Question

已知x/y=1/2 ，求x²+2xy+y²/x²-xy-2y²的值

Answer 1

解：
因为x/y=1/2
所以y=2x
∴x²+2xy+y²/x²-xy-2y²
=(x+y)²/(x-2y)(x+y)
=(x+y)/(x-2y)
=(x+2x)/(x-4x)
=3x/(-3x)
=-1

Answer 2

你好，如果/前是一个整体即（x²+2xy+y²）/（x²-xy-2y²）的话就很简单，把Y=2X带入然后上下同时除以X平方得 -1
如果不是的话则.....

Answer 3

x/y=1/2，则y=2x
x²+2xy+y²/x²-xy-2y²
=(x+y)²/(x+y)(x-2y)
=(x+y)/(x-2y)
=(x+2x)/(x-4x)
=3x/(-3x)
=-1

追问

x²+2xy+y²/x²-xy-2y²
=(x+y)²/(x+y)(x-2y)
这步是如何得来的？

Answer 4

分子分母同时除以y^2,
分子变为：（x/y）^2+2*x/y+1;
分母变为：（x/y）^2-x/y-2;
将 x/y=1/2代入得：
原式=-1 .

Answer 5

(x²+2xy+y²)/(x²-xy-2y²)
=(x+y)²/(x-2y)(x+y)
=(x+y)/(x-2y)
=(x/y+y/y)/(x/y-2y/y)
=(x/y+1)/(x/y-2)
=(1/2+1)/(1/2-2)
=(3/2)/(-3/2)
=-1