Appearance
question:## Zadatak B-2.3.Neka jex=frac{2}{(sqrt{3}+1)(sqrt[4]{3}+1)(sqrt[8]{3}+1)(sqrt[16]{3}+1)}Odredite (x+1)^{64}.
answer:## Rješenje.Pomnožimo brojnik i nazivnik s sqrt[16]{3}-1.begin{aligned}x & =frac{2(sqrt[16]{3}-1)}{(sqrt{3}+1)(sqrt[4]{3}+1)(sqrt[8]{3}+1)(sqrt[16]{3}+1)(sqrt[16]{3}-1)}= & =frac{2(sqrt[16]{3}-1)}{(sqrt{3}+1)(sqrt[4]{3}+1)(sqrt[8]{3}+1)(sqrt[8]{3}-1)}= & =frac{2(sqrt[16]{3}-1)}{(sqrt{3}+1)(sqrt[4]{3}+1)(sqrt[4]{3}-1)}= & =frac{2(sqrt[16]{3}-1)}{(sqrt{3}+1)(sqrt{3}-1)}= & =frac{2(sqrt[16]{3}-1)}{3-1} & =sqrt[16]{3}-1end{aligned}Tada je (x+1)^{64}=(sqrt[16]{3}-1+1)^{64}=(sqrt[16]{3})^{64}=3^{4}=81. 2 boda
question:Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.text{(A)} sqrt 2 qquad text{(B)} sqrt 3 qquad text{(C)} sqrt{frac{3}{2}} qquad text{(D)} frac{2}{sqrt{3}} qquad text{(E)} 2
answer:We assume the side length of the cube is 1. The side length of the tetrahedron is sqrt2, so the surface area is 4timesfrac{2sqrt3}{4}=2sqrt3. The surface area of the cube is 6times1times1=6, so the ratio of the surface area of the cube to the surface area of the tetrahedron is frac{6}{2sqrt3}=boxed{sqrt3}.-aopspandy
question:4. In the plane, there are 1001 points. Three of them are vertices of a triangle in which the remaining 998 points are located. All points are arranged such that no three of them are collinear. How many non-overlapping triangles form the given triangle?
answer:Solution. Let A, B, C be the vertices of the triangle in which the remaining 998 points are located, and let point M be one of the interior points. Point M is a vertex of a certain number of non-overlapping triangles, so the sum of the interior angles of all these triangles is 360^{circ}. We have 998 points like point M, so the sum of the interior angles of the triangles whose vertices are these points is 998 cdot 360^{circ}. If we add 180^{circ} to this sum, we get the sum of the interior angles of all non-overlapping triangles whose vertices are the given 1001 points. Therefore, the number of all triangles isfrac{998 cdot 360 + 180^{circ}}{180^{circ}} = 1997
question:The function f(x) = begin{cases} 2, & text{if } x < 1 x^{2} + ax, & text{if } x geq 1 end{cases}, if f(f(0)) = 4a, then the value of the real number a is ( )A: 0 B: 1 C: 2 D: 4
answer:For the function f(x) = begin{cases} 2, & text{if } x < 1 x^{2} + ax, & text{if } x geq 1 end{cases}, we have f(0) = 2, Since f(f(0)) = 4a, we get f(2) = 4a, which means 2^2 + 2a = 4a, solving this gives a = 2. Therefore, the correct choice is boxed{C}. By setting up equations using the piecewise function and solving them, we can find the solution. This problem tests the application of piecewise functions, how to find the value of a function, and computational skills.
question:3. It is known that for some positive coprime numbers m and n, the numbers m+1947 n and n+1947 m have a common prime divisor d>9. Find the smallest possible value of the number d under these conditions.
answer:Answer: d_{min }=139.For example,begin{aligned}& m=138, n=1 rightarrow 1947 m+n=1947 cdot 138+1=139 cdot 1933 & m+1947 n=138+1947=139 cdot 15end{aligned}
question:12. Let the equation x y=6(x+y) have all positive integer solutions left(x_{1}, y_{1}right),left(x_{2}, y_{2}right), cdots,left(x_{n}, y_{n}right), then sum_{k=1}^{n}left(x_{k}+y_{k}right)=qquad .
answer:x y-6(x+y)+36=36 Rightarrow(x-6)(y-6)=1 cdot 36=2 cdot 18=3 cdot 12=4 cdot 9=6 cdot 6 text {, }Then (x, y)=(7,42),(8,24),(9,18),(10,15),(12,12),(15,10),(18,9),(24,8),(42,7). So sum_{k=1}^{n}left(x_{k}+y_{k}right)=2(49+32+27+25)+24=290.