x+y+z=4 x-2y+z=-2 x+2y+3z=0求解用2元一次方程消元法求解
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 23:47:55
x+y+z=4 x-2y+z=-2 x+2y+3z=0求解用2元一次方程消元法求解
x+y+z=4 x-2y+z=-2 x+2y+3z=0求解
用2元一次方程消元法求解
x+y+z=4 x-2y+z=-2 x+2y+3z=0求解用2元一次方程消元法求解
x+y+z=4 (1)
x-2y+z=-2 (2)
x+2y+3z=0 (3)
(1)-(2)
3y=6
y=2
代入(1),(3)
x+z=2 (4)
x+3z=-6 (5)
(4)-(5)
-2z=8
z=-4
x=2-z=6
所以
x=6,y=2,z=-4
x+2y=3x+2z=4y+z 求x:y:z
x+y/2=z+x/3=y+z/4 x+y+z=27
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
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?
(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)
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z
已知:x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值.
x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值
2x+z=10 x+y-z=4 3x-y-z=0
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
x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,
若x,y,z成等差数列,则(z-x)^2-4(x-y)(y-x)=
(4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=?
(x+y-z)^2-(x-y+z)^2=?