2-[-3(-3/1-3/2xy)-2y²]-2(x²-xy+2y²) x=2/1 y=-1是不是-3又2/1
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2-[-3(-3/1-3/2xy)-2y²]-2(x²-xy+2y²)x=2/1y=-1是不是-3又2/12-[-3(-3/1-3/2xy)-2y²]-2(x&
2-[-3(-3/1-3/2xy)-2y²]-2(x²-xy+2y²) x=2/1 y=-1是不是-3又2/1
2-[-3(-3/1-3/2xy)-2y²]-2(x²-xy+2y²) x=2/1 y=-1
是不是-3又2/1
2-[-3(-3/1-3/2xy)-2y²]-2(x²-xy+2y²) x=2/1 y=-1是不是-3又2/1
1-2x-2y
1/2-2=-3/2
好像不对
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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