x*x+y*y+2z*z-2x+4y+4z+7=0,求xyz的值用公式法
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 10:55:35
x)ЪЮԪ6Ҫ56665y
{OYu=6OI*'RX)^;[
JD
55@@F@@@a:]C8#"m#
T
JWTTBD*ll@ s
x*x+y*y+2z*z-2x+4y+4z+7=0,求xyz的值用公式法
x*x+y*y+2z*z-2x+4y+4z+7=0,求xyz的值
用公式法
x*x+y*y+2z*z-2x+4y+4z+7=0,求xyz的值用公式法
x*x+y*y+2z*z-2x+4y+4z+7=0
(x*x-2x+1)+(y*y+4y+4)+2(z*z+2z+1)=0
(x-1)^2+(y+2)^2+2(z+1)^2=0
x=1,y=-2,z=-1
xyz=2
(x-y)^2+4z(x-y)+4z^2
化简-(-3x-y+2z)-{-x+【4x-(x-y-z)-3x+2z}
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
因式分解Z^2(x-y)-4(x-y)-3Z(y-z)
求化简(x+3y+2z)(x-3y+6z)(x+3y+4z-2z)(x-3y+4z+2z)
已知整数x、y、z,满足x≤y<z,且|x+y|+|y+z|+|z+x|=4,|x-y|+|y-z|+|z-x|=2,求x^2+y^2+z^2的值.
整数x、y、z满足x≤y<z|x+y|+|y+z|+|z+x|=4|x-y|+|y-z|+|z-x|=2求x²+y²+z²
已知整数x,y,z满足x≤y<z,且|x+y|+|y+z|+|z+x|=4 |x-y|+|y-z|+|z-x|=2 那么x²+y²+z²的值
若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-x)=
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)
(4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=?
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).