若3x+2y=0,求[1-x/(x+y)+2y²/(x²-y²)]÷[1+2y/(x-y)]的值
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最简单的 令x=2 y=-3
代入计算就可以了 [1-2/(2-3)+2*9/(4-9)]÷[1-6/(2+3)]=-3/5 ÷ (-1/5)=3
化简得结果也是可以的
[1-x/(x+y)+2y²/(x²-y²)] =((x²-y²)-x(x-y)+2y²)/(x²-y²)=y(x+y)/[(x+y)(x-y)]=y/(x-y)
1+2y/(x-y)=(x+y)/(x-y)
[1-x/(x+y)+2y²/(x²-y²)]÷[1+2y/(x-y)]= y/(x+y)
y=-1.5x
y/(x+y) = -1.5x /(x-1.5x)=3
代入计算就可以了 [1-2/(2-3)+2*9/(4-9)]÷[1-6/(2+3)]=-3/5 ÷ (-1/5)=3
化简得结果也是可以的
[1-x/(x+y)+2y²/(x²-y²)] =((x²-y²)-x(x-y)+2y²)/(x²-y²)=y(x+y)/[(x+y)(x-y)]=y/(x-y)
1+2y/(x-y)=(x+y)/(x-y)
[1-x/(x+y)+2y²/(x²-y²)]÷[1+2y/(x-y)]= y/(x+y)
y=-1.5x
y/(x+y) = -1.5x /(x-1.5x)=3
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