x+y+z=10 x+y+z=4 x+2y+z=7 解三元一次方程
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x+y+z=10 x+y+z=4 x+2y+z=7 解三元一次方程
x+y+z=10 x+y+z=4 x+2y+z=7 解三元一次方程
x+y+z=10 x+y+z=4 x+2y+z=7 解三元一次方程
你这题目抄错了吧,第一式和第二式就矛盾
题目抄错了吧?x+y+z=10 x+y+z=4怎么可能?x-y+z=10 x+y+z=4 4x+2y+z=7x-y+z=10..... (1) x+y+z=4...... (2) 4x+2y+z=7.. (3) (2)-(1)得 2y=-6 y=-3 (3)-(2)得 3x+y=3..... (4) 将y=-3代入(4)式,得: 3x-3=3 x=2 将x=...
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题目抄错了吧?x+y+z=10 x+y+z=4怎么可能?
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第一式和第二式有矛盾。无解。
2x+z=10 x+y-z=4 3x-y-z=0
(4x-2y-z)-{5x-[8y-2z-(x-2y)]-x-(3y-10z)}=?
(x+y-z)(x-y+z)=
x+y/2=z+x/3=y+z/4 x+y+z=27
x+2y=3x+2z=4y+z 求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
(4x-2y-z)-{5x-[8y-2z-(x+y)]-2(3y-10z)}=?好难啊
{2x+y=10 {x-y+z=4 {3x-y-z=0
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)
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian
已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,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^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
X=3z X+y+z=12 X+2y-2z=10
x+2y+3z=10,x-y+4z=10,x+3y+2z=2