(2x+3y-z)(2x-3y+z) 平方差公式
2个回答
(2x+3y-z)(2x-3y+z)
=[2x+(3y-z)][(2x-(3y-z)]
=4x^2-(3y-z)^2
=4x^2-(9y^2-6yz+z^2)
=4x^2-9y^2+6yz-z^2
相关问题
(x+2y+3z)(x-2y-3z)用平方差公式
证明:(y+z-2x)3+(z+x-2y)3+(x+y-2z)3=3(y+z-2x)(z+x-2y)(x+y-2z).
(-x+y-z)(x+y+z)等于什么(用平方差公式)
x^2(y+z)+y^2(z+x)+z^2(x+y)-(x^3+y^3+z^3)-2xyz
化简 [x^3(y^2-z^2)+y^3(z^2-x^2)+z^3(x^2-y^2)]/[x^3(y-z)+y^2(z-
因式分解X2(Y+Z)+Y2(Z+X)+Z2(X+Y)-(X3+Y3+Z3)-2XYZ
设x,y,z为正数,证明:2(x^3+y^3+z^3)≥(x^2)(y+z)+(y^2)(x+z)+(z^2)(x+y)
一.{x+2y+3z=14 2x-y-z=-3 3x+y+z=8 二.{3X-2Y-2Z=3 2X+2Y+3Z=12 X
x+2z=3 x+y+z=2 2x+y=2 3x-y-4z=5 2y+z=7 2x+3y-2z=0
x-5y-3z=0 y+z-3x=3 x+y+3z=14 y+4z=3 z+x-3y=5 2x+y+z=7 2x-z=1