x+2y=0(x≠0),求分式x²-xy分之2xy+y²
问题描述:
x+2y=0(x≠0),求分式x²-xy分之2xy+y²
答
x=-2yx²-xy分之2xy+y²=【2(-2y)y+y²】/【(-2y)²-(-2y)y】=-3y²/6y²=-0.5
x+2y=0(x≠0),求分式x²-xy分之2xy+y²
x=-2yx²-xy分之2xy+y²=【2(-2y)y+y²】/【(-2y)²-(-2y)y】=-3y²/6y²=-0.5