已知xy+x-y+1=6,(1-x²)(1-y²)+4xy=48 求xy-x+y+1的值
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已知x+y=6,xy=1,且x
已知1/x=1/y=6,则3x-2xy+3y/3xy+x+y=?已知1/x+1/y=6,则3x-2xy+3y/3xy+x+y=?
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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
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已知xy满足x>=1,x+y=