Question

初一数学题分式运算.

已知3x^2+xy-2y^2=0，求[（x+y)/(x-y) + （4xy)/(y^2-x^2) ]÷(x^2+2xy-3y^2)/(x^2-9y^2)的值。

Answer 1

3x²+xy-2xy=(x+y)(3x-2y)=0
x+y=0或3x-2y=0

原式=[(x+y)/(x-y)-4xy/(x+y)(x-y)]÷[(x-y)(x+3y)/(x+3y)(x-3y)]
={[(x+y)²-4xy]/(x+y)(x-y)}÷[(x-y)/(x-3y)]
=[(x²-2xy+y²)/(x+y)(x-y)]×[(x-3y)/(x-y)]
=[(x-y)²/(x+y)(x-y)]×[(x-3y)/(x-y)]
=[(x-y)/(x+y)]×[(x-3y)/(x-y)]
=(x-3y)/(x+y)
x+y在分母，不等于0
所以3x-2y=0
y=3x/2
所以原式=(x-9x/2)/(x+3x/2)
=(-7x/2)/(5x/2)
=-7/5