🧠:[<think>Okay, let's see. I need to figure out how many vertices and edges a convex polyhedron with 100 edges will have after all its original vertices are cut off by planes, and none of the cutting planes intersect inside or on the boundary of the polyhedron. Hmm, so this is like truncating each vertex, right? But since it's convex and the cutting planes don't interfere with each other, each original vertex is replaced by a new face, probably a triangular one if it's similar to truncation.First, I remember that when you truncate a polyhedron, each vertex is replaced by a new face, and each original face is modified into a new face with more edges. Also, each original edge is replaced by a new edge connecting the truncated parts. But I need to recall the exact changes in vertices, edges, and faces.Let me start by recalling Euler's formula: V - E + F = 2 for convex polyhedrons (since they are topologically equivalent to spheres). The original polyhedron has 100 edges. Let me note that if I can find the original number of vertices and faces, that might help. But I only know the number of edges here. Wait, maybe I need to use some relationships between the edges, vertices, and faces when truncating.In the original polyhedron, let me denote V as the original number of vertices, E = 100 edges, and F as the original number of faces. Then, Euler's formula tells us that V - 100 + F = 2, so V + F = 102. But I need more information. If I can find another equation relating V and F, maybe through the average number of edges per face or something.In polyhedrons, each edge is shared by two faces. So if I denote the average number of edges per face as, say, m, then total edges would be (m*F)/2 = E. Similarly, each vertex has a certain number of edges meeting there, say n on average, so total edges would be (n*V)/2 = E. But without knowing whether it's a regular polyhedron or not, maybe I can't proceed like this.Wait, but maybe the problem doesn't require the original number of vertices and faces. Maybe there's a general formula for truncation. Let me recall that truncating a polyhedron replaces each vertex with a new face (typically a polygon), and each original face becomes a larger polygon (since each corner is cut off). Each original edge is also replaced by a new edge.Specifically, when you truncate a vertex, you're replacing it with a face. The number of edges added per vertex would be equal to the number of edges meeting at that vertex. For example, in a cube, each vertex has 3 edges, so truncating a cube replaces each vertex with a triangular face, adding 3 new edges per vertex. But the original edges are each split into two edges. Wait, but in the problem statement, it's mentioned that all vertices are cut by planes such that no two cutting planes intersect inside or on the boundary of the polyhedron. So perhaps each truncation is such that the new faces don't intersect, meaning that the cuts are shallow enough. In such a case, each original edge would be intersected by two planes (from the adjacent vertices), thereby dividing the original edge into three parts? Wait, no. If you cut off a vertex with a plane, each edge adjacent to that vertex is cut once. So if two adjacent vertices are both truncated, their cutting planes will intersect the common edge at two different points, thereby replacing the original edge with a new edge between the two cuts. So each original edge is divided into a single new edge, right? Wait, maybe not. Let me visualize this.Suppose you have an edge connecting two vertices, A and B. When you truncate vertex A, you cut the edge near A, and when you truncate vertex B, you cut the edge near B. So the original edge AB is replaced by a new edge between the two cuts. So instead of one edge AB, you have one edge between the two truncated points. So the number of edges per original edge remains the same? Wait, but actually, when you truncate a vertex, each edge meeting at that vertex is cut once, and the truncation replaces the vertex with a polygon. Each edge is shared between two vertices, so each original edge is cut twice (once at each end), thereby creating a new edge in between the two cuts. Wait, so each original edge is split into three parts: the part near vertex A (which is removed), the part near vertex B (which is removed), and the middle part that connects the two truncated parts. But since we're removing the parts near the vertices, maybe the original edge is replaced by a single new edge where the middle part is. Wait, no, actually, when you truncate each vertex, the original edges are each truncated at both ends, but the middle part remains as an edge. However, the truncation at each vertex adds new edges around the newly formed face.Let me think of a simple example: a cube. A cube has 12 edges, 8 vertices, and 6 faces. When you truncate the cube's vertices, each vertex (which is 3-valent) becomes a triangular face. Each original edge is replaced by a new edge connecting the new triangular faces from adjacent vertices. Additionally, each original face (which was a square) becomes an octagon, since each corner is cut off. Wait, no, actually, truncating a cube moderately results in truncated cubes, where each original square face becomes a regular octagon and each original vertex becomes a regular triangle. But in that case, the number of edges can be calculated as follows: each original edge is replaced by a new edge between the triangle and the octagon. Each original edge is split into three parts? Wait, no. Let's count the edges.Original cube: 12 edges, 8 vertices, 6 faces.After truncation: Each vertex becomes a triangular face, so 8 new triangular faces. Each original face (a square) becomes an octagonal face. So total faces are 8 + 6 = 14. Each triangular face has 3 edges, each octagonal face has 8 edges. Total edges would be (8*3 + 6*8)/2 = (24 + 48)/2 = 72/2 = 36 edges. Wait, but original edges? Each original edge is adjacent to two octagons and connects two triangles? Wait, maybe each original edge is now replaced by a new edge where the octagons meet. Wait, actually, in the truncated cube, each original edge is replaced by a new edge that's part of the octagon. Wait, perhaps it's better to use Euler's formula.In the truncated cube, V' - E' + F' = 2.We know that F' = 14 (8 triangles + 6 octagons). Let's find E'. Each triangle has 3 edges, each octagon has 8 edges, total edges counted per face: 8*3 + 6*8 = 24 + 48 = 72. But each edge is shared by two faces, so E' = 72/2 = 36. Then, using Euler's formula: V' - 36 + 14 = 2 => V' = 24. So the truncated cube has 24 vertices, 36 edges, 14 faces. Original cube had 8 vertices, 12 edges, 6 faces. So truncating each vertex (8 vertices) added 8 triangular faces, and modified original square faces into octagons. The number of edges went from 12 to 36. That seems like a big jump. Alternatively, perhaps each original vertex is removed and replaced with a new face, and each original edge is split into three edges? Wait, but in the cube truncation, each original edge is between two vertices. Truncating each vertex cuts the edge at both ends, so the original edge is replaced by a single new edge connecting the two new faces. Wait, but in the calculation above, edges increased from 12 to 36. So 3 times as many edges. Maybe each original edge is split into three edges? No, in reality, each original edge is adjacent to two octagons and connects two triangles. Wait, no, each original edge is replaced by a single new edge in the truncated polyhedron. Wait, maybe the edges come from the original edges and the new edges around the truncated vertices.Wait, each original vertex is replaced by a triangular face. Each triangular face has three edges. These edges are new. Additionally, each original face (which was a square) now has each corner truncated, turning into an octagon with eight edges. But those edges are a combination of the original edges (which have been shortened) and the new edges from the truncation.Alternatively, perhaps the number of edges can be calculated as follows: each original edge is replaced by one new edge (the middle part between the two truncations), and each original vertex of degree n (with n edges) contributes n new edges from the truncation. But since each truncation at a vertex adds a face which is a polygon, the number of edges added per vertex is equal to the number of edges originally meeting at that vertex. However, these edges are shared between adjacent vertices.Wait, perhaps in general, truncating a polyhedron replaces each vertex with a new face, which is a regular polygon if the truncation is uniform, but here it's just a generic truncation where planes are cutting off the vertices without the planes intersecting each other inside or on the boundary. So the new face at each vertex is a polygon with as many edges as the degree of the vertex.In that case, for each original vertex of degree n, truncating it adds a new n-gonal face. Each original edge is connected to two vertices, so truncating both ends of the edge replaces the original edge with a new edge that connects the two new faces. Additionally, each original face, which had some number of edges, now has each of its edges truncated at both ends, so the original face becomes a polygon with twice as many edges? Wait, no.Wait, let's try to formalize this.Let the original polyhedron have V vertices, E edges, F faces.After truncation:- Each original vertex is replaced by a new face. The number of edges of this new face is equal to the degree (number of edges) of the original vertex. Let’s denote the degree of vertex i as d_i. Then the total number of edges added by these new faces is the sum over all vertices of d_i, but each such edge is shared between two new faces (from adjacent vertices). Wait, no. Each edge of the new face at vertex i corresponds to an original edge incident to vertex i. But truncating the vertex adds a new face whose edges correspond to the edges that were incident to the original vertex. So each original edge, which was incident to two vertices, will correspond to one edge of the new face at each end. Therefore, each original edge is associated with two new edges in the truncated polyhedron? Hmm, maybe not. Let me think.When you truncate a vertex, you're cutting off the tip, which creates a new face in place of the vertex. The edges of this new face correspond to the original edges that were meeting at that vertex. So each original edge, which connected two vertices, will now have a portion near each vertex replaced by an edge of the new face at that vertex. Therefore, each original edge is divided into three parts: the part near vertex A (removed), the part near vertex B (removed), and the middle part that becomes a new edge in the truncated polyhedron. Wait, but actually, when you truncate the vertices, each original edge is intersected by two planes (one at each end), resulting in the original edge being split into a single new edge between the two truncation planes. So each original edge is replaced by one new edge in the truncated polyhedron. Additionally, each vertex truncation adds a new face, which is a polygon with as many edges as the degree of the vertex. Each edge of this new polygon is adjacent to another new polygon from an adjacent vertex. Wait, no. Each edge of the new polygon at vertex i is a new edge that connects to the adjacent vertex's truncation.Wait, perhaps it's better to think in terms of duals or using Euler's formula. Let me try to figure out how truncation affects the counts.In general, when you truncate a polyhedron:- Each original vertex is replaced by a new face. The number of new faces is equal to the original number of vertices, V.- Each original face is transformed into a new face. The original face, which was a polygon with, say, m edges, becomes a polygon with 2m edges (since each corner is truncated, adding a new edge at each corner). Wait, no. Truncating each corner of a polygon replaces each original edge with a new edge, and each original vertex with a new edge. So an original m-gon face, after truncation, becomes an m-gon with each original edge split into two edges? Wait, maybe not. Let's think of a cube face, which is a square. After truncation, each corner is cut off, turning the square into an octagon. So each original edge is now flanked by two new edges from the truncation. So the original 4 edges become 8 edges. So in general, truncating a face with m edges results in a 2m-edged face? Wait, no. If you truncate a polygon by cutting off each vertex, each original edge is replaced by a new edge, and each original vertex is replaced by a new edge. Wait, actually, truncating a polygon's vertex turns each vertex into a new edge. So an m-gon, when truncated, becomes a 2m-gon. For example, a triangle becomes a hexagon, a square becomes an octagon, etc. So yes, each original face with m edges becomes a 2m-edged face.But in the case of a polyhedron, truncating each vertex affects all adjacent faces. So each original face, which was a polygon with m edges, will have each of its vertices truncated, thereby replacing each original vertex with a new edge. Therefore, the original m-edged face becomes a face with 2m edges? Wait, but each original edge is adjacent to two faces. Hmm, perhaps the original face's edges are each split into two by the truncation, but since each edge is shared by two faces, maybe each original edge contributes one edge to the original face's new face. Wait, this is getting confusing. Let's approach it step by step.Original polyhedron: V vertices, E edges, F faces.After truncation:- Each original vertex is removed and replaced by a new face. The number of new faces is V.- Each original face is modified. Suppose an original face was an m-sided polygon. Each of its m vertices is truncated, so each corner is cut off. This turns the m-sided polygon into a 2m-sided polygon? Wait, no. If you cut off each corner of an m-gon, you replace each original vertex with a new edge, so each original edge is now adjacent to a new edge from the truncation. So an m-gon becomes an m + m = 2m-gon? Wait, actually, when you truncate a polygon's vertex, each original edge is split into two edges at the point where the truncation plane intersects it. But since the truncation plane is cutting off the vertex, the intersection with the original edge is a point. Therefore, each original edge is divided into two parts: one part that remains with the original face and another part that is part of the new face created by truncating the vertex. Wait, but in 3D, the truncation of a vertex affects all adjacent faces. So each original edge is shared by two original faces. When you truncate the edge's two endpoints, the original edge is intersected by two planes (one at each end), effectively shortening the original edge. The portion between the two truncation planes becomes a new edge in the truncated polyhedron. The parts of the original edge beyond the truncation planes are removed. So each original edge contributes one new edge in the truncated polyhedron. Additionally, each truncation at a vertex creates a new face, which is a polygon with as many edges as the degree of the vertex. Each edge of this new polygon is adjacent to an edge from the truncation of an adjacent vertex. Wait, this is getting complicated. Let me try to calculate the number of edges after truncation.Each original edge is replaced by one new edge in the truncated polyhedron. Additionally, each vertex truncation adds a new face which is a polygon. Each edge of this polygon corresponds to an original edge that was incident to the vertex. Since each original edge is incident to two vertices, each original edge will contribute one edge to each of the two new faces created by truncating its endpoints. Therefore, the number of edges added by the vertex truncations is equal to the sum over all vertices of the degree of the vertex. However, since each original edge is counted twice (once for each endpoint), this sum is 2E. Therefore, the total number of edges from the vertex truncations is 2E. However, each edge of the new face is shared between two adjacent vertex truncations? Wait, no. Each edge of the new face at a vertex is a new edge that connects to an adjacent truncated vertex. Wait, perhaps not. Let me think of a single vertex truncation. If I have a vertex where three edges meet (like in a cube), truncating that vertex replaces it with a triangular face. Each edge that was incident to the original vertex is cut, and the triangular face has three edges, each corresponding to one of the original edges. However, each of these three edges is shared with the adjacent vertex's truncation. So in general, each edge added by the vertex truncation is shared between two vertex truncations (since each original edge connects two vertices). Therefore, the total number of new edges from vertex truncations is equal to the sum of the degrees of all vertices divided by 2. But the sum of the degrees of all vertices is 2E (since each edge is counted twice). Therefore, the number of edges from vertex truncations is 2E / 2 = E.Therefore, the total number of edges in the truncated polyhedron is the original edges (each original edge becomes one edge) plus the edges from the vertex truncations (which is E). So total edges would be E + E = 2E? Wait, but in the cube example, original E was 12, and after truncation it's 36. Which is 3E. So my previous reasoning must be wrong.Wait, let's use the cube as a test case. Original cube: V=8, E=12, F=6. After truncation, we have V'=24, E'=36, F'=14. So edges tripled. So E' = 3E. How does that happen?In the cube, each original edge is adjacent to two vertices. Truncating each vertex adds a triangular face. Each original edge is split into three parts? Wait, no. Each original edge is between two vertices. When you truncate each vertex, you cut the edge near each end, replacing the original edge with a new edge that connects the two truncated parts. So each original edge becomes one new edge. But also, each vertex truncation (which is a triangular face) has three edges. Since there are 8 vertices, each contributing 3 edges, that's 24 edges, but each edge is shared between two vertices, so 24 / 2 = 12 edges from the vertex truncations. Then, the original edges, which are now 12 edges, plus the 12 edges from vertex truncations gives 24 edges. But wait, in reality, the truncated cube has 36 edges, so my reasoning is flawed.Alternatively, perhaps each original edge is divided into three parts: two parts near the vertices (which become edges of the triangular faces) and one middle part (which remains as the edge between the truncated parts). So each original edge contributes three edges. Therefore, total edges would be 3E. In the cube, 3*12=36, which matches. But why three? Let's see: truncating each vertex adds a new edge for each original edge incident to it. Each original edge is incident to two vertices, so when both ends are truncated, you get two new edges per original edge? Wait, no. If you take an original edge, and you cut it once near each end, you divide it into three segments. The middle segment remains as a "central" edge, and the two segments near the vertices become part of the new triangular faces. However, each of those two segments is shared with another original edge. Wait, this is getting too vague.Alternatively, let's model it as follows:When you truncate a polyhedron, each original vertex is replaced by a new face. The number of edges of this face is equal to the number of edges incident to the original vertex (its degree). Each original edge is replaced by two new edges: one that is part of the new face at each end. Wait, but in reality, truncating the vertex cuts the original edge into a single new edge that connects the two new faces. Hmm.Alternatively, perhaps each original edge gives rise to two new edges in the truncated polyhedron: one for each adjacent face. Wait, but in the cube example, each original edge is part of two original faces. When you truncate the cube, each original edge is replaced by a single new edge where the two octagons meet. Additionally, the triangular faces at each vertex have their own edges. Each triangular face has three edges, which are shared with adjacent triangular faces? No, each edge of a triangular face is adjacent to an octagon. Wait, in the truncated cube, each triangular face (from the vertex truncation) is adjacent to three octagons. Each octagon is adjacent to eight edges. Let me count edges again: Each triangle has three edges, each octagon has eight edges. Total edges: (8 triangles * 3 + 6 octagons * 8)/2 = (24 + 48)/2 = 72/2 = 36 edges. So in this case, original edges (12) are replaced by something else.Wait, how do the original edges factor into this? In the original cube, each edge is part of two square faces. After truncation, each original edge is part of two octagonal faces. But the octagons have eight edges each, which would mean each original edge corresponds to one edge of the octagon. Since each octagon was originally a square with four edges, now it's an octagon with eight edges. So each original edge is split into two edges? But in reality, truncating each corner of the square adds a new edge at each corner, so each original edge is flanked by two new edges in the octagon. So each original edge is effectively divided into three parts: the original edge is now the middle part, and two new edges are added at the truncated corners. Wait, but that would make each original edge correspond to three edges in the octagon. But an octagon has eight edges, and the original square had four edges. So 4 original edges become 8 edges in the octagon, meaning each original edge is split into two. So each original edge is divided into two edges by the truncation at its two ends. Therefore, each original edge contributes two edges to the resulting polyhedron. But in the cube example, original 12 edges would become 24 edges, but the total edges after truncation are 36. So there must be another 12 edges coming from the vertex truncations. Since each vertex truncation adds a triangular face with three edges, and there are eight vertices, that would be 8*3=24 edges, but since each edge is shared between two faces, it's 24/2=12 edges. Therefore, total edges: 24 (from original edges) + 12 (from vertex truncations) = 36, which matches. So in this case, each original edge becomes two edges, and each vertex truncation adds three edges, but shared between two vertices.So generalizing, truncating a polyhedron:- Each original edge is split into two edges by truncations at its two endpoints. So original E edges become 2E edges.- Each original vertex of degree d (with d edges) is replaced by a d-sided face, which contributes d edges. However, each edge from this face is shared with an adjacent vertex's face. So total edges from vertex truncations is sum of degrees / 2 = (2E)/2 = E.Therefore, total edges after truncation: 2E (from original edges) + E (from vertex truncations) = 3E.In the cube example, 3*12=36 edges, which works. Therefore, in the problem, original E=100, so edges after truncation would be 3*100=300 edges.Wait, but let me verify this with another example. Take a tetrahedron: 4 vertices, 6 edges, 4 faces. Truncating it would result in a truncated tetrahedron with 12 faces: 4 triangular (original vertices) and 4 hexagonal (original faces). Wait, no. Wait, truncating a tetrahedron replaces each vertex with a triangular face and each original triangular face becomes a hexagon. So total faces: 4 + 4 = 8. Edges: each original edge is split into three parts? Wait, let's compute using the formula. Original E=6. Using the above logic, 3*6=18 edges. Let's check with Euler's formula. Truncated tetrahedron: 12 vertices, 18 edges, 8 faces. V - E + F = 12 - 18 + 8 = 2. Correct. So yes, the formula holds here as well. So it seems that truncating a polyhedron in this way triples the number of edges. So in general, edges after truncation = 3E.Therefore, if the original polyhedron has 100 edges, the resulting polyhedron will have 300 edges.Now, what about vertices? Let's think.Original polyhedron has V vertices. After truncation, each original vertex is replaced by a face. Each original face is modified, and each original edge is... Hmm, to find the number of vertices, let's consider that each original vertex is removed and replaced by a polygonal face. The vertices of this new face correspond to the original edges that met at the vertex. Each original edge, which connected two vertices, will have a new vertex at each end where it was truncated. Therefore, each original edge contributes two new vertices (one at each end). But actually, when you truncate an edge, you introduce a new vertex at the point where the truncation plane intersects the edge. Since each edge is truncated at both ends (by truncating its two vertices), each edge will have two new vertices? Wait, no. Each edge is intersected by two truncation planes (one at each end), creating one new vertex per edge end. However, each truncation plane at a vertex intersects all edges incident to that vertex, creating a new vertex for each such edge. Therefore, each original edge, which connects two vertices, will have two new vertices: one where it's intersected by the truncation plane at each end. Therefore, the total number of new vertices is 2E. However, each truncation at a vertex creates a polygonal face with as many vertices as the degree of the original vertex. Each of these vertices is a new vertex. Wait, but how does this relate?Wait, let's think of the cube again. Original cube has 8 vertices, 12 edges, 6 faces. After truncation, the truncated cube has 24 vertices. Each original vertex is replaced by three new vertices (the corners of the triangular face). Since each original vertex had three edges, truncating it adds three new vertices. Therefore, 8 vertices * 3 new vertices per vertex = 24 vertices. But each new vertex is shared between adjacent original vertices? Wait, no. Each original edge is between two original vertices. When you truncate each end of the edge, you get a new vertex on that edge. So each original edge gives rise to one new vertex? Wait, in the cube, original 12 edges, each truncated at both ends, but each truncation at a vertex affects three edges. So for each original edge, truncating at both ends adds two new vertices? Wait, but in the truncated cube, there are 24 vertices. Each triangular face has three vertices, and there are eight triangular faces. 8*3=24, but each vertex is shared between three triangular faces? Wait, no. Each vertex of the truncated cube is where one triangular face and two octagonal faces meet. Wait, how many vertices does the truncated cube have? It's 24, as per earlier. Each original edge is replaced by a new edge, and each original vertex is replaced by three new vertices. Wait, maybe each original edge's two truncations result in two new vertices. So 12 edges * 2 = 24 vertices. Yes, that makes sense. Each original edge, when truncated at both ends, contributes two new vertices. Therefore, the number of vertices in the truncated polyhedron is 2E.In the cube, 2*12=24, which matches. In the tetrahedron, original E=6, truncated tetrahedron has V'=12, which is 2*6=12. So that works. Therefore, in general, truncating a polyhedron in this way results in a polyhedron with 2E vertices.But wait, let me verify with another example. Take an octahedron: original V=6, E=12, F=8. Truncating it would result in a truncated octahedron, which has 36 edges, 24 vertices, and 14 faces. Wait, no: a truncated octahedron has 12 faces: 8 hexagonal and 6 square. Wait, original octahedron has 8 triangular faces, 6 vertices. Truncating each vertex (degree 4, since octahedron vertices are where four edges meet), so each vertex is replaced by a square face. Each original triangular face becomes a hexagon. So faces are 6 squares + 8 hexagons = 14 faces. Edges: each original edge is split into three parts? Wait, using the formula: 3E = 3*12=36 edges. Checking with Euler's formula: V' - 36 + 14 = 2 => V' = 24. Which matches. So V' = 24, which is 2E = 2*12=24. Correct. So yes, the number of vertices after truncation is 2E.Therefore, in general, when you truncate all vertices of a convex polyhedron such that no two cutting planes intersect inside or on the boundary, the resulting polyhedron has:- 3E edges- 2E verticesBut let me check again with the tetrahedron. Original E=6. Truncated tetrahedron has 18 edges and 12 vertices. 3*6=18, 2*6=12. Correct. So seems like this formula holds.Therefore, applying this to the problem: original polyhedron has 100 edges. After truncation, edges become 3*100=300, vertices become 2*100=200. So the answer should be 200 vertices and 300 edges.But wait, let me check if there are any conditions where this might not hold. The problem states that all vertices are cut by planes such that no two cutting planes intersect inside or on the boundary of the polyhedron. This implies that the truncation is such that the cutting planes only intersect at points outside the original polyhedron, which would mean that each truncation is a "shallow" cut, just removing the vertex without affecting the other parts. This is consistent with the truncation process we've considered, where each original edge is split into two parts (but in our earlier cube example, actually, each edge is split into three parts, but counted as one edge in the middle and two edges near the vertices). Wait, but according to our general formula, it's 3E edges and 2E vertices.But how is that possible? Let me think of a single edge. Original edge between two vertices. Truncate each vertex, cutting the edge near each end. The part of the edge that remains is the middle part, which is a new edge. The parts near the vertices are removed. So each original edge becomes one new edge. But according to our earlier reasoning, edges become 3E. There's a contradiction here.Wait, perhaps my initial assumption is wrong. Maybe when truncating each vertex, each original edge is indeed replaced by one new edge (the middle part), and the truncation at each vertex adds new edges. Let me re-examine the cube. Original edges: 12. After truncation, edges: 36. So 3 times as many. But according to this, each original edge is only one edge. How do we get the other 24 edges?Ah, perhaps the edges from the vertex truncations. Each vertex truncation adds edges around the new face. For the cube, each vertex truncation adds a triangular face with three edges. But each of those edges is shared between two adjacent vertex truncations. Wait, no. Each edge of the triangular face is adjacent to an octagonal face. Wait, in the truncated cube, each edge is either part of an octagon or part of a triangle. The edges from the triangles are new edges, and the edges from the octagons are modified original edges.Wait, maybe the total edges are composed of the edges from the original faces (now modified) and the edges from the new faces (vertex truncations). Each original face, which was an m-gon, becomes a 2m-gon? Wait, no, in the cube each square becomes an octagon (8 edges), which is 2*4. So each original edge is split into two parts by the truncations at its two ends. Therefore, each original edge is now two edges in the octagon. Wait, but that would mean each original edge becomes two edges, so 12 original edges become 24 edges. Then, the triangular faces from the vertex truncations add 8 vertices * 3 edges each = 24 edges, but since each edge is shared between two triangles (no, actually each edge of the triangle is shared with an octagon). Wait, each edge of the triangular face is adjacent to an octagon. Therefore, the edges from the triangles are new edges not counted in the original edges. So in total, the octagons contribute 6 faces * 8 edges = 48 edges, and the triangles contribute 8 faces * 3 edges = 24 edges. Total edges: (48 + 24)/2 = 36. So of these 36 edges, 48 edges come from octagons and 24 from triangles. The octagons' edges are derived from the original edges and the new edges added by truncation. Each original edge contributes two edges to the octagons (since each original edge is split into two by truncation). So original 12 edges * 2 = 24 edges for octagons. Then, the remaining edges of the octagons (48 - 24 = 24 edges) must come from the vertex truncations. Wait, but vertex truncations contribute edges to the triangles as well. Wait, this is confusing.Alternatively, when you truncate a vertex, the edges of the new face (triangle in the cube case) are new edges that connect to the adjacent truncated edges. So each edge of the triangular face is a new edge that didn't exist before. Similarly, the octagons have edges that are a combination of the original edges (split into two) and the new edges from the vertex truncations.But perhaps a better way is to recognize that truncating a polyhedron can be seen as a type of geometric dual operation combined with expansion. However, I need a more mathematical approach.Let me recall that when you truncate a polyhedron, the number of vertices becomes the original number of vertices plus the original number of edges. Wait, no. In the cube, original V=8, E=12. Truncated cube has V'=24. Which is 8 + 12*2? No. Wait, in the cube example, V'=24, which is 3 times original V=8. 24=3*8. But in the tetrahedron, original V=4, truncated V'=12, which is 3*4. But in the octahedron, original V=6, truncated V'=24=4*6. So that pattern doesn't hold.Alternatively, in the cube, each original vertex is replaced by three new vertices (the corners of the triangular face). So V'=original V * degree of each vertex. But in cube, each vertex has degree 3, so 8*3=24. In the octahedron, each vertex has degree 4, so 6*4=24. In the tetrahedron, each vertex has degree 3, so 4*3=12. So in general, if the original polyhedron is regular, then V'=V*d, where d is the degree of each vertex. However, in a non-regular polyhedron, vertices can have different degrees. So the number of vertices after truncation would be the sum over all vertices of their degrees. But each original edge is shared by two vertices, so the sum of degrees over all vertices is 2E. Therefore, the number of vertices after truncation is 2E.Wait, in the cube, sum of degrees is 8*3=24, which is 2E=24, since E=12. So yes, V'=24=2E. Similarly, in the tetrahedron, sum of degrees is 4*3=12=2E=12 (since E=6). And the octahedron: sum of degrees is 6*4=24=2E=24 (since E=12). Therefore, in general, for any polyhedron, regular or not, the number of vertices after truncation is equal to twice the original number of edges. Therefore, the answer for the number of vertices is 2E=200 when E=100. For edges, as we saw in the regular polyhedron examples, it's 3E. Therefore, edges become 3*100=300.Hence, the resulting polyhedron has 200 vertices and 300 edges.Final AnswerThe resulting polyhedron has boxed{200} vertices and boxed{300} edges.</think>To determine the number of vertices and edges of the resulting polyhedron after truncating all vertices of a convex polyhedron with 100 edges, we can use the following reasoning:1. Truncation Effects: - Each original vertex is replaced by a new face. The number of edges of this new face is equal to the degree of the original vertex. - Each original edge is intersected by two truncation planes (one at each end), resulting in the original edge being replaced by a new edge in the truncated polyhedron. - Each original face is transformed into a new face with more edges due to the truncation of its vertices.2. Edge Calculation: - Each original edge is replaced by one new edge in the truncated polyhedron. - Additionally, each vertex truncation adds new edges corresponding to the degree of the original vertex. However, each edge added by the vertex truncation is shared between two adjacent vertices. - The total number of edges after truncation is three times the original number of edges (3E). This is because each original edge contributes one edge, and each vertex truncation adds edges equal to twice the original number of edges (but shared between two vertices).3. Vertex Calculation: - Each original edge is split into two parts by the truncation planes, resulting in two new vertices for each original edge. - The total number of vertices after truncation is twice the original number of edges (2E).Using these principles, we find:- The original polyhedron has 100 edges.- After truncation, the number of edges becomes (3 times 100 = 300).- The number of vertices becomes (2 times 100 = 200).Thus, the resulting polyhedron has (boxed{200}) vertices and (boxed{300}) edges.,<think>Okay, let me try to figure out this problem. So, we have a convex polyhedron with 100 edges. Then, all the vertices are cut by planes in such a way that no two cutting planes intersect inside or on the boundary of the polyhedron. We need to find out how many vertices and edges the resulting polyhedron will have. Hmm, interesting. Let me start by recalling some basics about polyhedrons and their properties.First, a convex polyhedron has vertices, edges, and faces. I know Euler's formula relates these: V - E + F = 2, where V is vertices, E is edges, and F is faces. Here, the original polyhedron has 100 edges. So, E_original = 100. But we don't know the original number of vertices or faces. Maybe we can find some relationships here.The problem says all vertices are cut by planes, and no two cutting planes intersect inside or on the boundary of the polyhedron. So, each vertex is being truncated individually, and the cuts don't interfere with each other. That probably means each vertex is replaced by a new face, right? When you cut off a vertex with a plane, you replace the vertex with a new face. For a convex polyhedron, each vertex is a point where multiple edges meet. Truncating a vertex would replace that vertex with a flat face, usually a polygon, depending on how many edges met there.But wait, the problem says that the cutting planes do not intersect inside or on the boundary of the original polyhedron. So, each plane only affects one vertex and doesn't interfere with the others. Therefore, truncating each vertex independently. Let me think. For each vertex, truncating it would replace the vertex with a new face. The number of edges added per vertex would depend on the number of edges meeting at that vertex. In a typical truncation, each original edge is replaced by a new edge, and new edges are added around the new face. However, maybe in this case, since we are just cutting once per vertex, perhaps each vertex truncation adds a certain number of edges and vertices.Wait, perhaps the key here is that when you truncate a vertex, you are effectively replacing it with a new face. The number of edges around this new face would be equal to the degree of the original vertex (the number of edges meeting at that vertex). Each original edge connected to the vertex will be split into two edges by the truncation plane. But since each edge is shared by two vertices, truncating both ends would split the original edge into two parts. However, the problem states that all vertices are cut, so every edge will have both of its vertices truncated. Therefore, each original edge will be split into two parts, effectively doubling the number of edges? Wait, but maybe each edge is split into two, so each original edge becomes two edges. But also, each truncation adds a new face per vertex, which is a polygon with as many edges as the degree of the vertex.Wait, this might get complicated. Let me recall that when you truncate a polyhedron, you can think of it as cutting off each vertex, which turns each vertex into a new face. The number of new edges added per vertex is equal to the degree of the vertex. So, if a vertex has degree n, truncating it would add n edges. However, each original edge is split into two edges. So, for each original edge, we have two new edges. Therefore, the total number of edges after truncation would be original_edges * 2 + sum over all vertices of degree(vertex). But in a polyhedron, the sum over all vertices of the degrees is equal to 2E, because each edge is connected to two vertices. So, sum(degree(v)) = 2E. Therefore, the total edges after truncation would be 2E + 2E = 4E? Wait, that can't be right. Wait, no.Wait, let's take a simple example. Let's take a cube. A cube has 12 edges, 8 vertices, each of degree 3. If we truncate each vertex, turning each vertex into a triangular face. The truncated cube (which is a truncated cube) has 36 edges. Let's see how that works. Original edges: 12. Each original edge is split into three parts? Wait, no. Wait, when you truncate a cube, each original edge is truncated at both ends, so the original edge becomes a shorter edge between two new faces. Each original vertex is replaced by a triangular face. Each original edge is adjacent to two vertices, so truncating both ends of the edge would replace the original edge with a new edge connecting the two new triangular faces. Wait, actually, in the truncated cube, each original edge is replaced by a new edge, and each original vertex is replaced by a triangular face with three new edges. But the total edges are 36. Original edges: 12. Each original edge becomes one edge in the truncated cube. Each original vertex (8) adds three edges. So total edges would be 12 + 8*3 = 36. Wait, that works. So, edges after truncation: E' = E + 3V.But in the cube, V=8, E=12. Then E'=12 + 24 = 36. That's correct.But hold on, in this case, each original edge is not split into two, but perhaps the truncation replaces the original edge with a single edge between the two new faces, and each vertex contributes edges equal to its degree.But let's check with another example. Take a tetrahedron: 4 triangular faces, 4 vertices, each of degree 3, 6 edges. If we truncate it, the truncated tetrahedron has 18 edges. Let's see: original edges: 6. Each original edge is replaced by a new edge (connecting two new hexagonal faces?), Wait, no. Wait, truncating a tetrahedron: each vertex is replaced by a triangular face. Each original edge is adjacent to two vertices. So truncating each edge would replace each original edge with a shorter edge between two triangular faces. But maybe each original edge is split into three parts? Hmm, perhaps not. Wait, the truncated tetrahedron has 18 edges. Original edges: 6. Each original edge is divided into three edges? Wait, 6*3=18. Hmm, but that might not be the correct way to think about it.Wait, maybe the formula is different. Let me check. For a truncated tetrahedron, the number of edges is indeed 18. According to the formula, if we take E' = E + 3V, with E=6 and V=4, we get 6 + 12 = 18. That works. Similarly, for the cube, E' = 12 + 24 = 36. So that seems like a general formula: edges after truncation = original edges + 3 times original vertices. But wait, in the cube, each vertex has degree 3, so 3 edges per vertex, so adding 3 edges per vertex. For the tetrahedron, each vertex has degree 3, same thing. So maybe in general, when you truncate each vertex, replacing each vertex with a face that has as many edges as the degree of the vertex, and each original edge is adjacent to two vertices. So the total new edges added would be sum over all vertices of degree(v), which is 2E. So E' = original E + 2E = 3E? But in the cube example, original E=12, so 3*12=36, which matches. For the tetrahedron, original E=6, 3*6=18, which matches. So maybe the general formula is E' = 3E.Wait, but that seems too simple. Then, in the problem, original E=100, so E'=300? But wait, in the cube example, edges after truncation are 36, which is 3*12=36. For the tetrahedron, 3*6=18. So seems that way. But why does that happen?Because when you truncate each vertex, you add a number of edges equal to the degree of the vertex. So, sum over all vertices of degree(v) = 2E. So the total edges after truncation would be original edges (which are each split into one edge) plus 2E, so total 3E. Wait, but in the cube, original edges are 12. After truncation, each original edge is replaced by a single edge between two new faces. So, original edges: 12. Then, each vertex (8) adds 3 edges, which is 24 edges. So total edges: 12 + 24 = 36, which is 3*12=36. So yes, that's the case.Therefore, in general, truncating each vertex (such that the cutting planes do not intersect each other) would result in a new polyhedron with 3 times the original number of edges. So if the original has 100 edges, the new one would have 300 edges. Wait, but is this always the case?Wait, let me verify with another example. Take an octahedron, which has 12 edges, 6 vertices, each of degree 4. Truncating each vertex would replace each vertex with a square face. The truncated octahedron has 36 edges. So original edges: 12. New edges: each vertex (6) adds 4 edges, so 24. Total edges: 12 + 24 = 36, which is 3*12=36. Yes. So same formula. So seems like truncating each vertex, replacing each vertex with a face of degree equal to the original vertex's degree, adds edges equal to 2E, so total edges become 3E. So for the problem, if original E=100, then new E=300. So edges would be 300.But wait, the problem says "no two cutting planes intersect inside or on the boundary of the polyhedron." So maybe the truncation is such that the cutting planes only affect the vertex and do not intersect the edges or other cutting planes. Therefore, each vertex is truncated independently, which is the case in the standard truncation. So perhaps this formula applies here.Now, what about the vertices? Let's think. When you truncate a vertex, the original vertex is removed and replaced by a new face. The number of vertices added per truncation is equal to the number of edges meeting at that vertex. For example, truncating a cube vertex (degree 3) replaces the vertex with a triangular face, which has three new vertices. Similarly, truncating a tetrahedron vertex (degree 3) adds three new vertices. In the cube, original V=8, each vertex truncation adds 3 vertices, so total vertices after truncation: 8*3=24. Original vertices are removed. But wait, in the truncated cube, the number of vertices is actually 24. Yes. The original 8 vertices are gone, replaced by 8 triangular faces, each with 3 vertices. But also, each original face is modified. Wait, original cube has 6 faces. After truncation, each original face becomes a hexagon, because each corner is cut off. So each original face, which was a square, now has its corners cut, turning into an octagon? Wait, no. Wait, in a truncated cube, each original square face becomes an octagon, and each vertex is replaced by a triangle. Wait, but the truncated cube has 8 triangular faces and 6 octagonal faces. The number of vertices is calculated by Euler's formula: V - E + F = 2. For truncated cube: E=36, F=8+6=14. So V= E - F + 2= 36 -14 +2=24. So vertices are 24. Which comes from each original edge contributing a vertex? Wait, original edges:12. Each edge is now adjacent to two new faces (the triangular ones). So each original edge corresponds to a new vertex? Hmm.Wait, maybe there's a better way to calculate the number of vertices after truncation. Let's think. Each original vertex is replaced by a face with as many edges as the degree of the vertex. Each edge of this new face corresponds to a new vertex. However, each original edge is adjacent to two vertices, so truncating both ends of an original edge would create a new edge between the two new faces. The point where the original edge was truncated becomes a new vertex. Wait, maybe each original edge becomes a new vertex? Wait, in the cube, original edges:12, new vertices:24. Hmm, that doesn't add up. Wait, in the truncated cube, each original edge is replaced by a new edge, and the two ends are connected to the new triangular faces. The vertices of the truncated cube are the ones where the new edges meet. Each triangular face has three vertices, and each octagonal face has eight vertices. But each vertex is shared by one triangular face and two octagonal faces. Let me think.Alternatively, perhaps the number of vertices after truncation is equal to the original number of edges multiplied by 2. For the cube, original edges:12, 12*2=24. That matches. For the tetrahedron, original edges:6, truncated tetrahedron has 12 vertices. 6*2=12. For the octahedron, original edges:12, truncated octahedron has 24 vertices. 12*2=24. So that seems consistent. So the number of vertices after truncation is 2 times the original number of edges. Therefore, if original E=100, then new V'=2*100=200.But let's check with Euler's formula. For the truncated cube: V=24, E=36, F=14 (8 triangles +6 octagons). Then, 24 -36 +14=2, which works. So, 24=2*12 (original edges). So, seems that after truncation, the number of vertices is 2E.So general formula: If you truncate all vertices of a convex polyhedron, resulting in a new polyhedron where each original vertex is replaced by a face, then the new number of vertices is 2E (original edges), and the new number of edges is 3E (original edges). Let me verify with another example. Take a dodecahedron. Original E=30, V=20, F=12. Truncating it turns it into a truncated dodecahedron. Let's see, according to the formula, new V=2*30=60, new E=3*30=90. Let's check with Euler's formula. Truncated dodecahedron has 90 edges, 60 vertices, and 12+20=32 faces (12 decagons and 20 triangles). Then, 60 -90 +32=2. Correct. So yes, it works. Similarly, truncated icosahedron (soccer ball pattern) has original E=90 (for the original icosahedron), truncated to E'=270, V'=180. Wait, but the truncated icosahedron has 90 edges? Wait, original icosahedron has 30 edges. So truncating it would result in 3*30=90 edges. Wait, but the actual truncated icosahedron (soccer ball) has 90 edges, 60 vertices, and 32 faces (12 pentagons and 20 hexagons). Wait, according to Euler's formula: 60 -90 +32=2. Correct. So, the vertices there are 60=2*30 (original edges). So yes, that formula holds. So seems like when truncating all vertices, the number of vertices becomes 2E, and the number of edges becomes 3E, where E is the original number of edges.Therefore, in the given problem, original E=100. So, resulting polyhedron would have 2*100=200 vertices, and 3*100=300 edges. So the answer should be 200 vertices and 300 edges.But let me make sure this is applicable here. The problem says "all vertices are cut by planes such that no two cutting planes intersect inside or on the boundary of the polyhedron". So, in other words, each cutting plane only affects one vertex and doesn't intersect any other cutting planes. That is exactly the standard vertex truncation, where each vertex is cut off independently, and the truncation planes don't intersect each other except possibly outside the original polyhedron. So, the standard truncation process applies here. Therefore, the formulas we derived should hold.Therefore, the answer is vertices:200, edges:300.Final AnswerThe resulting polyhedron has boxed{200} vertices and boxed{300} edges.</think>Given a convex polyhedron with 100 edges, we need to determine the number of vertices and edges after truncating all vertices such that no two cutting planes intersect inside or on the boundary of the polyhedron.1. Understanding Truncation: - Truncating a vertex replaces it with a new face. The number of edges added per vertex is equal to the degree of the vertex. - Each original edge is split into two parts by the truncation planes at its endpoints, effectively replacing each original edge with a new edge.2. Key Formulas and Examples: - For a convex polyhedron, truncating each vertex results in a new polyhedron where the number of edges triples (3E) and the number of vertices doubles (2E) based on the original edges (E). - Examples such as the cube, tetrahedron, and octahedron confirm these formulas. For instance, truncating a cube (12 edges) results in 36 edges and 24 vertices, which fits the formulas 3E and 2E respectively.3. Application to the Given Problem: - Original edges (E) = 100. - New edges after truncation = 3E = 3 * 100 = 300. - New vertices after truncation = 2E = 2 * 100 = 200.Thus, the resulting polyhedron has boxed{200} vertices and boxed{300} edges.]

🔑:Solution. On every edge of the given polyhedron, there lie two vertices of the obtained one, with three edges emanating from each vertex of the obtained polyhedron. Therefore, the obtained polyhedron has 2 cdot 100 = 200 vertices and frac{200 cdot 3}{2} = 300 edges.

🧠:[<think>Okay, so I have this problem here: Given the expansion of (1 + x)^6 multiplied by (1 - 2x)^5, which equals a polynomial with coefficients a_0 through a_11. I need to find two things: 1) The sum of all coefficients from a_1 up to a_11. 2) The sum of all even-indexed coefficients, that is, a_0 + a_2 + a_4 + ... + a_10.Hmm, let me start with the first part. I remember that in polynomial expansions, the sum of all coefficients can be found by plugging in x = 1. So if I substitute x = 1 into the polynomial, I should get the sum of all coefficients. But wait, the problem asks for the sum from a_1 to a_11, not including a_0. So maybe I can first find the sum of all coefficients (including a_0) and then subtract a_0 to get the desired sum.Let me check that approach. If I substitute x = 1 into (1 + x)^6(1 - 2x)^5, that should give me a_0 + a_1 + a_2 + ... + a_11. Then, a_0 is just the constant term, which can be found by substituting x = 0 into the polynomial. So if I compute the value at x = 1 and subtract the value at x = 0, I should get the sum from a_1 to a_11. That makes sense.So, let's compute that. First, compute (1 + 1)^6 * (1 - 2*1)^5. That's 2^6 * (-1)^5. 2^6 is 64, and (-1)^5 is -1. So 64 * (-1) = -64. That's the sum of all coefficients including a_0.Now, find a_0. Substitute x = 0: (1 + 0)^6 * (1 - 2*0)^5 = 1^6 * 1^5 = 1*1 = 1. So a_0 is 1.Therefore, the sum from a_1 to a_11 is -64 - 1 = -65. Hmm, but wait, is that right? Let me confirm. If substituting x = 1 gives the total sum, which is -64, and a_0 is 1, then subtracting 1 from -64 gives -65. That seems straightforward. Let me check with an example. Suppose I have a simpler polynomial, like (1 + x)^2 = 1 + 2x + x^2. The sum of coefficients from a_1 to a_2 would be 2 + 1 = 3. If I substitute x = 1, I get 4, subtract a_0 (1), gives 3. That works. So I think this method is correct. So part (1) is -65.Now part (2), the sum of even-indexed coefficients. I remember that to find the sum of coefficients with even indices, you can use the substitution x = 1 and x = -1. Specifically, [f(1) + f(-1)] / 2 gives the sum of even coefficients, and [f(1) - f(-1)] / 2 gives the sum of odd coefficients. Let me recall why. When you substitute x = 1, you get the sum of all coefficients. When you substitute x = -1, you get the sum of coefficients multiplied by (-1)^n, where n is the exponent. So adding f(1) and f(-1) cancels out the odd terms because the odd terms will have opposite signs and the even terms will add up. Similarly, subtracting f(-1) from f(1) cancels the even terms and doubles the odd terms. So yes, [f(1) + f(-1)] / 2 should give the sum of even-indexed coefficients.So let's compute that. We already have f(1) = -64. Now compute f(-1):(1 + (-1))^6 * (1 - 2*(-1))^5 = (0)^6 * (1 + 2)^5 = 0 * 3^5 = 0. So f(-1) = 0.Therefore, the sum of even coefficients is [f(1) + f(-1)] / 2 = (-64 + 0)/2 = -32. Wait a second, but the question asks for a_0 + a_2 + ... + a_10. However, the original polynomial is degree 11, so the highest even index is 10. So our answer should be -32. But let me check if this is correct.But hold on, when we calculated f(-1), it was zero. Because (1 + (-1))^6 is zero. So regardless of the other term, the entire product is zero. Therefore, f(-1) is indeed zero. So the formula gives [ -64 + 0 ] / 2 = -32. But let me verify this with another method to be sure. Maybe expand the polynomials a bit.Alternatively, notice that (1 + x)^6(1 - 2x)^5 can be expanded as the product of two polynomials. The even coefficients in the product might be related to the sum of products of coefficients with even exponents. But that seems complicated.Alternatively, maybe compute a_0, a_2, etc., individually and sum them up? But that might be tedious, but perhaps for verification.Alternatively, perhaps there's another approach. Since (1 + x)^6 is symmetric, and (1 - 2x)^5 is not. But maybe we can use generating functions. Let's see.Alternatively, note that the sum of even coefficients is also equal to [f(1) + f(-1)] / 2 as we did. So given that f(1) = -64 and f(-1) = 0, so sum is -32. So that should be the answer. Let me confirm again. If the entire f(-1) is zero, then the even terms would be [f(1) + 0]/2 = -64/2 = -32. But wait, but in our first calculation, a_0 was 1. So how come the sum of even terms is -32? That would mean that a_0 + a_2 + ... + a_10 = -32. But a_0 is 1, so the rest of the even terms sum to -33? That seems possible? But let me check with another method.Wait, maybe we can compute f(1) and f(-1) correctly. Let me double-check:f(1) = (1 + 1)^6*(1 - 2*1)^5 = 2^6*(-1)^5 = 64*(-1) = -64. Correct.f(-1) = (1 + (-1))^6*(1 - 2*(-1))^5 = 0^6*(3)^5 = 0*243 = 0. Correct.So indeed, the formula gives [ -64 + 0 ] / 2 = -32. However, the problem statement says "a_0 + a_2 + ... + a_10". Let's recall that the original polynomial is of degree 11, so the even degrees go up to 10. So the formula includes a_0. So according to the formula, the sum is -32, which includes a_0. However, earlier we found that a_0 is 1, so if the sum of all even coefficients is -32, that would mean 1 + (sum of a_2 + a_4 + ... + a_10) = -32, so sum from a_2 is -33. But maybe that's correct? Wait, is this possible?Wait, but maybe my initial assumption is wrong here. Let me check with a small example. Let's take a simpler case where we can compute manually. Suppose we have (1 + x)^1*(1 - 2x)^1 = (1 + x)(1 - 2x) = 1 - 2x + x - 2x^2 = 1 - x - 2x^2. So coefficients: a0=1, a1=-1, a2=-2.Sum of even coefficients: a0 + a2 = 1 + (-2) = -1. Using the formula: f(1) = (1+1)(1-2*1) = 2*(-1) = -2. f(-1) = (1 + (-1))*(1 - 2*(-1)) = 0*3 = 0. Then [f(1) + f(-1)] /2 = (-2 + 0)/2 = -1. Which matches. So the formula works here.In this case, even though a0 is 1, the sum of even coefficients is -1. So even though a0 is positive, the other even coefficients can make the total negative. So in the original problem, getting -32 as the sum of even coefficients, including a0=1, seems plausible. So that would mean that the sum of a2 + a4 + ... +a10 is -33. But that's okay. The problem is just asking for a0 + a2 + ... +a10, which is -32.Therefore, part (2) is -32.But let me see if there's another way to cross-validate. Maybe compute a few coefficients and see.Alternatively, perhaps expand both polynomials and multiply them partially. Let's see.First, expand (1 + x)^6. The coefficients are from the binomial theorem: C(6,0), C(6,1), ..., C(6,6) which are 1, 6, 15, 20, 15, 6, 1.Then, expand (1 - 2x)^5. Using the binomial theorem, each term is C(5,k)*(1)^(5 -k)*(-2x)^k. So the coefficients are:For k=0: C(5,0)*(-2)^0 = 1k=1: C(5,1)*(-2)^1 = 5*(-2) = -10k=2: C(5,2)*(-2)^2 = 10*4 = 40k=3: C(5,3)*(-2)^3 = 10*(-8) = -80k=4: C(5,4)*(-2)^4 = 5*16 = 80k=5: C(5,5)*(-2)^5 = 1*(-32) = -32So the coefficients for (1 - 2x)^5 are [1, -10, 40, -80, 80, -32].Now, to find the product of the two polynomials:(1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6) * (1 -10x +40x^2 -80x^3 +80x^4 -32x^5)This will result in a polynomial of degree 6 + 5 = 11, with coefficients a0 to a11.To find a0 + a2 + a4 + ... + a10, which is the sum of even-indexed coefficients, maybe we can compute this sum without calculating each coefficient individually.Here's a trick: The sum of the even-indexed coefficients in a product of two polynomials is equal to the product of the sums of the even-indexed coefficients of each polynomial plus the product of the sums of the odd-indexed coefficients of each polynomial. Wait, maybe not. Alternatively, perhaps there's a generating function approach.Alternatively, note that the sum of even-indexed coefficients in the product is equal to [f(1) + f(-1)] / 2 as we did before, which gave us -32. But let's try to compute it by another method to verify.Alternatively, consider that when you multiply two polynomials, the even coefficients in the product come from either even-even or odd-odd combinations. For example, a2 in the product is the sum of products of coefficients from the first polynomial and the second polynomial where the indices add up to 2. So even indices can be formed by even + even or odd + odd. Therefore, the sum over all even indices in the product can be calculated as:Sum_{k even} (Sum_{i=0}^k A_i * B_{k-i}) )Which can be written as Sum_{i even} A_i * Sum_{j even} B_j + Sum_{i odd} A_i * Sum_{j odd} B_jWait, is that correct? Let me check with an example. Suppose A = [a0, a1, a2], B = [b0, b1, b2]. Then the product C will have c0 = a0b0, c1 = a0b1 + a1b0, c2 = a0b2 + a1b1 + a2b0, c3 = a1b2 + a2b1, c4 = a2b2.Sum of even coefficients: c0 + c2 + c4 = a0b0 + (a0b2 + a1b1 + a2b0) + a2b2.This can be written as a0(b0 + b2) + a1(b1) + a2(b0 + b2). Hmm, which is not exactly the same as Sum_{i even} A_i * Sum_{j even} B_j + Sum_{i odd} A_i * Sum_{j odd} B_j. Let's compute that:Sum_{i even} A_i = a0 + a2Sum_{j even} B_j = b0 + b2Sum_{i odd} A_i = a1Sum_{j odd} B_j = b1Then Sum_even_A * Sum_even_B + Sum_odd_A * Sum_odd_B = (a0 + a2)(b0 + b2) + a1*b1But in the actual sum c0 + c2 + c4, we have:= a0b0 + a0b2 + a1b1 + a2b0 + a2b2= a0(b0 + b2) + a1b1 + a2(b0 + b2)= (a0 + a2)(b0 + b2) + a1b1Which matches the expression above. So indeed, the sum of even coefficients in the product is equal to (Sum of even coefficients in A) * (Sum of even coefficients in B) + (Sum of odd coefficients in A) * (Sum of odd coefficients in B). Therefore, in our problem, to compute the sum of even coefficients in the product (1 + x)^6*(1 - 2x)^5, we can compute:[Sum of even coefficients in (1 + x)^6] * [Sum of even coefficients in (1 - 2x)^5] + [Sum of odd coefficients in (1 + x)^6] * [Sum of odd coefficients in (1 - 2x)^5]Let me compute these sums.First, (1 + x)^6:Sum of even coefficients: this is [f(1) + f(-1)] / 2, where f(x) = (1 + x)^6.f(1) = 2^6 = 64f(-1) = (1 - 1)^6 = 0Thus, sum of even coefficients = (64 + 0)/2 = 32Sum of odd coefficients: [f(1) - f(-1)] / 2 = (64 - 0)/2 = 32Similarly, for (1 - 2x)^5:Sum of even coefficients: [f(1) + f(-1)] / 2, where f(x) = (1 - 2x)^5Compute f(1) = (1 - 2*1)^5 = (-1)^5 = -1f(-1) = (1 - 2*(-1))^5 = (1 + 2)^5 = 3^5 = 243Thus, sum of even coefficients = (-1 + 243)/2 = 242/2 = 121Sum of odd coefficients: [f(1) - f(-1)] / 2 = (-1 - 243)/2 = (-244)/2 = -122Therefore, sum of even coefficients in the product is:(32)(121) + (32)(-122) = 32*(121 - 122) = 32*(-1) = -32Which matches our previous result. So that confirms that part (2) is indeed -32.So, even though the initial a0 is 1, the contributions from the higher even coefficients result in the total sum being -32. So the answers are:1) -652) -32But let me just make sure I didn't make any miscalculations here.First part: sum from a1 to a11 is f(1) - a0 = -64 - 1 = -65. That seems okay.Second part: Using two different methods, got -32. So that's probably correct. Alternatively, perhaps check a small coefficient. For example, compute a0, a1, a2, etc., manually and check.Compute a0: product of constant terms, which is 1*1 = 1. Correct.Compute a1: coefficient of x. It comes from (6x)*(1) + (1)*(-10x) = 6x -10x = -4x. So a1 = -4. Compute a2: coefficient of x^2. From (15x^2)*1 + (6x)*(-10x) + (1)*(40x^2) = 15x^2 -60x^2 +40x^2 = (15 -60 +40)x^2 = (-5)x^2. So a2 = -5.So a0 + a2 = 1 + (-5) = -4. But according to our total sum, the even coefficients sum to -32. So unless there are higher even coefficients that contribute more negatively. Let's compute a4 maybe.But this could take time, but let's see. Maybe compute a few more coefficients.a3: coefficient of x^3. From (20x^3)*1 + (15x^2)*(-10x) + (6x)*(40x^2) + (1)*(-80x^3) = 20x^3 -150x^3 +240x^3 -80x^3 = (20 -150 +240 -80)x^3 = (30)x^3. So a3 = 30.a4: coefficient of x^4. From (15x^4)*1 + (20x^3)*(-10x) + (15x^2)*(40x^2) + (6x)*(-80x^3) + (1)*(80x^4) = 15x^4 -200x^4 +600x^4 -480x^4 +80x^4= (15 -200 +600 -480 +80)x^4= (15 + 600 +80 -200 -480)x^4= (695 - 680)x^4= 15x^4. So a4 =15.a5: coefficient of x^5. Similarly, this would involve more terms, but let me check:From (6x^5)*1 + (15x^4)*(-10x) + (20x^3)*(40x^2) + (15x^2)*(-80x^3) + (6x)*(80x^4) + (1)*(-32x^5)=6x^5 -150x^5 +800x^5 -1200x^5 +480x^5 -32x^5= (6 -150 +800 -1200 +480 -32)x^5= (6 +800 +480 -150 -1200 -32)x^5= (1286 -1382)x^5= (-96)x^5. So a5 = -96.a6: Similarly, but this is getting tedious, but maybe a6 is the coefficient of x^6. Let's try:(x^6)*1 + (6x^5)*(-10x) + (15x^4)*(40x^2) + (20x^3)*(-80x^3) + (15x^2)*(80x^4) + (6x)*(-32x^5) + (1)*0 (since the second polynomial ends at x^5)=1x^6 -60x^6 +600x^6 -1600x^6 +1200x^6 -192x^6= (1 -60 +600 -1600 +1200 -192)x^6= (1 +600 +1200 -60 -1600 -192)x^6= (1801 -1852)x^6= (-51)x^6. So a6 = -51.Wait, but in our earlier formula, the even coefficients up to a10. So a0=1, a2=-5, a4=15, a6=-51, a8 and a10? Let's compute a8 and a10.But maybe this is getting too time-consuming, but let me try a8. a8: coefficient of x^8. Let's see, in the product, to get x^8, we need combinations where the exponents from (1 + x)^6 and (1 - 2x)^5 add up to 8. The degrees of (1 + x)^6 go up to 6, so the exponents from (1 + x)^6 can be from 0 to 6, and from (1 - 2x)^5 from 0 to 5. So to get x^8, possible combinations:Let me denote i as the exponent from (1 + x)^6 and j as the exponent from (1 - 2x)^5, so i + j =8. Since i <=6 and j <=5, the possible j is from 2 to 5, because if i >=3, j=5 gives i=3, etc.Wait:i + j =8, i <=6, j <=5.Thus, j can be from 3 to5 (since if j=5, i=3; j=4, i=4; j=3, i=5; j=2, i=6). Wait:Wait, j=3: i=5j=4: i=4j=5: i=3But also j=2: i=6 (since 6 +2=8)But (1 -2x)^5 has coefficients only up to j=5. So j=2 is allowed. So:i=6, j=2: coefficient C(6,6)*C(5,2)*(-2)^2 =1*10*4=40i=5, j=3: C(6,5)*C(5,3)*(-2)^3=6*10*(-8)= -480i=4, j=4: C(6,4)*C(5,4)*(-2)^4=15*5*16= 1200i=3, j=5: C(6,3)*C(5,5)*(-2)^5=20*1*(-32)= -640So summing these:40 -480 +1200 -640 = (40 +1200) - (480 +640) = 1240 -1120 = 120. So a8=120.Similarly, a10: coefficient of x^10. To get i + j=10. i <=6, j <=5. So possible j=5, i=5; j=4, i=6.Check:j=5, i=5: C(6,5)*C(5,5)*(-2)^5=6*1*(-32)= -192j=4, i=6: C(6,6)*C(5,4)*(-2)^4=1*5*16=80So total a10= -192 +80= -112.So even coefficients:a0=1, a2=-5, a4=15, a6=-51, a8=120, a10=-112Sum them up:1 + (-5) +15 + (-51) +120 + (-112) = 1 -5= -4-4 +15=1111 -51= -40-40 +120=8080 -112= -32So that matches our previous result of -32. Similarly, for part (1), the sum from a1 to a11. From the coefficients we computed:a1= -4, a2=-5, a3=30, a4=15, a5=-96, a6=-51, a8=120, a10=-112. But we need all a1 to a11. However, it's tedious to compute all, but since the total sum is -64, and a0=1, then the sum from a1 to a11 is -64 -1= -65. But let's check with the even and odd coefficients. If the sum of even coefficients is -32, then the sum of odd coefficients is [f(1) - f(-1)] / 2 = (-64 -0)/2 = -32. But the total sum should be even + odd = -32 + (-32) = -64, which matches. But the problem asks for a1 + a2 + ... +a11, which is total sum -a0 = -64 -1= -65. Alternatively, since sum of even coefficients is -32 and sum of odd coefficients is -32, then sum from a1 to a11 is (-32 + (-32)) - a0? Wait, no. Wait, the total sum is -64, which includes a0=1. Therefore, sum from a1 to a11 is -64 -1= -65.So even though some coefficients are positive and some negative, the total sum is -65.Therefore, the answers are:1) boxed{-65}2) boxed{-32}Final Answer(1) boxed{-65} (2) boxed{-32}</think>Given the expansion ((1 + x)^6(1 - 2x)^5 = a_0 + a_1x + a_2x^2 + ldots + a_{11}x^{11}), we need to find:1) The sum (a_1 + a_2 + a_3 + ldots + a_{11});2) The sum (a_0 + a_2 + a_4 + ldots + a_{10}). Key Steps and Calculations:1) Sum of coefficients from (a_1) to (a_{11}): - Substitute (x = 1) into the polynomial to get the sum of all coefficients: [ (1 + 1)^6(1 - 2 cdot 1)^5 = 2^6 cdot (-1)^5 = 64 cdot (-1) = -64 ] - The constant term (a_0) is found by substituting (x = 0): [ (1 + 0)^6(1 - 2 cdot 0)^5 = 1 ] - Subtract (a_0) from the total sum: [ -64 - 1 = -65 ]2) Sum of even-indexed coefficients (a_0 + a_2 + a_4 + ldots + a_{10}): - Use the substitution method for even-indexed coefficients: [ frac{f(1) + f(-1)}{2} ] - Calculate (f(1)) and (f(-1)): [ f(1) = -64 quad text{(from previous calculation)} ] [ f(-1) = (1 - 1)^6(1 - 2(-1))^5 = 0^6 cdot 3^5 = 0 ] - Compute the average: [ frac{-64 + 0}{2} = -32 ] Final Answers:1) (boxed{-65}) 2) (boxed{-32}),<think>Okay, let me try to figure out how to solve these two problems. So, we have the expansion of (1 + x)^6 multiplied by (1 - 2x)^5, and it's equal to a_0 + a_1x + a_2x^2 + ... + a_{11}x^{11}. We need to find two sums: the first one is the sum of all the coefficients from a_1 up to a_11, and the second one is the sum of the even-indexed coefficients, from a_0 to a_{10}.Hmm, for the first part, I remember that to find the sum of coefficients in a polynomial, you can substitute x = 1. That should give the sum of all coefficients. But here, they want the sum from a_1 to a_{11}, which means we need to subtract a_0 from the total sum. Similarly, for the second part, which asks for the sum of even-indexed coefficients, I recall that substituting x = 1 and x = -1 and averaging the results gives the sum of even coefficients. Let me verify that.First, let's tackle problem (1): a_1 + a_2 + ... + a_{11}. The total sum of all coefficients, including a_0, is obtained by plugging x = 1 into the polynomial. So let's compute that.(1 + 1)^6 * (1 - 2*1)^5 = (2)^6 * (-1)^5 = 64 * (-1) = -64. So the sum of all coefficients a_0 + a_1 + ... + a_{11} is -64. But we need a_1 + ... + a_{11}, so we subtract a_0. To find a_0, we can plug in x = 0 into the polynomial. That gives (1 + 0)^6*(1 - 0)^5 = 1*1 = 1. So a_0 is 1. Therefore, the sum from a_1 to a_{11} is -64 - 1 = -65. Wait, that seems straightforward. Let me check again.Wait, (1 + x)^6 at x=1 is 2^6 = 64, and (1 - 2x)^5 at x=1 is (1 - 2)^5 = (-1)^5 = -1. Multiplying them gives 64*(-1) = -64. So the sum of all coefficients is indeed -64. Then a_0 is 1, so subtracting that gives -64 - 1 = -65. That should be the answer for part (1). Hmm, seems right.Now part (2): a_0 + a_2 + a_4 + ... + a_{10}. This is the sum of the even-indexed coefficients. The standard trick for this is to use the fact that f(1) + f(-1) gives twice the sum of even coefficients. Let me recall why. If you have f(x) = a_0 + a_1x + a_2x^2 + ..., then f(1) = a_0 + a_1 + a_2 + a_3 + ... and f(-1) = a_0 - a_1 + a_2 - a_3 + ... Adding them together, the odd terms cancel out and the even terms double: 2(a_0 + a_2 + a_4 + ...). Therefore, the sum of even coefficients is [f(1) + f(-1)] / 2.So, let's compute f(1) and f(-1). We already know f(1) is -64. Let's compute f(-1).f(-1) = (1 + (-1))^6 * (1 - 2*(-1))^5 = (0)^6 * (1 + 2)^5 = 0 * 3^5 = 0. Wait, so f(-1) is 0? Hmm, that complicates things. Because then [f(1) + f(-1)] / 2 would be (-64 + 0)/2 = -32. So the sum of the even coefficients would be -32. But let me check again.Wait, (1 + x)^6 at x = -1 is (1 -1)^6 = 0^6 = 0. So yeah, the first term becomes 0, so the entire product is 0 regardless of the second term. Therefore, f(-1) = 0. So the average is (-64 + 0)/2 = -32. But wait, a_0 + a_2 + ... + a_{10} is supposed to be -32? Let me verify this another way, maybe by expanding the polynomials.Alternatively, maybe we can compute the even coefficients directly by considering the generating function. Let me see.Alternatively, maybe compute the sum another way. Since the product of (1 + x)^6 and (1 - 2x)^5 is the generating function for the coefficients a_0 to a_{11}, maybe we can use another substitution. But I think the standard method is f(1) + f(-1) over 2. However, in this case, since f(-1) is 0, the sum would be -32. But let's check if this makes sense.Wait, a_0 is 1. So if the sum of even coefficients is -32, then starting from a_0 which is 1, adding the rest of even coefficients gives a total of -32, which would mean the other even coefficients sum to -33. That seems possible, but let's see.Alternatively, maybe I made a mistake here. Let me check with a smaller case. Let's compute the product (1 + x)^1*(1 - 2x)^1 = (1 + x)(1 - 2x) = 1 - 2x + x - 2x^2 = 1 - x - 2x^2. So coefficients: a_0 =1, a_1=-1, a_2=-2. Then sum of even coefficients: a_0 + a_2 = 1 + (-2) = -1. Using the formula: f(1) = (1 +1)(1 -2) = 2*(-1) = -2. f(-1)=(1 -1)(1 +2)=0*3=0. So [f(1) + f(-1)]/2 = (-2 +0)/2 = -1. Which matches. So that works. So even in this case, when f(-1)=0, the formula still gives the correct answer.Therefore, even in our original problem, if f(-1) is 0, then the sum of even coefficients is (-64 +0)/2 = -32. So that's the answer for part (2). Therefore, part (1) is -65 and part (2) is -32.But wait, maybe I should check with another example where f(-1) is not zero, just to verify. Let's take (1 + x)^2*(1 - 2x)^1. Then expanding, (1 + 2x + x^2)(1 - 2x) = 1*(1 -2x) + 2x*(1 -2x) + x^2*(1 -2x) = 1 -2x + 2x -4x^2 + x^2 -2x^3 = 1 -3x^2 -2x^3. So coefficients: a_0=1, a_1=0, a_2=-3, a_3=-2. Sum of even coefficients: 1 + (-3) = -2. Using the formula: f(1) = (1+1)^2*(1-2*1) = 4*(-1) = -4. f(-1) = (1-1)^2*(1 -2*(-1)) = 0*(3) =0. Then [f(1) + f(-1)]/2 = (-4 +0)/2 = -2. Correct. So even though f(-1)=0, the formula works.Therefore, in our original problem, even if f(-1)=0, the method holds. So answer for part (2) is -32.Therefore, final answers: (1) -65, (2) -32.But wait, let me check another way for part (2). Maybe expand (1 + x)^6 and (1 - 2x)^5 separately and then multiply them. But that would be tedious. Alternatively, use generating functions with substitution.Alternatively, since (1 + x)^6(1 - 2x)^5 can be considered as the product of two polynomials. The sum of the even coefficients is equivalent to the sum over k even of (coefficient of x^k in the product). Alternatively, the generating function for even coefficients is [f(x) + f(-x)]/2. Therefore, evaluating at x=1 gives [f(1) + f(-1)]/2, which is exactly what we did. So that's correct. So as long as we compute [f(1) + f(-1)]/2, we get the sum of even coefficients.But in our case, f(-1) is 0, so that term doesn't contribute. So the sum is [f(1)]/2 = -64/2 = -32. Therefore, the answer is indeed -32. Therefore, both answers should be (1) -65 and (2) -32.Wait, but let me check with another approach for part (2). Suppose we let x = 1 and x = -1, but maybe in the original polynomials.Wait, (1 + x)^6 is the generating function for coefficients of x^k, where k is from 0 to 6, each term being C(6, k). Similarly, (1 - 2x)^5 is the generating function with coefficients C(5, k)*(-2)^k for k from 0 to 5.Therefore, the product will have coefficients a_n = sum_{i=0}^n C(6, i)*C(5, n - i)*(-2)^{n - i}, where n ranges from 0 to 11.But summing a_0 + a_2 + ... + a_{10} would be the same as summing over n even from 0 to 10 of a_n. That might be complicated, but maybe we can use generating functions.Alternatively, we can note that the sum of even coefficients is equal to [f(1) + f(-1)] / 2, which we already computed as -32. So unless there's a miscalculation here, that should be correct.Alternatively, maybe we can compute f(1) = -64 and f(-1) = 0, so indeed, the average is -32. So that seems right. Therefore, I think the answers are -65 and -32.Final Answer(1) boxed{-65} (2) boxed{-32}</think>Given the polynomial expansion ((1 + x)^6 (1 - 2x)^5 = a_0 + a_1x + a_2x^2 + ldots + a_{11}x^{11}), we need to find:1. The sum (a_1 + a_2 + a_3 + ldots + a_{11})2. The sum (a_0 + a_2 + a_4 + ldots + a_{10}) Key Steps:1. Sum of coefficients from (a_1) to (a_{11}): - The total sum of all coefficients (including (a_0)) is found by substituting (x = 1): [ (1 + 1)^6 (1 - 2 cdot 1)^5 = 2^6 cdot (-1)^5 = 64 cdot (-1) = -64 ] - The constant term (a_0) is found by substituting (x = 0): [ (1 + 0)^6 (1 - 2 cdot 0)^5 = 1 cdot 1 = 1 ] - Subtracting (a_0) from the total sum gives: [ -64 - 1 = -65 ]2. Sum of even-indexed coefficients (a_0 + a_2 + a_4 + ldots + a_{10}): - Using the property of polynomials, the sum of even-indexed coefficients is (frac{f(1) + f(-1)}{2}): - We already know (f(1) = -64). - Substituting (x = -1) into the polynomial: [ (1 - 1)^6 (1 - 2 cdot (-1))^5 = 0^6 cdot 3^5 = 0 ] - Calculating the average: [ frac{f(1) + f(-1)}{2} = frac{-64 + 0}{2} = -32 ] Final Answers:1. (boxed{-65})2. (boxed{-32})]

🔑:(1) Given (1+x)^6(1-2x)^5 = a_0 + a_1x + a_2x^2 + ldots + a_{11}x^{11}. Let x=1, we get a_0 + a_1 + a_2 + ldots + a_{11} = -2^6, Also, let x=0, we get a_0 = 1, Therefore, a_1 + a_2 + ldots + a_{11} = -2^6 - 1 = -65. So, the answer is boxed{-65}.(2) Then, let x=-1, we get a_0 - a_1 + a_2 - a_3 + ldots - a_{11} = 0 Adding this to the equation from part (1), we get a_0 + a_2 + ldots + a_{10} = (-2^6 + 0) = -32. So, the answer is boxed{-32}.

🧠:[<think>Okay, let's tackle this problem step by step. The question is asking me to calculate (-2)⁰ minus (1/2)⁻¹. Hmm, exponents can sometimes be tricky, especially with negative bases and negative exponents. Let me break it down.First, let's look at the first part: (-2)⁰. I remember that any non-zero number raised to the power of zero is 1. So, even though the base here is negative, as long as it's not zero, the result should still be 1. Let me confirm that. Yeah, the rule says a⁰ = 1 for any a ≠ 0. Since -2 is not zero, (-2)⁰ must be 1. Got that part down.Now, the second part is (1/2)⁻¹. Negative exponents always confuse me a bit. Wait, the rule is that a⁻ⁿ equals 1/(aⁿ), right? So, taking the reciprocal of the base and making the exponent positive. Let me apply that here. (1/2)⁻¹ would be the same as 1 divided by (1/2)¹. But (1/2)¹ is just 1/2. So, 1 divided by (1/2) is the same as multiplying by 2, which gives 2. So, (1/2)⁻¹ equals 2. Let me check that again. If I have a fraction like 1/2 and raise it to the power of -1, flipping the fraction gives 2/1, which is 2. Yeah, that makes sense. So that part checks out.Now, putting it all together: (-2)⁰ is 1, and (1/2)⁻¹ is 2. So the expression becomes 1 - 2. Subtracting 2 from 1 gives -1. Wait, is that all? Let me verify each step again to make sure I didn't make any mistakes.Starting with (-2)⁰: yes, any number to the zero power is 1, even negatives. Then (1/2)⁻¹: reciprocal of 1/2 is 2, and since the exponent is -1, that's correct. Then 1 - 2 is indeed -1. Hmm, seems straightforward. Maybe I was overcomplicating it, but better safe than sorry. Let me think if there's any alternative way to look at this.Alternatively, for the second term, (1/2)⁻¹. Another way to handle negative exponents is to write it as (2/1)¹, which is 2. So that also gives 2. So that's consistent. And the first term, (-2)⁰, is 1 because the exponent is zero. So, subtracting 2 from 1 gives -1. Yeah, that's correct.I don't see any errors here. Both parts of the expression are calculated correctly, and the operations are straightforward. So the final answer should be -1. Let me just write that in the required format.Final Answerboxed{-1}</think>To calculate ((-2)^{0} - left(frac{1}{2}right)^{-1}):1. Evaluate ((-2)^{0}): - Any non-zero number raised to the power of 0 is 1. Therefore, ((-2)^{0} = 1).2. Evaluate (left(frac{1}{2}right)^{-1}): - A negative exponent indicates the reciprocal of the base raised to the positive exponent. Thus, (left(frac{1}{2}right)^{-1} = frac{1}{left(frac{1}{2}right)} = 2).3. Subtract the results: - (1 - 2 = -1).Thus, the final answer is (boxed{-1}).,<think>Okay, let's see. I need to calculate (-2)⁰ minus (1/2)⁻¹. Hmm, exponents can sometimes be tricky, especially with negative bases and negative exponents. Let me break this down step by step.First, let's tackle (-2)⁰. Wait, any non-zero number raised to the power of 0 is 1, right? So even if the base is negative, as long as it's not zero, the result should be 1. So (-2)⁰ equals 1. That part seems straightforward.Now, the next term is (1/2)⁻¹. Negative exponents can be confusing. I remember that a negative exponent means taking the reciprocal of the base. So, for example, a⁻ⁿ is equal to 1/aⁿ. So applying that here, (1/2)⁻¹ would be the reciprocal of (1/2). The reciprocal of 1/2 is 2/1, which is 2. Let me verify that. If I have (1/2)⁻¹, that's the same as 1 divided by (1/2), right? And dividing by a fraction is the same as multiplying by its reciprocal, so 1 ÷ (1/2) = 1 * 2/1 = 2. Yeah, that checks out.So putting it all together, the original expression is (-2)⁰ minus (1/2)⁻¹, which simplifies to 1 - 2. Now, 1 minus 2 is -1. That seems pretty straightforward. Let me double-check each part to make sure I didn't make a mistake.First part: (-2)⁰. As I thought, anything to the power of 0 is 1. Negative numbers are okay here; it's only 0⁰ that's undefined. So that part is definitely 1.Second part: (1/2)⁻¹. We did the reciprocal, which turned 1/2 into 2. So subtracting 2 from 1 gives us -1. Wait, is there a chance I messed up the negative sign somewhere? Let me see. The exponent is -1, so applying the negative exponent rule, yes, reciprocal. So 1/(1/2) is 2. So subtracting 2, not negative 2. So 1 - 2 is indeed -1. Hmm, is there another way to look at (1/2)⁻¹? Maybe using the general formula a⁻ⁿ = (1/a)ⁿ. So (1/2)⁻¹ is the same as (2/1)¹, which is 2. So that's the same result. Okay, that's consistent.Alternatively, if I remember that (a/b)⁻ⁿ = (b/a)ⁿ, then (1/2)⁻¹ is (2/1)¹ = 2. Yep, same thing. So that's solid.So combining both parts: 1 (from the first term) minus 2 (from the second term) equals -1. That seems correct. Let me check if there's any order of operations I might have messed up. The problem is written as (-2)⁰ minus (1/2)⁻¹. There's no parentheses that would change the order here. The exponents are handled first, then the subtraction. So yes, compute each term separately and then subtract. No issues there.Wait, just to be thorough, let me consider if the first term was maybe -2⁰ instead of (-2)⁰. But no, the problem has parentheses around the -2, so it's definitely (-2)⁰. If it were written as -2⁰, that would be interpreted as -(2⁰) = -1, but here it's (-2)⁰, which is 1. Good thing the parentheses are there to clarify.So, no mistakes detected in the reasoning. The answer should be -1.Final Answerboxed{-1}</think>To calculate ((-2)^{0} - left(frac{1}{2}right)^{-1}):1. Evaluate ((-2)^{0}): - Any non-zero number raised to the power of 0 is 1. Therefore, ((-2)^{0} = 1).2. Evaluate (left(frac{1}{2}right)^{-1}): - A negative exponent indicates the reciprocal of the base. Thus, (left(frac{1}{2}right)^{-1} = 2).3. Subtract the results: - (1 - 2 = -1).Thus, the final answer is (boxed{-1}).]

🔑:To solve the given expression (-2)^{0} - (frac{1}{2})^{-1}, we break it down into two parts and solve each part step by step.1. For the first part, (-2)^{0}, any non-zero number raised to the power of 0 is 1. Therefore, we have:[(-2)^{0} = 1]2. For the second part, (frac{1}{2})^{-1}, when you have a negative exponent, you can take the reciprocal of the base and change the exponent to positive. So, we have:[(frac{1}{2})^{-1} = 2^1 = 2]Putting it all together, we substitute the results back into the original expression:[(-2)^{0} - (frac{1}{2})^{-1} = 1 - 2]Simplifying the expression gives us:[1 - 2 = -1]Therefore, the answer is: boxed{-1}.

🧠:[<think>Okay, let's see. I need to solve the equation 111,111222222 minus 333,333 equals x squared, and then find x from the options given. Hmm, first, let me make sure I understand the numbers correctly. The problem is written with spaces as thousand separators maybe? Let me check the original problem again.The problem says: 111222222 - 333333 = x² and x > 0. Then x is which of the options. The options are (A) 332233, (B) 223377, (C) 333333, (D) 331177. Alright, so the numbers are written without commas, but perhaps the spaces in the problem statement are just for readability? Wait, 111222222 minus 333333. Let me parse those numbers correctly.First number: 111222222. That is 111,222,222 if we separate the digits. Second number is 333,333. So subtract 333,333 from 111,222,222. Let me compute that subtraction.111,222,222 minus 333,333. Let me write it down:111,222,222 -     333,333 =?Starting from the rightmost digits:2 - 3: Can't do, borrow. But the next digit is also 2, so we need to keep borrowing. Let's see, maybe it's easier to subtract step by step.Alternatively, maybe factor the numbers or see if there's a pattern. Let me check the options. The options are all 6-digit numbers except (C) is 333,333. Let me compute the subtraction first.111,222,222 minus 333,333. Let's subtract:111,222,222 -     333,333 ------------Starting from the right:2 - 3: Can't do, borrow 1 from the next digit. But the next digit is 2, so we have to borrow multiple times. Let's do this step by step.Starting with the units place:The units digit of the first number is 2, and we subtract 3. So we need to borrow 1 from the tens place. But the tens place is 2, so it becomes 1, and we borrow 1 from the hundreds place. The hundreds place is 2, so it becomes 1, borrow from thousands place. Thousands place is 2, becomes 1, borrow from ten thousands place. Ten thousands place is 2, becomes 1, borrow from hundred thousands place. Hundred thousands place is 2, becomes 1, borrow from millions place. The millions place: Wait, the number is 111,222,222.Breaking it down:Digits: 1 1 1 2 2 2 2 2 2So positions (from right to left): units, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, hundred millions.Wait, actually, the number is 111,222,222. So:Hundred millions: 1Ten millions: 1Millions: 1Hundred thousands: 2Ten thousands: 2Thousands: 2Hundreds: 2Tens: 2Units: 2So when subtracting 333,333, which is 0 0 0 3 3 3 3 3 3 (but aligned to the right). Wait, actually:333,333 is 000333,333 if we consider the same number of digits as 111,222,222. Wait, 111,222,222 is 9 digits, and 333,333 is 6 digits, so it's 000,333,333 when aligned. So:111,222,222 - 000,333,333 =?So subtract digit by digit:Starting from the rightmost digit (units):2 - 3: Can't do, need to borrow. The next digit (tens) is 2, so borrow 1 from there, making it 1, and the units becomes 12 - 3 = 9.Next, tens place: Now the tens digit is 1 (after borrowing). 1 - 3: Can't do, borrow from hundreds place. Hundreds digit is 2, becomes 1, tens becomes 11 - 3 = 8.Hundreds place: Now 1 (after borrowing). 1 - 3: Can't do, borrow from thousands place. Thousands digit is 2, becomes 1, hundreds becomes 11 - 3 = 8.Thousands place: Now 1 (after borrowing). 1 - 3: Can't do, borrow from ten thousands place. Ten thousands digit is 2, becomes 1, thousands becomes 11 - 3 = 8.Ten thousands place: Now 1 (after borrowing). 1 - 3: Can't do, borrow from hundred thousands place. Hundred thousands digit is 2, becomes 1, ten thousands becomes 11 - 3 = 8.Hundred thousands place: Now 1 (after borrowing). 1 - 3: Can't do, borrow from millions place. Millions digit is 1, becomes 0, hundred thousands becomes 11 - 3 = 8.Millions place: Now 0. But we need to check the next digits. The ten millions place is 1, and the hundred millions place is 1. Let's see:After borrowing for the hundred thousands place, the millions digit was 1, now it's 0. Then, when we go to the millions place, since we have 0 - 0 (since the subtrahend has 0 in millions place), it's 0. Then ten millions place: 1 - 0 = 1. Hundred millions place: 1 - 0 = 1.Putting it all together, the result is:111,222,222 - 000,333,333 = 110,888,889Wait, let me verify. Let's compute 111,222,222 minus 333,333.Alternatively, 111,222,222 - 333,333 can be written as 111,222,222 - 300,000 - 33,333.111,222,222 - 300,000 = 110,922,222Then subtract 33,333: 110,922,222 - 33,333 = 110,888,889.Yes, that's correct. So the result is 110,888,889. Therefore, x² = 110,888,889. So we need to find x such that x² = 110,888,889 and x is positive. The options are (A) 332233, (B) 223377, (C) 333333, (D) 331177.So, we need to compute the square roots of these options or square the options and see which one equals 110,888,889.Alternatively, maybe there's a pattern here. Let me check the options:Option C is 333,333. Let me compute 333,333². Maybe that's a quick check. 333,333 squared.But 333,333 squared is equal to (333,000 + 333)^2 = 333,000² + 2*333,000*333 + 333².Compute 333,000²: 333² = 110,889, so 333,000² = 110,889,000,000.2*333,000*333 = 2*333,000*333 = 2*333*333,000 = 2*110,889*1,000 = 221,778,000.333² = 110,889.So total is 110,889,000,000 + 221,778,000 + 110,889 = Let's add them up:110,889,000,000 +     221,778,000 +         110,889 = 110,889,000,000 + 221,778,000 = 111,110,778,000 + 110,889 = 111,110,888,889.Wait, so 333,333 squared is 111,110,888,889. But our target is 110,888,889, which is way smaller. So option C is way too big. So option C is out.Then let's check the other options. Let's try option D: 331,177.Compute 331,177 squared. That might take some time. Alternatively, maybe there's a smarter approach. The number 110,888,889. Let me see if this number is a perfect square. Let me try to compute its square root.Let me try to estimate the square root of 110,888,889. Since 10,000 squared is 100,000,000. 105,000 squared is 11,025,000,000, which is way too big. Wait, wait, 10,000 squared is 100,000,000. 100,000,000 is 8 zeros. Wait, 110,888,889 is a 9-digit number. So its square root should be a 5-digit number, because 10,000²=100,000,000 (9 digits). Wait, actually, 10000² is 100,000,000 (9 digits). So sqrt(110,888,889) should be a bit more than 10,000. Wait, but 10,000² is 100,000,000. 11,000² is 121,000,000. So 10,000²=100,000,000, 15,000²=225,000,000. Wait, but 110,888,889 is between 10,000² and 15,000². Wait, but 10,000² is 100,000,000. 11,000² is 121,000,000. 10,500² is 110,250,000. So 10,500²=110,250,000. Our number is 110,888,889. So subtract 110,250,000 from 110,888,889: 638,889. So the square root is 10,500 plus some. Let's see how much.Let x = 10,500 + y. Then x² = (10,500 + y)² = 10,500² + 2*10,500*y + y². We need this to equal 110,888,889. Since 10,500² = 110,250,000, the difference is 638,889 = 2*10,500*y + y². Let's approximate:2*10,500*y = 21,000*y. So 21,000*y ≈ 638,889. Then y ≈ 638,889 / 21,000 ≈ 30.42. So y is approximately 30.42. Let's check 10,500 + 30 = 10,530. 10,530² = ?10,530²: Compute (10,500 + 30)² = 10,500² + 2*10,500*30 + 30² = 110,250,000 + 630,000 + 900 = 110,250,000 + 630,000 = 110,880,000 + 900 = 110,880,900.Compare to target 110,888,889. Difference is 110,888,889 - 110,880,900 = 7,989. So we need a bit more. Let's try y = 31. Then x = 10,531.10,531² = (10,530 + 1)² = 10,530² + 2*10,530*1 + 1 = 110,880,900 + 21,060 + 1 = 110,901,961. That's higher than 110,888,889. Wait, that's not right. Wait, 10,530² is 110,880,900. Adding 21,060 gives 110,901,960, plus 1 gives 110,901,961. But our target is 110,888,889, which is actually less than 110,880,900 + 21,060. Wait, this seems contradictory. Wait, maybe my initial estimation is wrong.Wait, perhaps the square root is not around 10,530 but higher? Wait, 10,530² is 110,880,900. 10,531² is 110,901,961. But our target is 110,888,889, which is between 10,530² and 10,531². But how? Wait, 10,530² = 110,880,900Then 10,530 + 1 = 10,531, squared is 110,901,961But 110,888,889 is between those two. Wait, but that can't be, because consecutive squares don't overlap. Wait, there must be a mistake here.Wait, perhaps my initial assumption is wrong. Wait, 10,000² = 100,000,00020,000² = 400,000,000Wait, but 110,888,889 is 110 million, which is between 10,000² and 20,000². Wait, 10,000² is 100,000,00015,000² is 225,000,000So 10,000 to 15,000. Wait, but 10,500² is 110,250,000Wait, 10,500² = (10,000 + 500)² = 100,000,000 + 2*10,000*500 + 500² = 100,000,000 + 10,000,000 + 250,000 = 110,250,000Then 10,530² = 110,250,000 + 2*10,500*30 + 30² = 110,250,000 + 630,000 + 900 = 110,880,900But the target is 110,888,889, which is 110,880,900 + 7,989. So between 10,530 and 10,531. But since we can't have a fraction in the integer square root, that would suggest that 110,888,889 is not a perfect square. But the problem says it is, so maybe my approach is wrong.Alternatively, perhaps the original problem has a typo, but assuming not, maybe I need to check the options again.Wait, the options given are all 6-digit numbers. Wait, 332,233; 223,377; 333,333; 331,177. Wait, these are all 6-digit numbers, so x is a 6-digit number, so x² would be around 10^12? Wait, no, wait. Wait, a 6-digit number is up to 999,999. Squaring that gives up to approximately 10^12, but our result is 110,888,889, which is a 9-digit number. So there's a discrepancy here. Wait, but the options are 6-digit numbers. So x is a 6-digit number, so x² would be a 12-digit number? But the result here is 9-digit. That can't be. So something is wrong.Wait, hold on. The problem states: 111222222 - 333333 = x². Let me check the numbers again.Wait, 111222222 minus 333333. Maybe I miscounted the digits. Let's count:111222222: 1,1,1,2,2,2,2,2,2 – nine digits.333,333: six digits. So 111,222,222 minus 333,333 is 110,888,889, which is eight digits? Wait, 110,888,889 is nine digits: 1,1,0,8,8,8,8,8,9.But x is supposed to be a six-digit number (the options are all six digits). But x squared is 110,888,889, which is nine digits, so x should be roughly sqrt(110,888,889) ≈ 10,530 as we saw earlier, which is a five-digit number. But none of the options are five digits. Wait, this is confusing. Let me check the problem statement again.Original problem: "111222222-333333=x^{2} and x>0. Then x=( ). (A) 332233 (B) 223377 (C)333 333 (D) 331177"Wait, maybe the numbers are written without commas. So 111222222 is 111222222, and 333333 is 333333. Then subtracting them: 111222222 - 333333.But let's compute that. 111,222,222 minus 333,333 is 110,888,889. But if the numbers are written without separators, is there another way to parse them? Wait, the problem is written as "111222222-333333=x^{2}". So it's 111222222 minus 333333 equals x squared.Wait, 111222222 minus 333333 is:Let me compute 111,222,222 - 333,333. As before, that is 110,888,889. Then x squared is 110,888,889, so x is sqrt(110,888,889). Which is approximately 10,530. But none of the options are close to that. The options are all six-digit numbers starting with 223, 331, 332, 333. So maybe there's a miscalculation here.Wait, hold on. Maybe the numbers are not in base 10? But that seems unlikely. Alternatively, maybe the problem is written with digits grouped differently. For example, in some countries, numbers are written with spaces as thousand separators, but here maybe the numbers are written with each group representing something else.Wait, let me check the original problem again: "111222222-333333=x^{2} and x>0. Then x=(quad). (A) 332233 (B) 223377 (C)333 333 (D) 331177"Wait, the first number is 111222222, which is 111,222,222. The second number is 333,333. The result is x², where x is one of the options. The options are 332,233; 223,377; 333,333; 331,177. Wait, but 333,333 squared we saw is 111,110,888,889, which is way larger than 110,888,889. So even if we consider the problem might have a typo, like maybe the first number is 111,222,222,222 instead of 111,222,222, making it 12 digits, then subtracting 333,333,333, but the problem as written is 111222222 (nine digits) minus 333333 (six digits) equals x squared. So x squared is nine digits, x is five digits. But the options are six digits. This is conflicting.Wait, unless there's a misinterpretation of the numbers. Let me check once again:If the problem is written as "111222222 - 333333 = x²", maybe "111222222" is meant to be 11,122,2222? But that doesn't parse. Alternatively, if it's 1,112,22222? But that also doesn't make sense. Alternatively, maybe the numbers are in a different base? For example, base 8 or base 16. But the options include digits like 7 and 8, so base 8 is out. In base 10, digits are 0-9, so that's okay.Alternatively, maybe the numbers are concatenated. Wait, 111222222 minus 333333. Wait, 111222222 is a concatenation of 111, 222, 222. Similarly, 333333 is three 3s, three 3s. Maybe there's a pattern here.Alternatively, perhaps the problem is presented with some missing commas or formatting issues. For example, maybe the first number is 111,222,222 and the second is 333,333, but written without commas. Then x is a six-digit number. Wait, but x squared is a nine-digit number. A six-digit number squared is a 12-digit number. So 100,000 squared is 10,000,000,000 (11 digits). 999,999 squared is 999,998,000,001 (12 digits). So x squared being 110,888,889 is a nine-digit number, so x must be a five-digit number (since 10,000 squared is 100,000,000 (nine digits)), but options given are all six-digit numbers. This is a contradiction. Therefore, there must be a mistake either in the problem statement or in my interpretation.Wait, perhaps the original problem uses Chinese numerals or some other formatting? But given the options are all numbers with digits 1-3 and 7, perhaps there's a different approach.Wait, maybe the numbers are in a different numeral system. For example, if the numbers are in base 10, but when subtracted, the result is a square in another base. But the problem doesn't specify that. Alternatively, maybe the numbers are written with each digit representing a different place value.Alternatively, maybe there's a factorization approach. Let me try to factor 110,888,889. Let's see.First, check if it's divisible by small primes. 110,888,889.Sum of digits: 1+1+0+8+8+8+8+8+9 = 1+1+0+8*5 +9 = 2 + 0 +40 +9=51. 51 is divisible by 3, so 110,888,889 is divisible by 3. Let's divide by 3.110,888,889 ÷ 3. Let's compute:3 | 110,888,8893 goes into 11 three times (3*3=9), remainder 2. Bring down 0: 20.3 into 20 is 6 (18), remainder 2. Bring down 8: 28.3 into 28 is 9 (27), remainder 1. Bring down 8: 18.3 into 18 is 6, remainder 0. Bring down 8: 08.3 into 8 is 2 (6), remainder 2. Bring down 8: 28.3 into 28 is 9 (27), remainder 1. Bring down 8: 18.3 into 18 is 6, remainder 0. Bring down 9: 09.3 into 9 is 3, remainder 0. So the result is 36,962,963.So 110,888,889 = 3 * 36,962,963. Let's check if 36,962,963 is divisible by 3: 3+6+9+6+2+9+6+3= 3+6=9, +9=18, +6=24, +2=26, +9=35, +6=41, +3=44. 44 is not divisible by 3. So next prime is 5. Ends with 3, not 0 or 5. Next is 7. Let's check 36,962,963 ÷7.7 into 36 is 5 (35), remainder 1. Bring down 9: 19.7 into 19 is 2 (14), remainder 5. Bring down 6: 56.7 into 56 is 8, remainder 0. Bring down 2: 02.7 into 2 is 0, remainder 2. Bring down 9: 29.7 into 29 is 4 (28), remainder 1. Bring down 6: 16.7 into 16 is 2 (14), remainder 2. Bring down 3: 23.7 into 23 is 3 (21), remainder 2. So no, not divisible by 7.Next prime 11. Let's apply the divisibility test for 11: alternating sum of digits.(3 + 9 + 2 + 6) - (6 + 6 + 9 + 3) = (20) - (24) = -4. Not divisible by 11.Next 13. This is getting tedious. Maybe 110,888,889 is a square. Wait, if x is 10,533? Wait, let's check 10,533 squared.10,533² = ?We can compute (10,500 + 33)² = 10,500² + 2*10,500*33 + 33² = 110,250,000 + 693,000 + 1,089 = 110,250,000 + 693,000 = 110,943,000 + 1,089 = 110,944,089. Still higher than 110,888,889.Wait, 10,500² = 110,250,00010,530² = 110,880,90010,530² is 110,880,900. The target is 110,888,889, which is 7,989 more.So 10,530 + y squared = 110,888,889. So 110,880,900 + 2*10,530*y + y² = 110,888,889. Then 2*10,530*y + y² = 7,989. Let's approximate y.2*10,530*y ≈ 21,060*y ≈ 7,989. So y ≈ 7,989 / 21,060 ≈ 0.379. So y is approximately 0.379, so x ≈ 10,530.379. So not an integer. Therefore, 110,888,889 is not a perfect square. But the problem states it is. Therefore, there must be a miscalculation or misinterpretation.Wait, going back to the original problem: "111222222-333333=x^{2}". Maybe the numbers are not in base 10? Let's check. If we assume base 8.Wait, in base 8, the digits would be 0-7. The numbers given are 111222222 and 333333, which in base 8 would translate to:111222222 (base 8) = 1*8^8 + 1*8^7 + 1*8^6 + 2*8^5 + 2*8^4 + 2*8^3 + 2*8^2 + 2*8 + 2. That's a huge number. Similarly, 333333 (base 8) = 3*8^5 + 3*8^4 + 3*8^3 + 3*8^2 + 3*8 + 3. Then subtracting them, but x would also be in base 8, but the options are in base 10 digits. This seems too convoluted.Alternatively, maybe the numbers are in base 4 or something. But the digits include 3, which is okay in base 4. However, the options include digits like 7, which would not be valid in base 4. So probably not.Alternatively, maybe the problem is presented with a typo. For example, maybe the first number is 111222222222 (twelve digits) minus 333333 (six digits) equals x squared, where x is a six-digit number. Then x squared would be a twelve-digit number. For example, 333,333 squared is 111,110,888,889, which is twelve digits. If the first number is 111,222,222,222 minus 333,333 equals x². Then x² would be 111,222,222,222 - 333,333 = 111,221,888,889. Is that a square? Maybe similar to 333,333 squared. 333,333 squared is 111,110,888,889. So 111,221,888,889 is larger. Maybe 333,333 plus 111,111 squared? No, that's not.Alternatively, maybe the original problem had a different first number. But given the problem as stated, there is a contradiction because x must be a five-digit number but the options are six-digit numbers.Wait, perhaps the original problem had the equation 111222222 - 333333 = x², but in reality, the numbers are written with Chinese numeral groupings or something else. But without more context, it's hard to tell.Alternatively, maybe the numbers are being treated as strings. For example, 111222222 minus 333333 as strings. But that doesn't make much sense.Alternatively, maybe the problem is a sequence puzzle. For example, 111222222 is a sequence of three 1s followed by six 2s. Subtract 333333 (three 3s, three 3s), and the result is x squared. Maybe x is a number with some pattern. The options are (A) 332233, (B) 223377, (C)333333, (D)331177. These all have repeating digits. Maybe the square has a pattern similar to 111222222 - 333333 = 110,888,889. Looking at 110,888,889, the digits are 1,1,0,8,8,8,8,8,9. Maybe this number is a palindrome or has some pattern. Let's see, 110,888,889. It starts with 11, then 0, then five 8s, then 9. Not a palindrome.Alternatively, perhaps this number is a square of a number consisting of 3s and other digits. Let me check 10,533 squared again: 10,533² = 110,944,089. Not matching. 10,529 squared: 10,529² = (10,500 + 29)² = 110,250,000 + 2*10,500*29 + 29² = 110,250,000 + 609,000 + 841 = 110,859,841. Still lower than 110,888,889. Difference is 110,888,889 - 110,859,841 = 29,048. 10,529 + 1 = 10,530: 10,530² = 110,880,900. Difference 110,888,889 - 110,880,900 = 7,989. 10,530 + 8 = 10,538: 10,538² = ?10,538 squared:Compute (10,500 + 38)² = 10,500² + 2*10,500*38 + 38² = 110,250,000 + 798,000 + 1,444 = 111,049,444. Too high. So no, not matching.Alternatively, maybe the answer is not obtained by direct computation but by some pattern. The given options are (A) 332233, (B) 223377, (C)333333, (D)331177.If we look at option A: 332,233. Squared:Compute 332,233². Let's see. This would be (330,000 + 2,233)² = 330,000² + 2*330,000*2,233 + 2,233². 330,000² = 108,900,000,000. 2*330,000*2,233 = 660,000*2,233 = 660,000*2,000 + 660,000*233 = 1,320,000,000 + 153,780,000 = 1,473,780,000. 2,233² = 4,986, but exact value: 2,233 * 2,233. Let's compute:2,233 * 2,233:First compute 2,200 * 2,200 = 4,840,000Then 2,200 * 33 = 72,60033 * 2,200 = 72,60033 * 33 = 1,089So total: 4,840,000 + 72,600 + 72,600 + 1,089 = 4,840,000 + 145,200 + 1,089 = 4,986,289So total is 108,900,000,000 + 1,473,780,000 = 110,373,780,000 + 4,986,289 = 110,378,766,289. Which is way larger than 110,888,889. So option A is too large.Option B: 223,377². Similarly, this would be a huge number. Option C we saw is 333,333²=111,110,888,889. Option D: 331,177².Compute 331,177²:Again, this would be (330,000 + 1,177)² = 330,000² + 2*330,000*1,177 + 1,177². 330,000²=108,900,000,000. 2*330,000*1,177=660,000*1,177= Let's compute 1,177*600,000=706,200,000 and 1,177*60,000=70,620,000. Wait, no, 660,000 * 1,177:First, 1,177 * 600,000 = 706,200,0001,177 * 60,000 = 70,620,000Total: 706,200,000 + 70,620,000 = 776,820,000Then 1,177²: 1,177*1,177. Let's compute:1,177*1,000 = 1,177,0001,177*100 = 117,7001,177*70 = 82,3901,177*7 = 8,239Adding up: 1,177,000 + 117,700 = 1,294,700 + 82,390 = 1,377,090 + 8,239 = 1,385,329So total square is 108,900,000,000 + 776,820,000 = 109,676,820,000 + 1,385,329 = 109,678,205,329. Still way larger than 110,888,889.So all options squared give numbers way larger than 110,888,889. So there must be a problem with the question or my interpretation.Wait, maybe the original problem uses a different operator? Instead of subtraction, maybe division or something else. But the problem states minus. Alternatively, maybe there's a missing comma in the problem. For example, if the first number is 1,112,222,222 minus 333,333, which would be 1,112,222,222 - 333,333 = 1,111,888,889. Is that a square? Let's check sqrt(1,111,888,889). Hmm, 33,333 squared is 1,111,088,889. Which is close. 33,334 squared is 1,111,155,556. So 1,111,888,889 is between those two. Not a perfect square. Alternatively, maybe the first number is 111,222,222,223 - 333,333, which would be 111,222,222,223 - 333,333,333 = 110,888,888,890. Not sure.Alternatively, maybe the problem is in base 10, but the numbers are split differently. For example, 111-222-222 minus 333-333. But 111-222-222 is not a standard number.Alternatively, maybe the original problem was presented with some formatting where 111222222 is actually 11 * 12^7 + ... but that seems too complex.Alternatively, maybe the problem is a riddle. For example, 111222222 - 333333: if we subtract 333333 from 111222222, the result is 110888889. If we write this number as 110888889, perhaps it can be divided into parts that reflect the options. For example, 11 08 88 89, but the options don't match. Alternatively, maybe it's a palindrome or has repeating digits. The number 110888889 is not a palindrome.Alternatively, maybe the problem is designed such that 111222222 - 333333 equals a square that is the square of one of the options, but due to a miscalculation in the problem design, none of the options fit. Alternatively, perhaps the problem is a trick question where the answer is (C) 333333, even though the math doesn't add up. But that seems unlikely.Alternatively, maybe there's a mistake in the problem's numbers. Let me check if 111222222 minus 3333333 (with an extra 3) would be a square. 111,222,222 - 3,333,333 = 107,888,889. What's sqrt(107,888,889)? Approximately 10,387. Not matching options.Alternatively, if the equation was 111,222,222 minus 333,333 equals (333,333 - 22,111)^2 or something. But I don't see a connection.Wait, maybe the problem is in base 6. Let's try converting the numbers to base 10. If 111222222 is in base 6, converting to base 10:1*6^8 + 1*6^7 + 1*6^6 + 2*6^5 + 2*6^4 + 2*6^3 + 2*6^2 + 2*6 + 2.Similarly, 333333 in base 6 is 3*6^5 + 3*6^4 + 3*6^3 + 3*6^2 + 3*6 + 3.Compute 111222222 (base6) in base10:6^0 = 1: 2*1=26^1 = 6: 2*6=126^2 = 36: 2*36=726^3 = 216: 2*216=4326^4 = 1296: 2*1296=25926^5 = 7776: 2*7776=15,5526^6 = 46,656: 1*46,656=46,6566^7 = 279,936: 1*279,936=279,9366^8 = 1,679,616: 1*1,679,616=1,679,616Adding them up:1,679,616 + 279,936 = 1,959,5521,959,552 + 46,656 = 2,006,2082,006,208 + 15,552 = 2,021,7602,021,760 + 2,592 = 2,024,3522,024,352 + 432 = 2,024,7842,024,784 + 72 = 2,024,8562,024,856 + 12 = 2,024,8682,024,868 + 2 = 2,024,870So 111222222 (base6) = 2,024,870 (base10).333,333 (base6) in base10:3*6^5 + 3*6^4 + 3*6^3 + 3*6^2 + 3*6 + 36^5=7776: 3*7776=23,3286^4=1296: 3*1296=3,8886^3=216: 3*216=6486^2=36: 3*36=1086^1=6: 3*6=186^0=1: 3*1=3Adding them up:23,328 + 3,888 = 27,21627,216 + 648 = 27,86427,864 + 108 = 27,97227,972 + 18 = 27,99027,990 + 3 = 27,993So 333,333 (base6) = 27,993 (base10).Subtract: 2,024,870 - 27,993 = 1,996,877.Now, is 1,996,877 a perfect square? Let's check sqrt(1,996,877). Approximately 1413, since 1400²=1,960,000 and 1413²=1,996,877. Exactly. Wait, 1413 squared is:1400²=1,960,0002*1400*13=36,40013²=169So 1,960,000 + 36,400 = 1,996,400 + 169 = 1,996,569. Close but not 1,996,877. Let's compute 1413*1413:1400*1400 = 1,960,0001400*13 = 18,20013*1400 = 18,20013*13 = 169Total: 1,960,000 + 18,200 + 18,200 + 169 = 1,960,000 + 36,400 = 1,996,400 + 169 = 1,996,569.So 1,996,569 vs 1,996,877. Difference is 308. So not a square. Therefore, even in base6, it's not a perfect square.This is getting frustrating. Maybe the problem is a trick question. Let's look at the options again. The options are (A) 332233, (B) 223377, (C) 333333, (D)331177.If we consider that 111222222 - 333333 = 110,888,889. Now, if we write this number as 110,888,889, maybe we can split it into 11 08 88 89, which could relate to option A: 332233 or D:331177. Not sure.Alternatively, maybe the digits of x correspond to the counts of digits in the original numbers. For example, 111222222 has three 1s and six 2s, minus 333333 which has six 3s. But I don't see how that would translate to x.Alternatively, maybe the answer is derived by some form of concatenation. For example, 111222222 - 333333: subtract 3 from 2, but that would be negative, so borrow, leading to a series of borrows changing the digits to 1,1,0,8,8,8,8,8,9. Maybe the number 110,888,889 can be split into 11 08 88 89, which doesn't match any options. Or 1 10 88 88 9. Not helpful.Alternatively, perhaps the square root is formed by taking digits from the original numbers. For example, 111222222 minus 333333: maybe subtracting digit-wise with some pattern. But this is too vague.Alternatively, perhaps the problem is in base 10, but there's a typo and the equation should be 111,222,222,222 - 333,333,333 = x², making x² = 110,888,888,889. Then check if that's a square. Let's see.Compute sqrt(110,888,888,889). Well, 333,333² = 111,110,888,889. So 333,333² is larger than 110,888,888,889. 333,000² = 110,889,000,000. Which is very close. 333,000² = 110,889,000,000. Then 333,000 squared is 110,889,000,000. Our number is 110,888,888,889. Difference is -111,111. So 333,000 - 111 = 332,889. Let's check 332,889². But this is random.Alternatively, maybe the square root is 333,333 - 333 = 333,000. But 333,000² = 110,889,000,000, which is not our number.Alternatively, the square root could be 333,333 - 444,444 = negative, which doesn't make sense.Alternatively, perhaps the answer is 333,333, but due to a miscalculation in the problem. But 333,333 squared is way larger than 110,888,889.Wait, if we ignore the discrepancy in the number of digits, and consider that maybe the problem expects a 6-digit answer regardless, even if the math doesn't add up, then the closest option might be (D) 331177. Let's check 331,177 squared.But earlier calculation showed 331,177² = 109,678,205,329, which is way off. None of the options work.Alternatively, maybe the problem is a prank and the correct answer is not listed, but given the options, perhaps there's a different approach. For example, the difference is 110,888,889. If we write this number in Chinese numerals or another system, it might relate to the options. But without knowledge of Chinese numerals, it's hard to say.Alternatively, maybe the problem is using the repeating digits. The original numbers have repeating digits: 111,222,222 and 333,333. The result is 110,888,889. Looking at the options, they also have repeating digits: 332233, 223377, 333333, 331177. Maybe there's a digit manipulation here. For example, subtracting the digits individually:111,222,222 -     333,333 =?But subtracting digit-wise, considering borrow:Starting from the right:2 - 3: Can't do, borrow 1 from the previous digit, which is 2. But that digit becomes 1, and the current digit becomes 12 - 3 = 9. Next digit: 1 (after borrow) - 3: Can't do, borrow again. Previous digit is 2, becomes 1, current digit becomes 11 - 3 = 8. Repeat this for the next four digits, each time becoming 8 after borrowing. The hundred thousands place: 1 (after borrowing) - 3: Can't do, borrow from millions place, which is 1, becomes 0, and current digit becomes 11 - 3 = 8. Then the millions place: 0 - 0 = 0, but since we borrowed, it's 0. Ten millions place: 1 - 0 = 1. Hundred millions place: 1 - 0 = 1.So the result is 110,888,889. The digits are 1,1,0,8,8,8,8,8,9. Maybe the square root is a number with digits 1,0,8,9 arranged somehow. The options given don't seem to reflect this.Alternatively, maybe the square root is 10,533 (but that's five digits), and the problem expects a different answer. But the options are six digits.This is really perplexing. Given that none of the options squared give the required result, and the calculation leads to x being a five-digit number, which isn't among the options, I think there might be a mistake in the problem. However, since this is a multiple-choice question, perhaps the intended answer is (C) 333333, despite the math not adding up, or there's a typo in the problem. Alternatively, maybe I made a mistake in the subtraction.Let me verify the subtraction one more time:111,222,222 minus 333,333:111,222,222 -     333,333 = 111,222,222 - 300,000 = 110,922,222 110,922,222 - 33,333 = 110,888,889. Yes, that's correct.So x² = 110,888,889. sqrt(110,888,889) ≈ 10,533. But none of the options are close to this.Wait, wait a second. Let me compute 10,533 squared again.10,533 * 10,533:Breakdown:10,533 = 10,000 + 500 + 33.So (10,000 + 500 + 33)² = 10,000² + 500² + 33² + 2*10,000*500 + 2*10,000*33 + 2*500*33.Compute each part:10,000² = 100,000,000500² = 250,00033² = 1,0892*10,000*500 = 10,000,0002*10,000*33 = 660,0002*500*33 = 33,000Now add them all:100,000,000 + 250,000 + 1,089 = 100,251,08910,000,000 + 660,000 + 33,000 = 10,693,000Total: 100,251,089 + 10,693,000 = 110,944,089Yes, 10,533² = 110,944,089, which is higher than 110,888,889. Therefore, no integer x satisfies the equation, which contradicts the problem statement.Given that, I think there's a high chance the problem has a typo. Perhaps the intended equation was 111,222,222,222 - 333,333,333 = x², leading to x = 333,333, as in option C. Because 333,333² = 111,110,888,889, and if the first number was 111,222,222,222 - 333,333,333 = 110,888,888,889, which is close to 333,333². But this is speculative.Alternatively, if the original problem was in a different language or used different notation, perhaps the numbers are different. Without further information, it's hard to resolve.However, given the options, and considering possible miscalculations or typos, the most plausible answer might be (C) 333333, even though the math doesn't add up as presented. Alternatively, maybe the problem expects us to recognize a pattern where the answer is composed of the digits 3 and 1 repeating, so option D: 331177. But without a clear pattern, this is guesswork.Alternatively, if we consider that 110,888,889 divided by 9 is 12,320,987.666..., not an integer. If we look for a number that when squared ends with 9, since 110,888,889 ends with 9, x must end with 3 or 7. Among the options, (A) ends with 3, (D) ends with 7. Let's check the last digit of the squares:If x ends with 3, x² ends with 9. If x ends with 7, x² also ends with 9. So both options A and D could be possible. Let's check the last two digits. 110,888,889 ends with 89. A number ending with 33 squared: 33²=1089, ends with 89. Similarly, 77²=5929, ends with 29. So if x ends with 33, x² ends with 89. So option A ends with 33, option D ends with 77. So only option A's last two digits squared would end with 89.Compute 332,233² and check the last few digits. 332,233²: last two digits 33²=1089, so ends with 89. Next, let's check the last four digits. 2233²= 2233*2233. Compute 2200²=4,840,000, 2200*33=72,600, 33*2200=72,600, 33²=1,089. Total: 4,840,000 + 72,600 + 72,600 + 1,089 = 4,840,000 + 145,200 + 1,089 = 4,986,289. So last four digits: 6289. But 110,888,889 has last four digits 8889. Doesn't match. Hmm.Alternatively, compute 332,233² modulo 10000 to get the last four digits. 332,233 mod 10000 is 2233. 2233²=4,986,289, so last four digits 6289. But the target is 8889. So not matching.Similarly, check if any of the options squared end with 8889.Take option D: 331,177. Last four digits 1177. Compute 1177²:1177*1177:Let's compute 1000² = 1,000,000177*1000*2 = 354,000177² = 31,329Total: 1,000,000 + 354,000 = 1,354,000 + 31,329 = 1,385,329. So last four digits 5329. Doesn't match.Option B: 223,377. Last four digits 3377. 3377²: 3377*3377. Last four digits: compute 77²=5,929, so ends with 2929. Not 8889.Option A: 332,233: last four digits 2233, squared ends with 6289.Option C: 333,333 squared ends with 8889. Wait, 333,333 squared is 111,110,888,889. Yes, the last four digits are 8889. But our target is 110,888,889, which ends with 8889. So if the problem had 111,110,888,889 as x², then x would be 333,333. But the problem states x² is 110,888,889. However, 333,333 squared is 111,110,888,889, which is different. But the last four digits are the same. So maybe the problem intended a different first number, leading to x=333,333. Given that option C's square ends with the same last four digits as the target, and the problem might have a missing digit, the intended answer is likely C) 333333.Therefore, despite the discrepancy, I think the answer is C) 333333.Final Answerboxed{C}</think>To solve the equation (111222222 - 333333 = x^2) and find (x) given the options, we start by performing the subtraction:[111222222 - 333333 = 110888889]We need to determine if (110888889) is a perfect square. The options provided are six-digit numbers, so we check if any of these options squared results in (110888889).1. Checking Option (C) 333333: - Compute (333333^2): [ 333333^2 = (333000 + 333)^2 = 333000^2 + 2 cdot 333000 cdot 333 + 333^2 ] [ 333000^2 = 110889000000 ] [ 2 cdot 333000 cdot 333 = 2 cdot 110889000 = 221778000 ] [ 333^2 = 110889 ] [ 333333^2 = 110889000000 + 221778000 + 110889 = 111110888889 ] - The result (111110888889) is close to (110888889) but not exact. However, considering the possibility of a typo or miscalculation in the problem statement, the closest match given the options and the repeating digit pattern, the intended answer is likely (333333).2. Verification of Other Options: - Other options were checked and found to result in much larger values when squared, which do not match (110888889).Given the options and the closest match in pattern and possible typo considerations, the answer is:[boxed{C}],<think>Alright, let me try to solve this problem step by step. So, the equation given is 111111222222 minus 333333 equals x squared, and we need to find the positive x value from the options provided. Hmm, okay.First, let me write down the numbers to make sure I have them right. The first number is 111111222222. Let me count the digits here. Starting from the left: 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2. So that's twelve digits in total. The first six digits are all 1s, and the last six digits are all 2s. Interesting. Then we subtract 333333 from this number, and the result should be a perfect square, x squared. We need to find x from the given options.The options are (A) 332233, (B) 223377, (C) 333333, (D) 331177. All of these are six-digit numbers. Since x is positive, we can ignore negative roots.Alright, so my plan is to first compute the subtraction: 111111222222 minus 333333. Then check if the result is a perfect square, and if so, see which of the options matches the square root. Alternatively, maybe there's a pattern or a factorization that can make this easier without brute-forcing the calculation.But given that these numbers are quite large, maybe there's a trick here. Let's see. Let me first note that 111111222222 is a twelve-digit number, and 333333 is a six-digit number. So subtracting them will affect the last six digits. Let me write them out:111111222222- 333333= ?Let me subtract digit by digit. Starting from the rightmost digit:The last six digits of the first number are 222222. Subtracting 333333 from that. But 222222 is less than 333333, so we'll need to borrow from the preceding digits. Let's handle this.The entire number is 111111 followed by 222222. So when we subtract 333333, we're essentially subtracting from the last six digits. Let me break it down:First, split the number into two parts: the first six digits (111111) and the last six digits (222222). When subtracting 333333, since 222222 - 333333 would be negative, we need to borrow 1 from the first part. So, borrowing 1 from the first part (which is 111111) reduces it by 1, making it 111110, and adds 10^6 (which is 1,000,000) to the second part. So the calculation becomes:First part: 111111 - 1 = 111110Second part: 222222 + 1,000,000 - 333333 = 222222 + 666667 = 888889So the result of the subtraction is 111110888889. Now, we need to check if this number is a perfect square.Given that x is one of the options, which are six-digit numbers, x squared would be a twelve-digit number. Let's see:Looking at the options:(A) 332233: Let's compute 332233^2.But before that, maybe check if there's a pattern or if 111110888889 is a known square. Let me see if the number 111110888889 has any properties. Let's check if it's a palindrome? 111110888889 – reading from left to right: 1,1,1,1,1,0,8,8,8,8,8,9. Not a palindrome. Hmm.Alternatively, maybe this number can be represented as something like (111111 - something)^2? Let's see. Let's think about the structure of the number. The first part after subtraction is 111110 and the second part is 888889. Let me write the entire number: 111,110,888,889.Wait, maybe this is a square of 333,333? Let's check option C: 333,333 squared. Let me compute that.333,333 * 333,333. Let's compute step by step.First, 300,000^2 = 90,000,000,000.Then, 33,333^2: but maybe it's better to compute 333,333^2.Alternatively, notice that 333,333 is 1/3 of 1,000,000 minus 1, so 333,333 = 1,000,000/3 - 1/3, but that might complicate.Alternatively, compute (300,000 + 33,333)^2 = 300,000^2 + 2*300,000*33,333 + 33,333^2.Compute each term:300,000^2 = 90,000,000,000.2*300,000*33,333 = 2*300,000*33,333 = 600,000*33,333.Compute 600,000 * 33,333:First, 600,000 * 30,000 = 18,000,000,000600,000 * 3,333 = 600,000 * 3,000 = 1,800,000,000600,000 * 333 = 600,000 * 300 = 180,000,000600,000 * 33 = 19,800,000So adding those:18,000,000,000 + 1,800,000,000 = 19,800,000,00019,800,000,000 + 180,000,000 = 19,980,000,00019,980,000,000 + 19,800,000 = 20,000, (wait, maybe my approach is getting messy here. Maybe better to use another method.)Alternatively, 33,333 is approximately 33.333 thousand. Wait, perhaps use the formula (a + b)^2.Alternatively, 333,333 * 333,333. Let's compute this as follows:333,333 * 333,333= (333,333)^2We can note that 333,333 is 333 thousand plus 333. Let me see:(333,000 + 333)^2 = 333,000^2 + 2*333,000*333 + 333^2Compute each term:333,000^2 = (333)^2 * (1,000)^2 = 110,889 * 1,000,000 = 110,889,000,0002*333,000*333 = 2*333*1,000*333 = 2*333^2*1,000333^2 = 110,889, so 2*110,889*1,000 = 221,778,000333^2 = 110,889So total is 110,889,000,000 + 221,778,000 + 110,889Adding them together:110,889,000,000 + 221,778,000 = 111,110,778,000111,110,778,000 + 110,889 = 111,110,888,889Wait a second! That's exactly the number we have after subtraction: 111,110,888,889. So 333,333 squared is 111,110,888,889. Therefore, x is 333,333, which is option C.Therefore, the answer is C.But let me verify again just to be sure.Original equation: 111111222222 - 333333 = x^2We computed 111111222222 - 333333 = 111110888889Then, 333,333^2 = 111,110,888,889, which matches the result. Therefore, x is 333,333, option C.But let me check if any other options could square to this number. Given that the options are all six-digit numbers, and 333,333 is the only one in the options that, when squared, gives a number close to 111,111,000,000 (since 333,333 is about a third of 1,000,000, so squaring it gives roughly (1/3 of a million)^2 which is about 1/9 of a trillion, but our number is 111 billion, which is roughly in line. The other options:(A) 332,233: Let's approximate. 330,000 squared is 108,900,000,000. 332,233 squared would be a bit more, but not 111 billion.(B) 223,377: This is about 223,000. Squared is around 223,000^2 = 49,729,000,000, way less than 111 billion.(D) 331,177: Similar to option A, around 330,000 squared is 108.9 billion. So 331,177 squared would be a bit more, maybe 109 billion, still less than 111 billion.Therefore, only option C) 333,333 squares to 111,110,888,889 which is exactly the result of the subtraction. So the answer must be C.Final Answerboxed{C}</think>Given the equation (111111222222 - 333333 = x^2) where (x > 0), we need to find the value of (x) from the given options.First, we perform the subtraction:[111111222222 - 333333]Breaking down the subtraction:- The number (111111222222) can be split into two parts: (111111) (first six digits) and (222222) (last six digits).- Subtracting (333333) from (222222) requires borrowing 1 from the first part. This reduces the first part by 1 (making it (111110)) and adds (1000000) to the second part.- The second part becomes (222222 + 1000000 - 333333 = 888889).Thus, the result of the subtraction is:[111110888889]Next, we check if this result is a perfect square. We consider the given options and compute the square of option (C) (333333):[333333^2 = (333000 + 333)^2 = 333000^2 + 2 cdot 333000 cdot 333 + 333^2]Calculating each term:- (333000^2 = 110889000000)- (2 cdot 333000 cdot 333 = 221778000)- (333^2 = 110889)Adding these together:[110889000000 + 221778000 + 110889 = 111110888889]This matches the result of the subtraction. Therefore, the value of (x) is (333333), which corresponds to option (C).[boxed{C}]]

🔑:begin{array}{l}text {-1.C. } 111111222222-333333 =111111 times(10^{6}+2)-111111 times 3 =111111 times(10^{6}-1)=frac{999999}{9}(10^{6}-1) =frac{10^{6}-1}{9}(10^{6}-1)=left(frac{10^{6}-1}{3}right)^{2}=333333^{2} .end{array}

🧠:[<think>Okay, let me try to figure out this problem step by step. So, we have a dataset: 2, 2x, y, 14. The problem says the only mode is 14, and the mean is 12. We need to find the median. Hmm, let's break this down.First, mode being 14 means that 14 appears more times than any other number in the set. Since there are four numbers in the set, and 14 is the only mode, it must appear at least twice. The original set has the numbers 2, 2x, y, and 14. So, unless one of the other numbers is equal to 14, the only way for 14 to be the mode is if it appears at least twice. But in the given set, 14 is already present once. So, either 2, 2x, or y has to be 14 to make it the mode. However, the problem states that the only mode is 14, which means no other number can appear more than once. Therefore, if 2 or y were 14, then 14 would still be the mode, but we need to check if that introduces another mode. Wait, actually, since the set is 2, 2x, y, 14. If two of these numbers are 14, then 14 is the mode. If three numbers are 14, then it's still the mode, but the problem says "the only mode is 14," so other numbers shouldn't be modes. So, let's see.First, let's note that the mean is 12. The mean of four numbers is (2 + 2x + y + 14)/4 = 12. So, let's write that equation:(2 + 2x + y + 14)/4 = 12Multiply both sides by 4:2 + 2x + y + 14 = 48Combine like terms:2x + y + 16 = 48Subtract 16:2x + y = 32So, that's our first equation: 2x + y = 32.Now, the mode is 14, which as we discussed, means that 14 must appear at least twice, and no other number appears more than once. Let's look at the original set: 2, 2x, y, 14. For 14 to be a mode, at least one of the other numbers (2, 2x, y) must be 14. Let's check possibilities.Case 1: 2 = 14. But 2 is given, and 14 is different. So that's not possible.Case 2: 2x = 14. Then x = 7. If 2x is 14, then the set becomes 2, 14, y, 14. So, 14 appears twice. Now, we need to ensure that no other number is a mode. The other numbers are 2 and y. So, y can't be 14 or 2, otherwise, there would be another mode. Wait, if y is 14, then 14 appears three times, which is still mode, but if y is 2, then 2 appears twice, making mode both 2 and 14. But the problem says the only mode is 14, so y cannot be 2. So, in this case, 2x =14, x=7, so 2x=14. Then, the set is 2,14,y,14. To have only 14 as mode, y must be different from both 2 and 14. So, y ≠2 and y≠14. Then, since we have the equation 2x + y =32, and x=7, substituting in:2*7 + y =32 =>14 + y=32 => y=18.So, y=18. Then the data set is 2,14,18,14. Let's check the modes. 14 appears twice, 2 and 18 once. So, mode is 14. Good. Then, in this case, the median would be... Let's order the numbers. Since it's four numbers, when ordered, the median is average of the second and third numbers.So, order: 2,14,14,18. So, second and third numbers are 14 and 14. So, median is (14 +14)/2=14.Alternatively, if we considered another case where y=14, then let's check.Case 3: y=14. Then, the set is 2, 2x,14,14. Then, 14 appears twice, but if 2x is also 14, then x=7, which is the same as Case 2. But if 2x is not 14, then 2x could be another number. Wait, but if y=14, then we have to ensure that 2x isn't equal to 14. Because if 2x=14, then again we have two 14s. But if y=14, and 2x is not 14, then 14 is still the mode. But in this case, let's see.If y=14, then using the equation:2x + y =32 =>2x +14=32 =>2x=18 =>x=9. So, 2x=18. Then the set becomes 2,18,14,14. So, same as previous case, sorted as 2,14,14,18. So, same median. So, in this case, even if y=14, but 2x=18, the set is same as previous. So, actually, whether 2x=14 or y=14, both lead to the same set. Wait, but in the first case, x=7, y=18, in the second case, x=9, y=14. But these result in the same dataset: 2,14,14,18. Wait, no. If x=9, then 2x=18, y=14. So, the numbers would be 2,18,14,14. So sorted: 2,14,14,18. If x=7, then 2x=14, y=18, so the numbers are 2,14,18,14. Sorted: same. So, same result. So, either way, same dataset. So, median is 14.But wait, but in the first case, when x=7, y=18, and in the second case, x=9, y=14, but in the second case, does y=14 make another mode? Wait, the original problem says the only mode is 14. If y=14, then the set is 2,18,14,14. So, 14 appears twice, others once. So, mode is 14. Similarly, if 2x=14 and y=18, same thing. So, both cases are valid. Therefore, the median is 14 in both cases.But wait, let's check if there are other possibilities. For instance, what if both 2x and y equal 14? Then, the set would be 2,14,14,14. Then, 14 is the mode, but the problem says "the only mode is 14", which is still true. But in this case, the median would be (14 +14)/2=14. But does this case hold?If both 2x=14 and y=14, then 2x=14 =>x=7, and y=14. Then, substituting into the equation:2x + y = 32 =>14 +14=28≠32. So, that doesn't satisfy the equation. Therefore, this case is invalid. So, both 2x and y can't be 14.Alternatively, if three numbers are 14, then sum would be 2 +14 +14 +14=44, but the total required is 48, so 44≠48. Therefore, impossible. So, in order to have 14 as the only mode, we must have exactly two instances of 14, and the other two numbers must be distinct and not equal to each other or 14.Therefore, the only possibilities are:Either 2x=14 and y=18, leading to the set 2,14,18,14,Or y=14 and 2x=18, leading to the set 2,18,14,14.Both of which sort to 2,14,14,18, so the median is 14.Wait, but let's check if there's another case where the original 2 is duplicated. If the original 2 is duplicated, meaning 2x=2, then x=1. Then, the set would be 2,2, y,14. Then, modes would be 2 and 14, which contradicts the problem's statement that only 14 is the mode. So, that's invalid. So, 2x cannot be 2. Similarly, y cannot be 2, because then the set would be 2,2x,2,14. If y=2, then 2 appears twice. If 2x is also 2, then x=1, which leads to three 2s and one 14. Then, mode is 2. But since the problem states mode is 14, this is invalid. If y=2 and 2x≠2, then the modes are 2 and 14 (each appearing twice). But the problem says the only mode is 14, so that's invalid. Therefore, y can't be 2, and 2x can't be 2.Therefore, the only possibilities are that either 2x=14 or y=14, leading to two instances of 14, and the other numbers being distinct and not 14 or 2. Therefore, in both cases, the sorted array is 2,14,14,18, giving the median of 14.But let me just check the math again. The equation 2x + y =32. If 2x=14, then x=7, so y=18. If y=14, then 2x=18, so x=9. Both are valid. So, in both scenarios, the data set becomes 2,14,14,18 when sorted. Therefore, median is (14 +14)/2=14.Wait, but what if y= something else? Wait, if neither y nor 2x is 14, then the number 14 appears only once, so mode would not exist or all numbers are modes? Wait, in the original set, if 14 appears only once, and all other numbers appear once, then there is no mode. But the problem says the only mode is 14, so 14 must appear more than once. Therefore, we must have either y=14 or 2x=14.Therefore, those are the only two possibilities. So, in both cases, median is 14. So, the answer is 14.But let me just verify once again. Suppose 2x=14, so x=7, then y=32 - 2*7=18. So, the set is 2,14,18,14. Sort it: 2,14,14,18. Median average of 2nd and 3rd terms: 14 and 14. So, 14.If y=14, then 2x=32 -14=18, so x=9. Then, the set is 2,18,14,14. Sort it: 2,14,14,18. Same as above. Median is 14. So, either way, median is 14. Therefore, the answer is 14.Final AnswerThe median of the data is boxed{14}.</think>Given the dataset: (2), (2x), (y), (14), where the only mode is (14) and the mean is (12), we need to find the median.First, we calculate the mean:[frac{2 + 2x + y + 14}{4} = 12]Multiplying both sides by 4:[2 + 2x + y + 14 = 48]Simplifying:[2x + y + 16 = 48 implies 2x + y = 32]Since (14) is the only mode, it must appear at least twice. The dataset has four elements: (2), (2x), (y), and (14). Therefore, either (2x = 14) or (y = 14).Case 1: (2x = 14)- Solving for (x): (x = 7)- Substituting (x = 7) into (2x + y = 32): (2(7) + y = 32 implies 14 + y = 32 implies y = 18)- The dataset becomes: (2), (14), (18), (14)- Sorted dataset: (2), (14), (14), (18)Case 2: (y = 14)- Substituting (y = 14) into (2x + y = 32): (2x + 14 = 32 implies 2x = 18 implies x = 9)- The dataset becomes: (2), (18), (14), (14)- Sorted dataset: (2), (14), (14), (18)In both cases, the sorted dataset is (2), (14), (14), (18). The median is the average of the second and third values:[text{Median} = frac{14 + 14}{2} = 14]Thus, the median of the data is (boxed{14}).,<think>Okay, so I have this problem here: In a set of numbers 2, 2x, y, 14, the only mode is 14, and the mean is 12. I need to find the median of the data. Hmm, let me try to break this down step by step.First, let's recall what each term means. The mode is the number that appears most frequently in the dataset. Since the problem states that the only mode is 14, that means 14 must appear more times than any other number in the set. The mean being 12 means the average of the four numbers is 12. And the median will be the middle value when the numbers are ordered from least to greatest. Since there are four numbers, the median will be the average of the second and third numbers after sorting.Alright, let's start with the mode. The set given is 2, 2x, y, 14. Currently, each number appears once. For 14 to be the only mode, it has to appear at least twice, right? Because the other numbers (2, 2x, y) can only appear once each unless there are duplicates. So, in order for 14 to be the mode, at least one of the other numbers must be 14, but not more than the number of times 14 appears. Wait, but if 14 is the only mode, that means all other numbers can appear at most once. But since there are four numbers in the set, and 14 needs to appear at least twice, that implies that two of the numbers in the set must be 14. So, either 2x is 14, or y is 14, or one of them is 14. But 2 is already in the set, so if 2x is 14, then x would be 7, and 2x would be 14. Then, we would have two 14s, which would make 14 the mode. Alternatively, if y is 14, then we have two 14s as well. However, if both 2x and y are 14, then we would have three 14s, which still makes 14 the mode. But the problem says the "only" mode is 14, so we have to make sure that no other number appears more than once. So, if 2x is 14, then x is 7, and 2x becomes 14, but we already have another 14. So that would give us two 14s. If y is 14, then we have two 14s. But if both 2x and y are 14, that's three 14s, but that's still okay. However, the problem states the "only" mode is 14, so we need to ensure that 2x and y aren't equal to any other numbers in the set except 14. So, 2x can't be 2 because then we would have two 2s, making 2 a mode as well, which is not allowed. Similarly, y can't be 2 for the same reason. Also, 2x and y can't be equal to each other unless they are equal to 14, because if they were equal to some other number, then that number would be a mode as well. So, 2x and y can either be 14 or unique numbers not equal to 2 or each other. So, possible scenarios:1. Either 2x = 14 or y = 14 (or both). Let's consider each case.Case 1: 2x = 14. Then, x = 7. So, the set becomes 2, 14, y, 14. Then, the numbers are 2, y, 14, 14. To ensure that 14 is the only mode, y cannot be 2 or 14. But y could be another number. However, we also have the mean condition. The mean is 12, so the sum of all numbers is 4 * 12 = 48. Let's check the sum in this case. The numbers would be 2, 14, y, 14. Sum is 2 + 14 + y + 14 = 30 + y. So, 30 + y = 48 => y = 18. So y would be 18. Then the set is 2, 14, 18, 14. When sorted: 2, 14, 14, 18. Median is average of 14 and 14, which is 14.Case 2: y = 14. Then, the set becomes 2, 2x, 14, 14. Then, similar reasoning. The sum is 2 + 2x + 14 + 14 = 30 + 2x. So, 30 + 2x = 48 => 2x = 18 => x = 9. Therefore, 2x is 18. So the set is 2, 18, 14, 14. When sorted: 2, 14, 14, 18. Median is again 14.Case 3: Both 2x = 14 and y = 14. Then, x = 7 and y = 14. The set becomes 2, 14, 14, 14. Sum is 2 + 14 +14 +14 = 44. But 4*12 = 48, so this sum is 44, which is insufficient. Therefore, this case is invalid because the sum is not 48. So, Case 3 is impossible.Therefore, only Cases 1 and 2 are possible. In both cases, the median is 14. Wait, but let me check again. In Case 1: 2, 14, 18, 14. When sorted: 2, 14, 14, 18. So median is average of second and third numbers: (14 + 14)/2 = 14. In Case 2: 2, 14, 14, 18. Same sorted order. So same median. So regardless of whether 2x is 14 or y is 14, the median is 14. Therefore, the answer should be 14. But let me just verify all constraints again.In both cases, 14 is the only mode. In Case 1: numbers are 2, 14, 14, 18. Modes: 14. In Case 2: same numbers. So mode is 14. So that's good. Also, the mean is 12 in both cases. So that works. So the median is 14.Wait, but hold on. Let me check if there's any other possibility. For instance, could 2x be equal to y, but not equal to 14? Then, we would have two of the same number, which would create another mode. But the problem says the only mode is 14, so that's not allowed. So 2x and y can't be equal unless they are 14. So if 2x = y, then that would mean y = 2x, but then if they are not 14, that would create two instances of another number, which is not allowed. Therefore, 2x can't equal y unless they are both 14. So the only possibilities are the ones considered above. So there are no other cases. Therefore, the median must be 14. So I think that's the answer. Let me just write it out again step by step to make sure.Given the set {2, 2x, y, 14} with mode 14 and mean 12.1. Since mode is 14, 14 must appear at least twice. So either 2x =14 or y=14 (or both, but both leads to sum issues).2. Calculate the sum: 2 + 2x + y +14 = 16 + 2x + y. The mean is 12, so total sum is 48. Therefore, 16 + 2x + y =48 => 2x + y =32.3. If 2x=14, then x=7, and y=32 -14=18. So the set becomes {2,14,18,14}, sorted {2,14,14,18}, median=14.4. If y=14, then 2x +14=32 =>2x=18 =>x=9. The set becomes {2,18,14,14}, sorted {2,14,14,18}, median=14.5. Both 2x=14 and y=14 leads to sum=2+14+14+14=44≠48, invalid.Thus, in all valid cases, the median is 14. Therefore, the answer is 14.Final AnswerThe median of the data is boxed{14}.</think>Given the set ({2, 2x, y, 14}) with the only mode being 14 and a mean of 12, we need to find the median.1. Mode Analysis: Since 14 is the only mode, it must appear at least twice. Therefore, either (2x = 14) or (y = 14) (or both, but both would lead to a sum issue).2. Mean Calculation: The mean is 12, so the total sum of the numbers is (4 times 12 = 48). The sum of the given numbers is (2 + 2x + y + 14 = 16 + 2x + y). Therefore, (16 + 2x + y = 48) which simplifies to (2x + y = 32).3. Case Analysis: - Case 1: If (2x = 14), then (x = 7) and substituting into (2x + y = 32) gives (14 + y = 32) leading to (y = 18). The set becomes ({2, 14, 18, 14}), which when sorted is ({2, 14, 14, 18}). The median is the average of the second and third numbers: (frac{14 + 14}{2} = 14). - Case 2: If (y = 14), substituting into (2x + y = 32) gives (2x + 14 = 32) leading to (2x = 18) and (x = 9). The set becomes ({2, 18, 14, 14}), which when sorted is ({2, 14, 14, 18}). The median is again the average of the second and third numbers: (frac{14 + 14}{2} = 14).4. Invalid Case: If both (2x = 14) and (y = 14), the sum would be (2 + 14 + 14 + 14 = 44), which is not 48, thus invalid.In both valid cases, the median is 14. Therefore, the median of the data is (boxed{14}).]

🔑:Given that the mean of the set 2, 2x, y, 14 is 12, we can write the equation for the mean as follows:[frac{1}{4} times (2 + 2x + y + 14) = 12]Expanding and simplifying this equation gives us:[frac{1}{4} times (16 + 2x + y) = 12 implies 16 + 2x + y = 48]Subtracting 16 from both sides, we find:[2x + y = 32]Given that the only mode is 14, this implies that 14 is the only number that repeats in the set, and since 14 is already one of the numbers, 2x or y must equal 14. This leads to two cases:1. If x = 7, then 2x = 14. Substituting x = 7 into the equation 2x + y = 32 gives us:[2(7) + y = 32 implies 14 + y = 32 implies y = 18]2. If y = 14, then we need to find x such that 2x + 14 = 32. Solving for x gives:[2x = 18 implies x = 9]However, since 14 is the only mode, and for x=9, 2x would be 18, not 14, the correct set of numbers is when x=7 and y=18. Thus, the numbers in ascending order are 2, 14, 14, 18.The median of this data set is the average of the two middle numbers, which are both 14. Therefore, the median is:[frac{14 + 14}{2} = 14]Hence, the median of the data is boxed{14}.

🧠:[<think>Okay, let's see. I need to solve this equation: √(2y + x - 2) + |x + 2| = 0. Hmm, both a square root and an absolute value. I remember that square roots are always non-negative, right? And absolute values are also non-negative. So the sum of two non-negative things equals zero. That means each of them must individually be zero. Because if one was positive and the other negative, their sum could still be zero, but since they can't be negative, the only way their sum is zero is if both are zero. Yeah, that makes sense.So, that gives me two equations:1. √(2y + x - 2) = 02. |x + 2| = 0Let me write those down. First, the square root equals zero. That implies that the expression inside the square root is zero. So:2y + x - 2 = 0And the absolute value of (x + 2) equals zero. Absolute value is zero only when the inside is zero. So:x + 2 = 0Therefore, from the second equation, x = -2. Then plug that into the first equation to find y.Substituting x = -2 into 2y + x - 2 = 0:2y + (-2) - 2 = 02y - 4 = 02y = 4y = 2So, the solution is x = -2 and y = 2. Let me check if that works in the original equation.First, compute √(2y + x - 2):2y + x - 2 = 2*2 + (-2) - 2 = 4 - 2 - 2 = 0. So √0 = 0.Then |x + 2| = |-2 + 2| = |0| = 0.Adding them together: 0 + 0 = 0. Yep, that checks out. So part (1) is solved: x = -2, y = 2.Now part (2): Find the value of √(3x² - 4y) + (x/y)^2033.First, let's substitute x = -2 and y = 2 into the expression.Compute 3x² - 4y: 3*(-2)² - 4*2 = 3*4 - 8 = 12 - 8 = 4. So √4 = 2.Then compute (x/y)^2033: (-2/2)^2033 = (-1)^2033. Since 2033 is an odd exponent, (-1)^2033 = -1.Therefore, adding those two parts: 2 + (-1) = 1.Wait, let me verify each step again to be sure. First, 3x²: x is -2, so (-2)^2 is 4, 3*4=12. Then 4y: y=2, so 4*2=8. 12 - 8=4. Square root of 4 is indeed 2. Then x/y: (-2)/2=-1. Raising -1 to the power of 2033, which is an odd number, so it's -1. So 2 + (-1)=1. That seems right.But let me double-check if I interpreted the original problem correctly. The square root is only over 2y + x - 2, right? The problem was written as √(2y + x - 2) + |x + 2|=0. So yes, separate terms. And part (2) is sqrt(3x² -4y) + (x/y)^2033. Yes, that's how it's written. So substituting x=-2, y=2 into those expressions. All steps check out.I think that's all. So the answers are x=-2, y=2, and the value of the second expression is 1.Final Answer(1) The values are ( boxed{x = -2} ) and ( boxed{y = 2} ).(2) The value of the expression is ( boxed{1} ).</think>Given the equation (sqrt{2y + x - 2} + |x + 2| = 0):1. Since both the square root and the absolute value are non-negative, each must be zero for their sum to be zero: - (sqrt{2y + x - 2} = 0) implies (2y + x - 2 = 0). - (|x + 2| = 0) implies (x + 2 = 0). Solving (x + 2 = 0) gives (x = -2). Substituting (x = -2) into (2y + x - 2 = 0): [ 2y - 2 - 2 = 0 implies 2y - 4 = 0 implies 2y = 4 implies y = 2 ] Therefore, the values are (x = -2) and (y = 2).2. Substitute (x = -2) and (y = 2) into the expression (sqrt{3x^2 - 4y} + left(frac{x}{y}right)^{2033}): - Compute (3x^2 - 4y): [ 3(-2)^2 - 4(2) = 3 cdot 4 - 8 = 12 - 8 = 4 ] The square root is (sqrt{4} = 2). - Compute (left(frac{x}{y}right)^{2033}): [ left(frac{-2}{2}right)^{2033} = (-1)^{2033} = -1 ] Adding these results: [ 2 + (-1) = 1 ] Final Answer(1) The values are ( boxed{x = -2} ) and ( boxed{y = 2} ).(2) The value of the expression is ( boxed{1} ).,<think>Okay, so I have this equation to solve: √(2y + x - 2) + |x + 2| = 0. Hmm, interesting. The problem has two parts: first, find the values of x and y, and second, compute some expression involving those values. Let me tackle part 1 first.Alright, so the equation is the sum of a square root and an absolute value, and they're saying that this sum equals zero. Both the square root and the absolute value are non-negative by definition, right? Because a square root can't be negative (since you can't take the square root of a negative number in real numbers) and the absolute value is always positive or zero. So if two non-negative things add up to zero, each of them must be zero individually. That makes sense. So, I can split the equation into two separate equations:1. √(2y + x - 2) = 02. |x + 2| = 0Let me verify that logic. If either one was positive, then the sum would be positive, which contradicts the equation being equal to zero. So yeah, both have to be zero. Cool.Starting with the second equation: |x + 2| = 0. The absolute value of something is zero only when the inside is zero. So x + 2 = 0 => x = -2. Got that. So x is -2.Now plug x = -2 into the first equation: √(2y + (-2) - 2) = 0. Simplify inside the square root: 2y - 2 - 2 = 2y - 4. So √(2y - 4) = 0. If the square root is zero, then the inside must be zero. So 2y - 4 = 0 => 2y = 4 => y = 2. Okay, so y is 2.Wait, let me check if these values actually satisfy the original equation. Let's substitute x = -2 and y = 2 into √(2y + x - 2) + |x + 2|. Calculate each part:First, 2y + x - 2 = 2*2 + (-2) - 2 = 4 - 2 - 2 = 0. So √0 = 0. Then |x + 2| = |-2 + 2| = |0| = 0. So 0 + 0 = 0, which matches the equation. Perfect, so x = -2 and y = 2 are the solutions. That's part 1 done.Now part 2: Find the value of √(3x² - 4y) + (x/y)^2033. Alright, let's substitute x = -2 and y = 2 into this expression.First, compute 3x² - 4y. Let's break it down:x² = (-2)² = 4. So 3x² = 3*4 = 12. Then, 4y = 4*2 = 8. So 3x² - 4y = 12 - 8 = 4. Then the square root of that is √4 = 2. So that part is 2.Next, compute (x/y)^2033. x is -2, y is 2, so x/y = (-2)/2 = -1. Then (-1)^2033. Since 2033 is an odd exponent, (-1)^odd = -1. So that part is -1.Therefore, adding them together: 2 + (-1) = 1.Wait, let me double-check each step to make sure I didn't make any mistakes.Starting with 3x² - 4y: x is -2, so squared is 4, multiplied by 3 gives 12. 4y is 4*2=8. 12-8=4. Square root of 4 is 2. That seems right.Then x/y is -2/2 = -1. Exponent 2033: since 2033 divided by 2 is 1016.5, so it's an odd exponent. So (-1)^2033 is -1. Correct. Then 2 + (-1) is 1. Yes, that seems right.So the value of the expression is 1. Let me just check once more if I substituted the values correctly.Yes, x = -2, y = 2. All substitutions look correct. Exponent is indeed odd, so negative one raised to an odd power is negative one. Square root of 4 is 2. So 2 + (-1) is 1. All checks out.Therefore, part 2 answer is 1.Final Answer(1) The values are boxed{x = -2} and boxed{y = 2}; (2) The value of the expression is boxed{1}.</think>Given the equation (sqrt{2y + x - 2} + |x + 2| = 0):1. To find the values of (x) and (y): - Since both the square root and the absolute value are non-negative, each must be zero individually. - From (|x + 2| = 0), we get (x + 2 = 0 implies x = -2). - Substituting (x = -2) into (sqrt{2y + x - 2} = 0), we get: [ sqrt{2y - 2 - 2} = 0 implies sqrt{2y - 4} = 0 implies 2y - 4 = 0 implies y = 2 ] - Verification: Substituting (x = -2) and (y = 2) into the original equation confirms it holds true.2. To find the value of (sqrt{3x^2 - 4y} + left(frac{x}{y}right)^{2033}): - Substitute (x = -2) and (y = 2): - Calculate (3x^2 - 4y): [ 3(-2)^2 - 4(2) = 3 cdot 4 - 8 = 12 - 8 = 4 ] - The square root is (sqrt{4} = 2). - Calculate (left(frac{x}{y}right)^{2033}): [ left(frac{-2}{2}right)^{2033} = (-1)^{2033} = -1 ] - Add the results: (2 + (-1) = 1). Final Answer(1) The values are (boxed{x = -2}) and (boxed{y = 2}); (2) The value of the expression is (boxed{1}).]

🔑:To solve the given problem, we proceed as follows:Part (1): Find the values of x and yGiven the equation sqrt{2y+x-2}+|x+2|=0, we can deduce two conditions from this equation because the square root and the absolute value must both be non-negative and their sum is zero. This means each must individually be equal to zero. Therefore, we have:1. sqrt{2y+x-2}=02. |x+2|=0From the first condition, squaring both sides gives us:[2y+x-2=0]From the second condition, since the absolute value of a number is zero only when that number itself is zero, we have:[x+2=0]Solving these equations gives us:- From x+2=0, we find x=-2.- Substituting x=-2 into 2y+x-2=0 gives 2y-2-2=0 Rightarrow 2y=4 Rightarrow y=2.Therefore, the values of x and y are x=-2 and y=2. Part (2): Find the value of sqrt{3x^2-4y}+(frac{x}{y})^{2033}Given x=-2 and y=2, we substitute these values into the expression:[sqrt{3x^2-4y}+left(frac{x}{y}right)^{2033} = sqrt{3(-2)^2-4(2)}+left(frac{-2}{2}right)^{2033}]Simplifying the expression inside the square root:[= sqrt{12-8}+left(-1right)^{2033}]Since 12-8=4 and (-1)^{2033}=-1, we have:[= sqrt{4}-1 = 2-1 = 1]Thus, the value of the expression is boxed{1}.