x^2+xy=2 y^2+xy=7,则2(x-y)(x+y)=
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x^2+xy=2 y^2+xy=7,则2(x-y)(x+y)=
x^2+xy=2 y^2+xy=7,则2(x-y)(x+y)=
x^2+xy=2 y^2+xy=7,则2(x-y)(x+y)=
x^2+xy=2--------1式
y^2+xy=7---------2式
用1式减去2式有x^2-y^2=(x-y)(x+y)=-5
x^2+xy=2
y^2+xy=7
两式相减得
x^2-y^2=-5
(x+y)(x-y)=-5
所以原式=2(x-y)(x+y)=-10
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x*x+xy=3,xy+y*y=2,则x*x+2xy+y*y=________?
已知1/x-1/y=2,求3x+7xy-3y/2x-3xy-2y答案是1/x-1/y=2(y-x)/xy=2 得:x-y=-2xy(3x+7xy-3y)/(2x-3xy-2y)=[3(x-y)+7xy]/[2(x-y)-3xy]=(-6xy+7xy)/(-4xy-3xy)=xy/(-7xy)=-1/7 谁能帮我解释一下=[3(x-y)+7xy]/[2(x-y)-3xy]=(-6xy+7xy)/(-4xy-3xy) 是怎
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