分式加减 (x-y)(z-y)/x+y - (y-x)(y-z)/x+z 不要跳步骤,

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分式加减 (x-y)(z-y)/x+y - (y-x)(y-z)/x+z 不要跳步骤,
分式加减 (x-y)(z-y)/x+y - (y-x)(y-z)/x+z 不要跳步骤,

分式加减 (x-y)(z-y)/x+y - (y-x)(y-z)/x+z 不要跳步骤,
(x-y)(z-y)/x+y - (y-x)(y-z)/x+z
=(x-y)(z-y)(x+z)/(x+y)(x+z) - (y-x)(y-z)(x+y)/(x+z)(x+y)
=[(x-y)(z-y)(x+z) - (y-x)(y-z)(x+y)]/(x+z)(x+y)
={(x-y)(z-y)(x+z) -[- (y-x)][-(y-z)](x+y)}/(x+z)(x+y)
=[(x-y)(z-y)(x+z) -(x-y)(z-y)(x+y)]/(x+z)(x+y)
=(x-y)(z-y)[(x+z) -(x+y)]/(x+z)(x+y)
=(x-y)(z-y)(x+z -x-y)/(x+z)(x+y)
=(x-y)(z-y)(z -y)/(x+z)(x+y)
=(x-y)(z-y)²/(x+z)(x+y)

(x-y)(z-y)/x+y - (y-x)(y-z)/x+z
=(x-y)(z-y)/x+y - (x-y)(z-y)/x+z
=(x-y)(z-y)(1/x+y-1/x+z)
=(x-y)(z-y)〔x+z-x-y/(x+y)(x+z)〕
=(x-y)(z-y)〔z-y/(x+y)(x+z)〕
=(x-y)(z-y)²/(x+y)(x+z)

原式=（x+y)/(x-y)(z-y)-(x+z)/{[-(x-y)][-(z-y)]
=(x+y)/(x-y)(z-y)-(x+z)/(x-y)(z-y)
=[(x+y)-(x+z)]/(x-y)(z-y)
=(x+y-x-z)/(x-y)(z-y)
=(y-z)/(x-y)(z-y)
=-1/(x-y)
=1/(y-x)

分式加减 (x-y)(z-y)/x+y - (y-x)(y-z)/x+z 不要跳步骤, 分式加减的几道题目求速度1 .(x+y)²/（x-z)(y-z) - (y+z)²/（x-z)(y-z) = 2.2x+3y/x-y+2y/y-x-3y/x-y= 化简:分式加减化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-y-x)/(z-y)(z-x) 分式约分:(y+z-x)/{x^2-(y+z)^2} x-y分之2减去（x-y）的平方分之x+y 分式加减咋做啊? x-y分之2减去(x-y)的平方分之x+y 分式的加减, 分式加减法：已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z 分式的加减:[(x/y-y/x)除以（x+y）+x（1/y-1/x)]除以1+x/y,其中x=5,y=2 初二分式难题,利用初二知识解决,已知（x+y)/z=(x+z)/y=(y+z)/x,求（x+y-z)/(x+y+z) 已知 ,x+y/z=x+z/y=y+z/x,则分式(x+y)(y+z)(z+x)/xyz的值 x-2y/x+y 减 x-y/x+y 加 -x-4y/x+4y初二上学期分式的加减. 用加减消元法解方程组x+y=10 y+z=22 z+x=18 xy/x-z-xy/y-z 分式化简分式加减计算5x／y-3＋2x／3-y 急分式加减计算5x／y-3＋2x／3-y 分式的加减(1/x+1/y)^2-(1/x-1/y)^2 y/x²-xy+x/y²-xy 分式的加减, 已知:(x+y)/z=(x+z)/y=(z+y)/x,且xyz不等于0,则分式(x+y)(x+z)(z+x)/xyz的值为?