x^2 2y^2 3z^2=12

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x^2 2y^2 3z^2=12
3x+2y+z=13 x+y+2z=7 2x+3y-z=12 x=?y=?z=?

3x+2y+z=133x+3y+6z=21得y+5z=812x+2y+4z=142x+3y-z=12得y-5z=-22由1.2得y=3z=1x=2

解方程组{3x+y-z=4,2x-y+3z=12,x+y+z=6}

{3x+y-z=4①,2x-y+3z=12②,x+y+z=6③}①+②得5x+2z=16④,②+③得3x+4z=18⑤④×2—⑤得7x=14,x=2所以z=3、y=1所以方程组的解为x=2、y=1、z

{3x+2y+z=13 x+y+2z=7 2x+3y-z=12,求x,y,z

答案为x=2,y=3,z=1;解答过程为:1式与3式相加可得5x+5y=25,算出x+y=5,代入2式得z=1,再把z=1代入可得x=2,y=3;要采纳哦!

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)

有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y

①x+y+z=6 3x-y+2z=12 x-y-3z=-4 ②x+y-z=2 4x-2y+3y+8=0 x+3y-2z-

(1)x+y+z=6①3x-y+2z=12②x-y-3z=-4③①+②4x+3z=18④②-③2x+5z=16⑤⑤×24x+10z=32⑥⑥-④7z=14解得z=2代入⑤2x+5×2=16解得x=3将

(1)y+3x+17=0 5y-2x+17=0 (2)x+y+z=12 x+2y+5z=22 x=4y (3)x+y+z

(1)方程1*5得5y+15x+85=0(3)(3)-(2)17x+68=0所以x=-4将x=-4代入1得y=-5(2x+y+z=12(1)x+2y+5z=22(2)x=4y(3)(3)分别代入(1)

如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y,

x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1

解下列方程:x-4y=0,x+2y+5z=22,x+y+z=12

x-4y=0①,x+2y+5z=22②,x+y+z=12③由①得:x=4y分别代入②③整理得:6y+5z=22④z=12-5y⑤把⑤代入④得:6y+60-25y=22,解得y=2把y=2代入⑤得:z=

1.x+y=16,y+z=12,z+x=102.3x-y+z=4,2x+3y-z=12,x+y+z=63.x+y+z=6

1.x+y=16①y+z=12②z+x=10③①-②x-z=4④③+④2x=14x=7⑤⑤代入①y=9⑥⑥代入②z=3x=7,y=9,z=3(2)3x-y+z=4①2x+3y-z=12②x+y+z=6

已知{x:y:z=1:2:3,x+y+z=12,求x、y、z的值

x:y:z=1:2:3,x=k,y=2k,z=3kx+y+z=k+2k+3k=6k=12k=2x=2,y=4,z=6

解方程{3x -y+z=4 2x+3y-z=12 x+y+z=6

x=2y=3z=13x-y+z=42x+3y-z=12消z得5x+2y=162x+3y-z=12x+y+z=6消z得3x+4y=185x+2y=163x+4y=18合并得x=2y=3代入x+y+z=6

3x-y+z=3 2x+y-3z=11 x+y+z=12 解方程

3x-y+z=3(1)2x+y-3z=11(2)x+y+z=12(3)(1)+(2)5x-2z=14(4)(1)+(3)4x+2z=15(5)(4)+(5)9x=29所以x=29/9z=(5x-14)

已知x+y/2=y+z/3=z+x/4,且x+2y+z=12,求x-2y+z的值

设(x+y)/2=(y+z)/3=(z+x)/4=kx+y=2ky+z=3kx+z=4k1式+3式2x+y+z=6k2x+3k=6k2x=3kx=3k/2代入1式得y=k/2代入3式得z=5k/2∵x

1.x+y+z=21,x-y=1,2x+z-y=13.2.3x+2y+z=13,x+y+2z=7 ,2z+3y-z=12

1.x=10,y=9,z=22.x=3,y=2,z=13.x=30,y=20,z=16.

已知x,y,z满足方程组x+2y-z=21 x-y+2z=12

x+2y-z=21①x-y+2z=12②①*2+②=3x+3y=54即x+y=18得出y=18-x代入②得x+z=15得出z=15-x代入186/x²+y²+z²得出18

x+2y+3z=12x+3y+z=23x+y+2z=3

x+2y+3z=1            ①2x+3y+z=2 &nb

x=y/z=z/3,x+y+z =12,求2x+3y+4z是多少,

3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.

解三元一次方程组 2x-2y+z=-9 3x+2y-2z=23 x+y+z=12

2x-2y+z=-9(1)3x+2y-2z=23(2)x+y+z=12(3)(1)×2+(2)7x-2y=5(4)(1)-(2)x-3y=-21(5)(4)×3-(5)×221x-2x=15+4219

已知x+y-7z=0 x-2y+5z=0(xyz不等于0),求x+2y-z/y-2x+12z

x+y-7z=0①x-2y+5z=0②①-②得:3y-12z=0,即y=4z,2①+②得:3x-9z=0,即x=3z所以x+2y-z/y-2x+12z=11/10,再来题难点的.

已知x+y/2=y+z/2=x+z/4,且x+2y+z=12,求x-2y+z

由(x+y)/2=(y+z)/2得x=z由x=z,(x+y)/2=(x+z/)4得y=0由x+2y+z=2x+0=12得2x=12则,x-2y+z=2x-0=12感觉你题目打错了,请检查一下