寻找前世之旅同款项链:(x+2y-z+3a)(x-2y+z+3a)
来源:百度文库 编辑:高考问答 时间:2024/04/29 06:34:17
=[(x+3a)+(2y-z)][(x+3a)-(2y-z)]
=(x+3a)^2-(2y-z)^2
=x^2+6ax+9a^2-4y^2+4yz-z^2
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(x+2y-z+3a)(x-2y+z+3a)
=[(x+3a)+2y-z)]*[(x+3a)-(2y-z)]
=(x+3a)^2-(2y-z)^2
(x+2y-z+3a)(x-2y+z+3a)
2a(y-x)-3b(z-y) 分解因式
下列式子不能用平方差计算公式算的是A(x-y-z)(-x+y+z) B(-2x+3y)(3y+2x) C(x+y+z)(x-y+z) D(-5-X)(X-5
已知3x-z=x+y+z=4x+2y-z,求x : y : z
x =2y,3y=4z,则x:y:z =?:?:?
因式分解x^3(y^2-z^2)+y^3(z^2-x^2)+z^3(x^2-y^2)
设x:y:z=2:3:5,且满足3x-3y+z=a,求2x+3y-4z的值
2x+Y+3z=23,x+4y+5z=36,求X+2Y+3z~~~~
X+Y+Z=6 X-3Y+2Z=1 3X+2Y-Z=4
3x-y+z=10 x+2y-z=6 3x-y+z=10