Question

若3x+2y=0,求[1-x/(x+y)+2y²/(x²-y²)]÷[1+2y/(x-y)]的值

Answer 1

最简单的 令x=2 y=-3
代入计算就可以了 [1-2/(2-3)+2*9/(4-9)]÷[1-6/(2+3)]=-3/5 ÷ (-1/5)=3

化简得结果也是可以的
[1-x/(x+y)+2y²/(x²-y²)] =((x²-y²)-x(x-y)+2y²)/(x²-y²)=y(x+y)/[(x+y)(x-y)]=y/(x-y)

1+2y/(x-y)=(x+y)/(x-y)

[1-x/(x+y)+2y²/(x²-y²)]÷[1+2y/(x-y)]= y/(x+y)

y=-1.5x

y/(x+y) = -1.5x /(x-1.5x)=3

Answer 2

由3x+2y=0可得y=-3x/2,x+y=-x/2,x-y=2x,x^2-y^2=-x^2,代入式中可得原式=3