x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/19 01:27:47
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyzx＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyzx＾2(y+z)+y
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz
=x^2(y+z)+y^2*z+y^2*x+z^2*x+z^2*y-x^3-y^3-z^3-2xyz
=x^2(y+z-x)+x(y^2+z^2-2yz)+(y^2*z+z^2*y-y^3-z^3)
=x^2(y+z-x)+x(y-z)^2+[y^2(z-y)-z^2(z-y)]
=x^2(y+z-x)+x(y-z)^2+(y+z)(y-z)(z-y)
=x^2(y+z-x)+x(y-z)^2-(y+z)(y-z)^2
=x^2(y+z-x)+(y-z)^2(x-y-z)
=(x-y-z)(z-y-x)(z-y+x)
哇楼上太耐心了。。我觉得这种题写步骤出来特别麻烦~
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
（x-2y+z）（-x+2y+z）
x+2y-z）（x+2y-z）
（x+2y-z）(x-2y+z)
计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答
化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z)
计算题(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z).
1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z)
(x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之（x-z)(y-z)=?第三部分那个是 （y+z-2x)(x-2y+z)分之（x-z)(y-z)
用行列式的性质证明：y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证?
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z）+z/(x+y+2z)
因式分解：(x+y-2z)立方+(y+z-2x)立方+(z+x-2z)立方
x＾2(y+z)+y＾2（z+x）+z＾2(x+y）-(x＾3+y＾3+z＾3）-2xyz
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)