约分 x²-(y-z)²除以(x+y)²-z² 这是个分式!
1个回答
原式=[(x-y+z)(x+y-z)]/[(x+y-z)(x+y+z)] 注:根据a²-b²=(a-b)(a+b)
=(x-y+z)/(x+y+z)
相关问题
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.
约分,(x+y)²-z² 分之 x² -(y-z)²
约分:(x+y)² -z² 分之 x²-(y-z)²
x:y:z=3:4:6,求分式x+y-z/x-y+z
X除以Y等于2,X除以Z等于4,Z等于1,求代数式(X+Y+Z)除以(X-Y+Z)的值
约分x^2-(y-z)^2/(x+y)^2-z^2
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)/(
已知(x+y)/z=(x+z)/y=(y+z)/x=2,且xyz≠0,则分式(x+y)(x+z)(y+z)/xyz的值为
分式方程4x*x---- =y1+4x*x4y*y---- =z1+4y*y4z*z---- =x1+4z*z----表
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-