since the groups are labeled) gives 6*6=36. Thus, same result. Therefore, the answer is 150 - 36 = 114. So yes, 114.So multiple approaches confirm the answer is 114. Therefore, the number of different allocation methods is 114.Final Answerboxed{114}</think>To determine the number of different allocation methods for dividing five students into three groups with each group having at least one person, and students A and B not being in the same group, we can use the following reasoning:1. Total number of ways to divide five students into three groups (each group non-empty): Using the principle of inclusion-exclusion: [ 3^5 - 3 cdot 2^5 + 3 cdot 1^5 = 243 - 96 + 3 = 150 ]2. Number of ways where A and B are in the same group: Treat A and B as a single entity, reducing the problem to dividing four entities (AB, C, D, E) into three groups: [ 3^4 - 3 cdot 2^4 + 3 cdot 1^4 = 81 - 48 + 3 = 36 ]3. Subtract the invalid cases from the total: [ 150 - 36 = 114 ]Thus, the number of different allocation methods is (boxed{114}).,<think>Okay, so the problem is about dividing five students into three groups for volunteer activities. Each group needs to have at least one person, and students A and B can't be in the same group. We need to find the number of different allocation methods. Hmm, let me think step by step.First, without any restrictions, how many ways can we divide five students into three groups where each group has at least one person? I remember that this is a problem related to partitions of a set. The formula for the number of ways to partition a set of n elements into k non-empty subsets is given by the Stirling numbers of the second kind, denoted S(n, k). Then, since the groups are doing different activities, we need to multiply by k! to account for the order of the groups. So the total number would be S(5, 3) multiplied by 3!.But wait, let me check if that's right. The Stirling number S(n, k) counts the number of ways to partition a set into k non-empty subsets, without considering the order of the subsets. However, in this problem, the groups are distinct because they're participating in different activities. So yes, we do need to multiply by 3! to assign each of the three groups to the three different activities. So the total number is S(5,3) * 3!.Now, let me recall the value of S(5,3). The formula for Stirling numbers of the second kind is recursive, but maybe I remember the value. Alternatively, I can calculate it. The formula is:S(n, k) = S(n-1, k-1) + k * S(n-1, k)So S(5,3) = S(4,2) + 3*S(4,3)Similarly, S(4,2) = S(3,1) + 2*S(3,2) = 1 + 2*S(3,2)S(3,2) = S(2,1) + 2*S(2,2) = 1 + 2*1 = 3So S(4,2) = 1 + 2*3 = 7Then S(4,3) = S(3,2) + 3*S(3,3) = 3 + 3*1 = 6Therefore S(5,3) = 7 + 3*6 = 7 + 18 = 25Therefore, the total number of ways without any restrictions is 25 * 6 = 150. Hmm, let me check this with another method to be sure.Alternatively, the number of onto functions from a set of 5 elements to 3 elements is 3! * S(5,3). So that's the same as above. So if S(5,3)=25, then 25*6=150. That seems right.Wait, but another way to calculate this is using inclusion-exclusion. The number of ways to assign 5 students to 3 groups where each group is non-empty is 3^5 - 3*2^5 + 3*1^5. Let's compute that:3^5 = 2432^5 = 32, so 3*32 = 961^5 =1, so 3*1=3So 243 - 96 +3 = 150. Yes, so that's the same answer. So that's correct. So without restrictions, there are 150 ways.But now, we have the restriction that A and B cannot be in the same group. So we need to subtract the number of allocations where A and B are in the same group from the total.Alternatively, maybe it's easier to compute the total number without restrictions (150) and subtract the number of allocations where A and B are in the same group. Let me try that approach.So first, total is 150. Now, how many allocations have A and B in the same group?To compute this, we can consider A and B as a single entity. So we're effectively dividing the remaining 3 students (C, D, E) along with the AB group into three different activities. But we need to ensure that each activity has at least one person. However, combining A and B into a single entity might complicate things since the AB group can be in any of the three activities. Wait, but perhaps a better approach is:If A and B are in the same group, then that group can be any of the three activities, and the other two activities must have at least one person each.So let's model this as:First, assign A and B to one of the three groups. Then, assign the remaining three students (C, D, E) to the three groups, with the condition that the other two groups (the ones not containing AB) must each have at least one person.Wait, but since each group must have at least one person, the group containing AB is already non-empty. The other two groups must each have at least one person from the remaining three students.So, the problem reduces to assigning C, D, E to the three groups, but the two groups that don't contain AB must each have at least one person. Wait, but since there are three groups, and we have one group already occupied by AB, the remaining two groups need to be assigned the three students such that each of these two groups has at least one student. But wait, we have three students to assign to two groups, each group must have at least one student. However, we have three groups in total. Wait, maybe I need to clarify.Wait, the three groups are for three different activities. So even if we fix AB in one group, we still need to assign the other three students (C, D, E) into the remaining two groups (since the third group is already occupied by AB). Wait, no, that's not correct. Wait, if we have three groups, each group corresponds to an activity, but the activities are different. So if we fix AB into one group (say group 1), then we still have to assign C, D, E into the three groups, but with the constraint that groups 2 and 3 must each have at least one person. Because each group must have at least one person. So group 1 has AB, which is two people, but groups 2 and 3 need to have at least one person each. However, we have three people left (C, D, E). So we need to distribute them into groups 2 and 3 such that each of these groups has at least one person.Wait, but if we have three people to assign to two groups (groups 2 and 3), with each group having at least one person, then the number of ways is 2^3 - 2 = 6. Because each person can go to group 2 or 3, that's 2^3=8, minus the two cases where all go to group 2 or all go to group 3. So 8-2=6. However, since the groups are distinguishable (they're different activities), each assignment is unique. So 6 ways.But wait, but in addition to that, we also need to consider which group AB is assigned to. Originally, the groups are three different activities. So if we fix AB into group 1, the activities are still different. However, AB could be in group 1, 2, or 3. So we need to multiply by 3 for the three possible groups that AB can be assigned to.Therefore, the total number of allocations where A and B are in the same group is 3 * (number of ways to assign C, D, E to the remaining two groups with each having at least one person). So that's 3 * 6 = 18.Wait, but hold on, if we fix AB into group 1, then we need to assign C, D, E to groups 1, 2, 3 such that groups 2 and 3 have at least one person each. Wait, no. Wait, if AB are in group 1, then group 1 already has AB, so groups 2 and 3 need to have at least one person each. However, we have three people (C, D, E) to assign to all three groups, but groups 2 and 3 must have at least one person. However, group 1 can have more people as well (since AB is already there). So actually, we need to assign C, D, E to the three groups, with the condition that groups 2 and 3 each have at least one person. Group 1 can have any number, including zero.Wait, but since each group must have at least one person, but group 1 already has AB. So group 1 is already satisfied. Groups 2 and 3 must have at least one person each from the remaining three. So how many ways can we assign C, D, E to groups 1, 2, 3 with the condition that groups 2 and 3 are not empty. Wait, group 1 can have any number (including zero) of the remaining three, but groups 2 and 3 must each have at least one.So this is equivalent to: assign C, D, E to three groups (group1, group2, group3), with the constraints that group2 and group3 each have at least one person. So the total number of assignments is equal to the number of onto functions from {C,D,E} to {group2, group3}, multiplied by the number of ways to assign to group1. Wait, maybe not. Wait, actually, the problem is that group1 can have some of the remaining three as well, but groups 2 and 3 need to have at least one each. So how to count this?Alternatively, think of it as assigning each of C, D, E to either group1, group2, or group3, with the condition that group2 and group3 each get at least one person. So the total number is equal to the number of assignments where group2 and group3 are not empty, regardless of group1. So that would be total assignments minus the assignments where group2 is empty or group3 is empty, plus the assignments where both group2 and group3 are empty (inclusion-exclusion). But since group1 already has AB, the other groups can have zero or more, but groups 2 and 3 need to have at least one each.So total assignments without restrictions: 3^3 = 27.Subtract assignments where group2 is empty: 2^3 = 8 (assign all to group1 or group3).Subtract assignments where group3 is empty: 2^3 =8 (assign all to group1 or group2).But then add back the assignments where both group2 and group3 are empty: 1 (assign all to group1).So total valid assignments: 27 - 8 -8 +1 = 12.So for each choice of the group that AB is assigned to (group1, group2, or group3), the number of ways to assign the remaining three students is 12. Wait, but if AB is assigned to group1, then groups 2 and 3 must be non-empty. If AB is assigned to group2, then groups1 and3 must be non-empty. Similarly, if AB is assigned to group3, then groups1 and2 must be non-empty. So regardless of which group AB is in, the number of assignments for the remaining students is 12. Wait, but 3 * 12 = 36. But earlier, I thought it was 3 *6=18. Which is correct?Wait, perhaps my previous reasoning was wrong. Let me clarify. Suppose AB is in group1. Then, we need to assign C, D, E to the three groups, but groups2 and3 must each have at least one person. So group1 can have 0,1,2, or3 of the remaining students, but groups2 and3 must have at least1 each. So the number of assignments is as calculated by inclusion-exclusion: 27 -8 -8 +1=12. Therefore, if AB is in group1, there are 12 ways. Similarly, if AB is in group2, then groups1 and3 must have at least one person each. The number of assignments would be the same:12. Similarly, AB in group3, 12. Therefore, total number of allocations where A and B are in the same group is 3*12=36.Therefore, the total number of allocations without restrictions is 150, subtract the 36 where A and B are together, gives 150 -36=114. So the answer is 114?Wait, but let me check this again. Alternatively, maybe there's a different approach.Another way: Since A and B cannot be in the same group, let's consider assigning A and B first, then assign the remaining students.First, assign A to a group. There are 3 choices. Then, assign B to a different group. There are 2 choices. Then assign the remaining three students (C, D, E) to the three groups. Each of the three groups can have any number of students, but each group must have at least one person. Wait, but groups already have A and B in different groups, so groups1, group2, and group3 each already have at least one person (A is in one group, B is in another, and the third group is empty). Wait, no. If A is in group1, B is in group2, then group3 is empty. But each group must have at least one person, so group3 must have at least one person. Therefore, after assigning A and B to different groups, we need to assign C, D, E to the three groups such that the third group (which is empty) gets at least one person. Wait, this seems complicated.Alternatively, perhaps after placing A and B into different groups, we need to distribute the remaining three students into the three groups, but making sure that the third group (the one not containing A or B) gets at least one person. So let's model this.Suppose we have three groups: Group1, Group2, Group3.Case 1: A is in Group1, B is in Group2. Then Group3 has no one. We need to assign C, D, E to the three groups, but Group3 must have at least one person. So each of C, D, E can go to any group, but Group3 needs at least one. The number of ways is total assignments minus the assignments where Group3 gets none. Total assignments: 3^3=27. Assignments where Group3 gets none: 2^3=8. Therefore, 27-8=19. But wait, but in this case, Group1 already has A, so adding more people to Group1 is allowed. Similarly for Group2. So actually, the remaining three students can be assigned to any groups, but Group3 must have at least one. So 19.But since A could be in any of the three groups, and B must be in a different group. Wait, perhaps the number of ways to assign A and B is 3 choices for A's group, then 2 choices for B's group, so 3*2=6. Then, for each such assignment, we have 19 ways to assign C, D, E? Wait, but that would lead to 6*19=114, which matches the previous result. Therefore, the total number is 114. So that seems correct.Wait, but let me verify. Let's take the first approach where total is 150, subtract the allocations where A and B are together (36) to get 114. The second approach directly counts 6*19=114. So both methods give the same result, which is reassuring.Therefore, the answer is 114. Let me just check once more.Alternatively, think of it as arranging the five students into three distinct groups (since the activities are different), each group non-empty, with A and B not in the same group.First, total number of allocations is 3! * S(5,3)=6*25=150, as before.Number of allocations where A and B are in the same group: Let's compute that.If A and B are in the same group, we can consider them as a single entity. Then, we have four entities: AB, C, D, E. But wait, no: since the groups are labeled (because the activities are different), so the group containing AB is a specific group. Wait, maybe another way.Alternatively, to compute the number of allocations where A and B are in the same group:First, choose which group (activity) AB is assigned to: 3 choices.Then, assign the remaining three students (C, D, E) to the three groups. However, the other two groups (the ones not containing AB) must each have at least one person. So, similar to the previous reasoning, the number of ways to assign C, D, E is the number of onto functions from {C,D,E} to the other two groups. Wait, no. Because the three groups are all distinct. Wait, if AB is in group1, then groups2 and group3 must each have at least one person. So we need to assign C, D, E to group1, group2, group3, with group2 and group3 each getting at least one person.As calculated before, that's 12 ways per group assignment of AB. So total is 3*12=36. Hence, 150-36=114.Alternatively, using inclusion-exclusion: The number of allocations where A and B are together is equal to:Number of ways where A and B are in group1 + group2 + group3.For each group, the number is: Assign A and B to that group, then assign the remaining three students to the three groups, ensuring the other two groups each have at least one student. As calculated, that's 12 per group. So total 3*12=36.Therefore, the answer is 150-36=114. Therefore, the number of different allocation methods is 114.Yes, this seems consistent across multiple approaches. So I think the answer is 114.Final AnswerThe number of different allocation methods is boxed{114}.</think>To solve the problem of dividing five students into three groups with each group having at least one person and ensuring that students A and B are not in the same group, we can use the following steps:1. Calculate the total number of ways to divide five students into three groups without any restrictions: - The number of ways to partition 5 students into 3 non-empty groups is given by the Stirling numbers of the second kind, ( S(5, 3) ), which is 25. - Since the groups are distinct (due to different activities), we multiply by ( 3! ) to account for the permutations of the groups. Thus, the total number of ways is ( 25 times 6 = 150 ).2. Calculate the number of ways where A and B are in the same group: - Consider A and B as a single entity. This entity can be in any of the three groups. - The remaining three students (C, D, E) need to be assigned to the three groups such that the other two groups (not containing A and B) each have at least one person. - Using inclusion-exclusion principle, the number of ways to assign C, D, E to three groups (with the constraint that two specific groups must be non-empty) is calculated as follows: - Total assignments: ( 3^3 = 27 ) - Subtract assignments where one group is empty: ( 2 times 2^3 = 16 ) - Add back the case where both groups are empty: ( 1 ) - Valid assignments: ( 27 - 16 + 1 = 12 ) - Since A and B can be in any of the three groups, multiply by 3: ( 3 times 12 = 36 ).3. Subtract the invalid cases from the total: - Total valid allocations: ( 150 - 36 = 114 ).Thus, the number of different allocation methods is (boxed{114}).]

🔑:Analysis: This problem can be broken down into two steps:Step 1: Divide the five students into three groups such that students A and B are not in the same group.1. If the groups are divided into 1, 2, and 2 students, there are frac { C_{ 5 }^{ 1 } C_{ 4 }^{ 2 } C_{ 2 }^{ 2 }}{ A_{ 2 }^{ 2 }} = 15 ways. However, among these, there are C32 = 3 ways in which students A and B are in the same group. Thus, there are 15 - 3 = 12 valid grouping methods in this case.2. If the groups are divided into 3, 1, and 1 students, there are frac { C_{ 5 }^{ 3 } C_{ 2 }^{ 1 } C_{ 1 }^{ 1 }}{ A_{ 2 }^{ 2 }} = 10 ways. Similarly, there are C31 = 3 ways in which students A and B are in the same group. Thus, there are 10 - 3 = 7 valid grouping methods in this case.Combining both cases, there are a total of 12 + 7 = 19 ways to divide the students into groups such that students A and B are not together.Step 2: Permute the 3 groups to correspond to the 3 different volunteer activities. There are A33 = 6 ways to do this.Therefore, the total number of different allocation methods is 19 × 6 = boxed{114}.This problem involves the application of permutations and combinations, as well as the principles of counting in stages and by classification. It can also be solved indirectly to avoid case analysis.

🧠:[<think>Okay, so I need to find the standard equation of hyperbola ( C_1 ) which has the same focal length as ( C_2: frac{x^2}{7} - y^2 = 1 ), and ( C_1 ) passes through the point (3,1). The options are given, so maybe after figuring it out, I can check which one matches. Let me start by recalling some hyperbola basics.First, the standard form of a hyperbola is either ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ) (horizontal transverse axis) or ( frac{y^2}{a^2} - frac{x^2}{b^2} = 1 ) (vertical transverse axis). The focal length is the distance between the two foci, which is ( 2c ), where ( c^2 = a^2 + b^2 ) for hyperbolas. So, the focal length for both ( C_1 ) and ( C_2 ) must be equal. Let me first find the focal length of ( C_2 ).For ( C_2 ): ( frac{x^2}{7} - frac{y^2}{1} = 1 ). So, ( a^2 = 7 ), ( b^2 = 1 ). Then, ( c^2 = a^2 + b^2 = 7 + 1 = 8 ), so ( c = 2sqrt{2} ). Therefore, the focal length is ( 2c = 4sqrt{2} ). So, ( C_1 ) must also have a focal length of ( 4sqrt{2} ), meaning its ( c ) is also ( 2sqrt{2} ).Now, ( C_1 ) could be either horizontal or vertical. Let me denote its standard equation as either:1. ( frac{x^2}{A^2} - frac{y^2}{B^2} = 1 ), with ( A^2 + B^2 = c^2 = 8 )or2. ( frac{y^2}{A^2} - frac{x^2}{B^2} = 1 ), with ( A^2 + B^2 = c^2 = 8 )Additionally, ( C_1 ) passes through the point (3,1), so substituting x=3 and y=1 into the equation should satisfy it.So, let's first consider case 1: horizontal transverse axis.Equation: ( frac{3^2}{A^2} - frac{1^2}{B^2} = 1 ) => ( frac{9}{A^2} - frac{1}{B^2} = 1 ).But we also know that ( A^2 + B^2 = 8 ). Let me set up variables: let ( A^2 = a ), ( B^2 = b ). So:Equation 1: ( frac{9}{a} - frac{1}{b} = 1 )Equation 2: ( a + b = 8 )So, we need to solve these two equations for a and b. Let me express b from equation 2: ( b = 8 - a ). Substitute into equation 1:( frac{9}{a} - frac{1}{8 - a} = 1 )Let me solve this equation for a. Multiply both sides by ( a(8 - a) ):9(8 - a) - a = a(8 - a)Calculate left side:72 - 9a - a = 72 - 10aRight side: 8a - a^2So equation becomes:72 - 10a = 8a - a^2Bring all terms to left side:72 -10a -8a + a^2 = 0 => a^2 -18a +72=0Factor the quadratic:Looking for two numbers that multiply to 72 and add to -18. Let's see: factors of 72: 1&72, 2&36, 3&24, 4&18, 6&12, 8&9. Hmm, 6 and 12 multiply to 72 and add to 18. But the signs would both be negative. So (a - 6)(a - 12) = 0. Therefore, a = 6 or a = 12. But since a + b =8, if a=6, then b=2. If a=12, then b=8 -12= -4, which is impossible because b^2 cannot be negative. So only a=6 and b=2. Therefore, the equation is ( frac{x^2}{6} - frac{y^2}{2} =1 ). That's option A, and also appears in options C and D.Now, let's consider case 2: vertical transverse axis.Equation: ( frac{y^2}{A^2} - frac{x^2}{B^2} =1 ). Passes through (3,1), so substituting gives:( frac{1^2}{A^2} - frac{3^2}{B^2} =1 ) => ( frac{1}{A^2} - frac{9}{B^2} =1 )Again, ( A^2 + B^2 =8 ). Let me use the same substitution: ( A^2 = a ), ( B^2 = b ). So:Equation 1: ( frac{1}{a} - frac{9}{b} =1 )Equation 2: ( a + b =8 )Express b =8 - a, substitute into equation1:( frac{1}{a} - frac{9}{8 - a} =1 )Multiply both sides by a(8 -a):(8 -a) -9a =a(8 -a)Left side: 8 -a -9a =8 -10aRight side:8a -a^2So equation:8 -10a =8a -a^2Bring all terms to left:8 -10a -8a +a^2=0 =>a^2 -18a +8=0Wait, let's check that again:Original equation after multiplying:Left side: 8 -10aRight side:8a -a^2Bring all terms to left:8 -10a -8a +a^2=0So that's 8 -18a +a^2=0, which is a^2 -18a +8=0Quadratic equation: a^2 -18a +8=0This doesn't factor nicely, so use quadratic formula:a = [18 ± sqrt(324 -32)] /2 = [18 ± sqrt(292)] /2 = [18 ± 2*sqrt(73)] /2 = 9 ± sqrt(73)So a =9 + sqrt(73) or a=9 - sqrt(73). But since a and b must be positive (they are squares), let's check:sqrt(73) is approximately 8.544. So 9 - sqrt(73) ≈ 0.456, which is positive. So a=9 - sqrt(73), which is approximately 0.456, then b=8 -a=8 - (9 - sqrt(73))= -1 + sqrt(73). sqrt(73)≈8.544, so b≈-1 +8.544≈7.544, which is positive. So that's valid.The other solution a=9 + sqrt(73)≈17.544, then b=8 -a≈8 -17.544≈-9.544, which is negative, so discard.Thus, a=9 - sqrt(73), and b= sqrt(73) -1. Therefore, the equation is ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} -1}=1 ). That's option B, and also appears in option C.So, combining both cases, ( C_1 ) could be either ( frac{x^2}{6} - frac{y^2}{2}=1 ) (horizontal) or ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} -1}=1 ) (vertical). So the correct answer should include both options. Looking at the choices:A: Only horizontalB: Only vertical (the specific one)C: Both horizontal and vertical (as I found)D: Horizontal and another vertical ( y^2 - frac{x^2}{3}=1 ). Wait, what is that?Wait, let's check option D: ( y^2 - frac{x^2}{3}=1 ). Let's see if this hyperbola has the same focal length.For ( y^2 - x^2/3 =1 ), this is vertical hyperbola. Here, ( a^2=1 ), ( b^2=3 ). Then ( c^2 =a^2 +b^2=1 +3=4 ), so c=2. Focal length is 4. But ( C_2 ) has focal length ( 4sqrt{2} ). So 4 vs 4√2. They are different. Therefore, this hyperbola in option D does NOT have the same focal length. Hence, option D is incorrect.But why is this presented in the options? Maybe a miscalculation.Wait, maybe I made a mistake when considering case 2. Let me check again.Wait, for case 2, vertical hyperbola, after solving, I obtained ( A^2 =9 - sqrt{73} ), which is approximately 0.456, so A is sqrt(9 - sqrt(73))≈0.675. Then ( B^2= sqrt(73)-1≈7.544 ), so B≈2.747.But the standard equation would be ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73}-1}=1 ). However, this looks a bit strange because the denominators are written with radicals. Let me check if that's the case.But according to the calculation, yes. So that's option B. However, the problem is that 9 - sqrt(73) is approximately 0.456, which is positive, so that's okay. But the answer in B is written as ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} -1} ). Wait, but ( sqrt{73} -1 ) is positive, so denominators are positive. So that equation is valid.Alternatively, maybe there is another solution? But when I solved case 2, the vertical hyperbola, I ended up with only one valid solution. So the answer should be C, right? Because C includes both the horizontal hyperbola and the vertical one with those denominators.But in the options, D is presented as ( frac{x^2}{6} - frac{y^2}{2}=1 ) or ( y^2 - frac{x^2}{3}=1 ). But the latter doesn't have the same focal length. Let's verify again.( y^2 - frac{x^2}{3} =1 ). Here, ( a^2 =1 ), ( b^2 =3 ), so ( c^2=1+3=4 ), ( c=2 ), focal length 4. But ( C_2 ) had focal length (4sqrt{2} ). So not the same. Therefore, D is incorrect.But then why is D presented as an option? Maybe a trap if someone miscalculates the focal length. So the correct answer must be C, which includes both hyperbola orientations with correct denominators.Wait, but looking back at the options:A: ( frac{x^2}{6} - frac{y^2}{2}=1 )B: ( frac{y^2}{9-sqrt{73}} - frac{x^2}{sqrt{73}-1}=1 )C: A or BD: A or ( y^2 - frac{x^2}{3}=1 )So since B is a valid answer for the vertical case, and A is valid for the horizontal case, then C is correct. However, looking at the problem statement: "the standard equation of the hyperbola ( C_1 ) is ( )". So there might be two possible hyperbolas, hence the answer is C.But let me check if option B is indeed a standard equation. The standard form requires positive denominators. Since (9 - sqrt{73}) is positive? Wait, sqrt(73) is approx 8.544, so 9 -8.544≈0.456, which is positive. Similarly, sqrt(73) -1≈7.544, positive. So yes, both denominators are positive, so option B is indeed a standard equation. Therefore, the answer is C.But wait, another thought: sometimes when hyperbola equations have such denominators with radicals, maybe they are simplified in another way? For example, rationalizing the denominators or something. But in the options, B is presented as is, so it's acceptable.Alternatively, maybe there is a miscalculation in case 2. Let me check the solving process again.Case 2: vertical hyperbola, equation ( frac{1}{a} - frac{9}{b} =1 ), with ( a + b=8 ).Expressed as:( frac{1}{a} - frac{9}{8 - a} =1 )Multiply by ( a(8 -a) ):(8 -a) -9a = a(8 -a)Left side: 8 -a -9a =8 -10aRight side:8a -a^2So equation: 8 -10a =8a -a^2Bring everything to left:8 -10a -8a +a^2=0 → a^2 -18a +8=0Solutions: [18 ± sqrt(324 -32)] /2 = [18 ± sqrt(292)] /2 = [18 ± 2*sqrt(73)] /2 =9 ± sqrt(73). Correct.So a=9 - sqrt(73)≈0.456, then b=8 -a=8 - (9 - sqrt(73))=sqrt(73)-1≈7.544. So denominators are positive, as above. So yes, equation is ( frac{y^2}{9 - sqrt(73)} - frac{x^2}{sqrt(73) -1}=1 ), which is option B.Therefore, the answer is C.But looking at the options again, D suggests another hyperbola, which is simpler but incorrect. So why does D include ( y^2 - x^2/3=1 )? Maybe someone might miscalculate the focal length? For example, thinking that since 6 +2=8, then another combination where a^2 +b^2=8, but in the case of ( y^2 -x^2/3=1 ), a^2=1, b^2=3, sum is 4, not 8. So incorrect.Thus, the correct answer is C.But let me check the options again. The answer options are:A: ( frac{x^2}{6} - frac{y^2}{2}=1 )B: ( frac{y^2}{9-sqrt{73}} - frac{x^2}{sqrt{73}-1}=1 )C: A or BD: A or ( y^2 - frac{x^2}{3}=1 )Yes, so since there are two valid hyperbolas (horizontal and vertical) with the same focal length passing through (3,1), the answer is C. The other answer in D is incorrect because it has a different focal length.So the final answer is C. But wait, let me check once more.Wait, the problem says "the standard equation of the hyperbola ( C_1 ) is ( )". So maybe there are two possibilities, hence the answer is C. But let me confirm the calculation for the vertical hyperbola.Substituting the vertical hyperbola equation ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} -1}=1 ) with the point (3,1):Left side: ( frac{1^2}{9 - sqrt{73}} - frac{3^2}{sqrt{73} -1} )Compute first term: ( frac{1}{9 - sqrt{73}} ). Let's rationalize denominator:Multiply numerator and denominator by (9 + sqrt{73}):( frac{9 + sqrt{73}}{(9)^2 - (sqrt(73))^2} = frac{9 + sqrt{73}}{81 -73} = frac{9 + sqrt{73}}{8} )Second term: ( frac{9}{sqrt{73} -1} ). Rationalize:Multiply numerator and denominator by ( sqrt{73} +1 ):( frac{9(sqrt{73} +1)}{(sqrt{73})^2 -1^2} = frac{9(sqrt{73} +1)}{73 -1} = frac{9(sqrt{73} +1)}{72} = frac{sqrt{73} +1}{8} )So left side becomes:( frac{9 + sqrt{73}}{8} - frac{sqrt{73} +1}{8} = frac{9 + sqrt{73} - sqrt{73} -1}{8} = frac{8}{8}=1 ). So yes, it satisfies the equation. Therefore, the vertical hyperbola is indeed valid.Therefore, both A and B are valid, so answer C.But the options have C as "A or B", and D as "A or another". So C is correct.Therefore, the answer is C.Final Answerboxed{C}</think>To find the standard equation of the hyperbola ( C_1 ) that has the same focal length as ( C_2: frac{x^2}{7} - y^2 = 1 ) and passes through the point ((3, 1)):1. Calculate the focal length of ( C_2 ): - For ( C_2 ), ( a^2 = 7 ) and ( b^2 = 1 ). - The focal length is ( 2c ) where ( c^2 = a^2 + b^2 = 7 + 1 = 8 ). - Thus, ( c = 2sqrt{2} ) and the focal length is ( 4sqrt{2} ).2. Determine the focal length for ( C_1 ): - ( C_1 ) must have the same focal length ( 4sqrt{2} ), so ( c = 2sqrt{2} ).3. Consider both horizontal and vertical hyperbola cases for ( C_1 ): - Horizontal Hyperbola: - Equation: ( frac{x^2}{A^2} - frac{y^2}{B^2} = 1 ) - Substitute ((3, 1)): ( frac{9}{A^2} - frac{1}{B^2} = 1 ) - With ( A^2 + B^2 = 8 ), solving gives ( A^2 = 6 ) and ( B^2 = 2 ). - Equation: ( frac{x^2}{6} - frac{y^2}{2} = 1 ) (Option A). - Vertical Hyperbola: - Equation: ( frac{y^2}{A^2} - frac{x^2}{B^2} = 1 ) - Substitute ((3, 1)): ( frac{1}{A^2} - frac{9}{B^2} = 1 ) - With ( A^2 + B^2 = 8 ), solving gives ( A^2 = 9 - sqrt{73} ) and ( B^2 = sqrt{73} - 1 ). - Equation: ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} - 1} = 1 ) (Option B).4. Verify options: - Option C includes both valid equations from the horizontal and vertical cases. - Option D includes an invalid hyperbola with incorrect focal length.Thus, the correct answer is (boxed{C}).,<think>Okay, so I need to find the standard equation of hyperbola C₁ which has the same focal length as C₂: x²/7 - y² = 1, and passes through the point (3,1). The options are given, so maybe I can check each one, but first, let's recall what focal length means for a hyperbola.For a hyperbola in standard form x²/a² - y²/b² = 1, the foci are located at (±c, 0) where c² = a² + b². The focal length is the distance between the two foci, which is 2c. So, if C₁ has the same focal length as C₂, then their c values must be equal.First, let's find the focal length of C₂. For C₂: x²/7 - y² = 1, so a² = 7 and b² = 1. Therefore, c² = a² + b² = 7 + 1 = 8, so c = 2√2. Hence, the focal length is 2c = 4√2. Therefore, for hyperbola C₁, we need 2c = 4√2, so c = 2√2 as well. So, for C₁, c² = (2√2)² = 8.Now, the standard equation of C₁ could be either of the two forms: (x²/A² - y²/B² = 1) or (y²/A² - x²/B² = 1), depending on whether it opens horizontally or vertically. Since the problem doesn't specify the orientation, we have to consider both possibilities.So, for C₁, if it's horizontal, then c² = A² + B² = 8. If it's vertical, then c² = A² + B² = 8 as well (but here, the roles of x and y are swapped). Additionally, C₁ passes through the point (3,1), so plugging x=3 and y=1 into the equation should satisfy it.Let me check the options given. The answer choices are A, B, C, D. Let's analyze each.Option A: x²/6 - y²/2 = 1. Let's check if this has c² = 6 + 2 = 8. Yes, c² = 8, so focal length is same. Then, does it pass through (3,1)? Plug in x=3, y=1: (9)/6 - (1)/2 = 1.5 - 0.5 = 1. Which equals 1. So yes, it passes through. So A is a valid option.Option B: y²/(9 - sqrt(73)) - x²/(sqrt(73) - 1) = 1. Hmm, this looks complicated. Let's compute c² here. For a vertical hyperbola, c² = A² + B², where A² is the denominator under y², and B² is the denominator under x². Wait, but in standard form, the denominators under y² and x² are A² and B². But in hyperbola equations, the terms are positive. Here, the denominators are written as (9 - sqrt(73)) and (sqrt(73) -1). Wait, 9 - sqrt(73) is approximately 9 - 8.544 = 0.456, which is positive. Similarly, sqrt(73) -1 ≈ 8.544 -1 = 7.544, which is positive. So both denominators are positive, so that's okay.So A² = 9 - sqrt(73), and B² = sqrt(73) -1. Then c² = A² + B² = (9 - sqrt(73)) + (sqrt(73) -1) = 9 -1 = 8. So yes, c²=8. So focal length is same. Now, check if (3,1) lies on it.Compute left side: (1)² / (9 - sqrt(73)) - (3)² / (sqrt(73) -1) = 1/(9 - sqrt(73)) - 9/(sqrt(73) -1).Let me rationalize the denominators to compute this.First term: 1/(9 - sqrt(73)) = (9 + sqrt(73)) / [(9 - sqrt(73))(9 + sqrt(73))] = (9 + sqrt(73))/(81 - 73) = (9 + sqrt(73))/8.Second term: -9/(sqrt(73) -1) = -9(sqrt(73) +1)/[(sqrt(73) -1)(sqrt(73)+1)] = -9(sqrt(73) +1)/(73 -1) = -9(sqrt(73) +1)/72 = - (sqrt(73) +1)/8.So total expression: (9 + sqrt(73))/8 - (sqrt(73) +1)/8 = [9 + sqrt(73) - sqrt(73) -1]/8 = (8)/8 = 1. So yes, it equals 1. Therefore, point (3,1) is on hyperbola B. So B is also valid.So both A and B are possible. But let me check the options. Option C says A or B, and D says A or another equation. Wait, let's check option D: it says x²/6 - y²/2=1 or y² - x²/3=1. Let's check the second equation in D: y² - x²/3 =1. Let's compute c² here. For this hyperbola, since it's y²/a² - x²/b²=1, then c² = a² + b². Here, a²=1, b²=3, so c²=1 +3=4, so c=2. Therefore, the focal length would be 4, but in original C₂, the focal length is 4√2, which is approximately 5.656. So c=2 vs c=2√2. So focal length is different, so equation in D's second option is invalid. Therefore, D is incorrect. But let me confirm.Original C₂ has focal length 4√2, so C₁ must also have focal length 4√2. The second equation in D has focal length 4, which is different, so it's invalid. Therefore, D is not correct.Option C says either A or B. Since we saw both A and B are valid, then C is the answer. But wait, let me check if there are any other possible hyperbolas. The problem says "the standard equation", so there are two possibilities depending on orientation.Wait, but the options given are A, B, C, D. Since C includes A and B, which we have confirmed, but let me check if there are other hyperbolas with same focal length passing through (3,1). Let's suppose that maybe we can derive it.Suppose the hyperbola C₁ is horizontal: x²/A² - y²/B²=1, with A² + B²=8. Then, plugging (3,1) into the equation: 9/A² - 1/B²=1. We also have A² + B²=8. So we need to solve these two equations.Let me set variables: let’s denote A² = a, B² = b. Then, a + b = 8, and 9/a - 1/b =1. So from the first equation, b =8 -a. Substitute into the second equation:9/a - 1/(8 - a) =1.Multiply both sides by a(8 -a):9(8 -a) -a = a(8 -a)=> 72 -9a -a =8a -a²=>72 -10a =8a -a²Bring all terms to left:72 -10a -8a +a²=0=>a² -18a +72=0Solve quadratic equation: a = [18 ± sqrt(324 - 288)]/2 = [18 ± sqrt(36)]/2 = [18 ±6]/2.So a= (18+6)/2=24/2=12 or a=(18-6)/2=12/2=6.Therefore, a=12 or a=6. Then, since a + b=8, if a=12, b=8 -12= -4, which is invalid because b must be positive. So discard a=12. Then, a=6, so b=8 -6=2. Therefore, A²=6, B²=2. So the equation is x²/6 - y²/2=1, which is option A. So only one horizontal hyperbola.Now, if the hyperbola is vertical: y²/A² -x²/B²=1, with A² + B²=8. Then, plugging (3,1) into equation: 1/A² -9/B²=1. Let’s denote A² = a, B² = b. Then, a + b =8, and 1/a -9/b=1.From a + b=8, so b=8 -a. Substitute into equation:1/a -9/(8 -a)=1.Multiply both sides by a(8 -a):(8 -a) -9a = a(8 -a)=>8 -a -9a =8a -a²=>8 -10a=8a -a²Bring all terms to left:8 -10a -8a +a²=0=>a² -18a +8=0Wait, solving:8 -10a =8a -a²=> -10a -8a +a² +8=0=>a² -18a +8=0Then, quadratic equation: a = [18 ± sqrt(324 -32)]/2= [18 ± sqrt(292)]/2= [18 ± 2*sqrt(73)]/2=9 ± sqrt(73).So a=9 + sqrt(73) or a=9 - sqrt(73). Now, check if these are positive.sqrt(73)≈8.544, so 9 - sqrt(73)≈0.456, which is positive. 9 + sqrt(73)≈17.544, which is positive. But since a + b=8, if a=9 + sqrt(73), then b=8 -a=8 -9 -sqrt(73)= -1 -sqrt(73), which is negative. So invalid. Therefore, only valid solution is a=9 - sqrt(73), so b=8 - (9 - sqrt(73))= -1 + sqrt(73). sqrt(73)≈8.544, so -1 +8.544≈7.544, which is positive. So a=9 - sqrt(73)≈0.456, and b≈7.544. So the equation is y²/(9 - sqrt(73)) - x²/(sqrt(73)-1)=1, which is option B. Therefore, there are two hyperbolas: A and B. So answer is option C.But wait, let's check the options again. Option C is A or B, and D is A or another equation which is invalid. So the correct answer should be C. But let's check if the problem specifies anything else. The hyperbola must be in standard form. Options A and B are both standard forms. So answer C is correct.But wait, the options given are:A: x²/6 - y²/2=1B: y²/(9−sqrt(73)) -x²/(sqrt(73)−1)=1C: A or BD: A or y² -x²/3=1Since we found that both A and B are valid, the correct answer is C. But let me check the problem statement again. It says "the standard equation of the hyperbola C₁ is". Since there are two possible standard equations depending on orientation, the answer is C. However, let me check if option D's second equation is valid. The equation in D is y² -x²/3=1, which can be written as y²/1 -x²/3=1. Then c²=1 +3=4, so c=2, which gives focal length 4, but original c=2√2≈2.828, so different. Therefore, D is wrong.Therefore, the correct answer is C. However, in the options, C is listed as "A or B", which is correct, so the answer should be C.Wait, but the original problem is given by the user. Let me check again the options:A: x²/6 - y²/2=1B: y²/(9−sqrt(73)) -x²/(sqrt(73)−1)=1C: A or BD: A or y² -x²/3=1Yes, so C is the correct answer. However, the original problem in Chinese might have different ordering, but according to the translation provided, the options are A, B, C, D as above. Therefore, the answer is C. So the user should choose option C.But wait, let me check again the calculation for B. When we plugged in (3,1) into B, we did get 1. So B is correct. The problem is that B looks complicated, but mathematically, it's valid. So both A and B are correct, hence C is the answer. So the answer is C.Final Answerboxed{C}</think>To find the standard equation of hyperbola ( C_1 ) which has the same focal length as ( C_2: frac{x^2}{7} - y^2 = 1 ) and passes through the point ( (3, 1) ):1. Determine the focal length of ( C_2 ): - For ( C_2 ), ( a^2 = 7 ) and ( b^2 = 1 ). - The focal length ( c ) is calculated as ( c^2 = a^2 + b^2 = 7 + 1 = 8 ), so ( c = 2sqrt{2} ). - The focal length is ( 2c = 4sqrt{2} ).2. Check hyperbola equations for ( C_1 ): - ( C_1 ) must have the same focal length, so ( c^2 = 8 ).3. Check option A: - Equation: ( frac{x^2}{6} - frac{y^2}{2} = 1 ). - Calculate ( c^2 = 6 + 2 = 8 ), which matches. - Check if it passes through ( (3, 1) ): ( frac{9}{6} - frac{1}{2} = 1.5 - 0.5 = 1 ), which is valid.4. Check option B: - Equation: ( frac{y^2}{9 - sqrt{73}} - frac{x^2}{sqrt{73} - 1} = 1 ). - Calculate ( c^2 = (9 - sqrt{73}) + (sqrt{73} - 1) = 8 ), which matches. - Check if it passes through ( (3, 1) ): After rationalizing and simplifying, it equals 1, which is valid.5. Check option D: - Equation: ( y^2 - frac{x^2}{3} = 1 ). - Calculate ( c^2 = 1 + 3 = 4 ), which does not match ( 8 ), so invalid.6. Conclusion: - Both options A and B are valid, so the correct answer is C.boxed{C}]

🔑:Given that the hyperbola C_2: frac{x^2}{7} - y^2 = 1 has a focal length of 4sqrt{2}, we can deduce that the hyperbola C_1 must also have a focal length of 4sqrt{2}. This leads us to the equation 2c = 4sqrt{2}, which simplifies to c = 2sqrt{2}. We will consider two cases based on the orientation of the hyperbola C_1.Case 1: Focus on the x-axisFor a hyperbola with its focus on the x-axis, we use the standard form frac{x^2}{a^2} - frac{y^2}{b^2} = 1 where a > 0 and b > 0. Substituting the point (3,1) into this equation, we get:[frac{3^2}{a^2} - frac{1^2}{b^2} = 1]This simplifies to:[frac{9}{a^2} - frac{1}{b^2} = 1 quad text{(Equation 1)}]Given that c^2 = a^2 + b^2 = 8, we can solve these equations simultaneously to find a^2 = 6 and b^2 = 2. Thus, the standard equation of the hyperbola C_1 in this case is:[frac{x^2}{6} - frac{y^2}{2} = 1]Case 2: Focus on the y-axisFor a hyperbola with its focus on the y-axis, we use the standard form frac{y^2}{a^2} - frac{x^2}{b^2} = 1 where a > 0 and b > 0. Substituting the point (3,1) into this equation, we get:[frac{1^2}{a^2} - frac{3^2}{b^2} = 1]This simplifies to:[frac{1}{a^2} - frac{9}{b^2} = 1 quad text{(Equation 2)}]Given that c^2 = a^2 + b^2 = 8, we can solve these equations simultaneously to find a^2 = 9 - sqrt{73} and b^2 = sqrt{73} - 1. Thus, the standard equation of the hyperbola C_1 in this case is:[frac{y^2}{9-sqrt{73}} - frac{x^2}{sqrt{73}-1} = 1]Combining the results from both cases, the standard equation of the hyperbola C_1 can be either frac{x^2}{6} - frac{y^2}{2} = 1 or frac{y^2}{9-sqrt{73}} - frac{x^2}{sqrt{73}-1} = 1.Therefore, the correct answer is boxed{C}.