(x+y-1)(x-y-1)-(x-y)的平方答案是多少
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(x+y-1)(x-y-1)-(x-y)的平方答案是多少
(x+y-1)(x-y-1)-(x-y)的平方答案是多少
(x+y-1)(x-y-1)-(x-y)的平方答案是多少
原式=[(x-1)+y]*[(x-1)-y]-(x-y)²
=(x-1)²-y²-(x-y)²
=x²-2x+1-y²-x²+2xy-y²
=-2y²+2xy-2x+1
好简单。。。
(x+y)(x-y)+4(y-1)
[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y
y'/y=1/x
计算:(1+ y/x-y )·(1- x/x+y)
感激不尽1.x,y满足(x-y-1)(x+y)
(1-y+x/y)除以y平方-x平方/x
证明:(y/x)+(x/y)+xy≥x+y+1
x+y分之1-(x+y)的平方x-y
1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z)
1/(x-y)/(1/y-/x)
因式分解:(1+x+y)(1-x-y)
(x+5y)(x-7y-1)
(x+y)²-4(x+y-1)
(x-y)(2x-y-1)
(x+y)(x+y-1)-12因式分解
因式分解:(x+y)-4(x+y-1)
y'-(1/x)y=xe^-x
{(x,y) |x|+|y|