(X^2-Y^2+Z^2)^2-4X^2Z^2
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 06:44:05
(X^2-Y^2+Z^2)^2-4X^2Z^2(X^2-Y^2+Z^2)^2-4X^2Z^2(X^2-Y^2+Z^2)^2-4X^2Z^2原式=[(x^2-2xz+z^2)-y^2][(x^2+2xz
(X^2-Y^2+Z^2)^2-4X^2Z^2
(X^2-Y^2+Z^2)^2-4X^2Z^2
(X^2-Y^2+Z^2)^2-4X^2Z^2
原式
=[(x^2-2xz+z^2)-y^2][(x^2+2xz+y^2)-y^2]
=[(x-z)^2-y^2][(x+z)^2-y^2]
=(x-z+y)(x-z-y)(x+y+z)(x-y+z)
狂用平方差公式
=x^4+y^4+z^4-2x^2y^2-2y^2z^2+2x^2z^2-4x^2z^2
=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2x^2z^2
这里是既约的吧?
x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2倒可以分:
(X^2+Y^2+Z^2)^2
=(x-z+y)(x-z-y)(x+y+z)(x-y+z)
化简-(-3x-y+2z)-{-x+【4x-(x-y-z)-3x+2z}
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-y)^2+4z(x-y)+4z^2
计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y)
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=3x+2z=4y+z 求x:y:z
x+y/2=z+x/3=y+z/4 x+y+z=27
数学--整式运算4(x-y+z)-2(x+y-z)-3(-x-y-z)
若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-z)=
(x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z
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
(X+Z)(X-Z)+x(X+2Y)怎么解?
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 *