z=6 - 2*x^2 - y^2

(y-z)^2+(z-x)^2+(x-y)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2[(y-z)^2-(y+z-2x)^2]+[(z-x)^2-(x+z-2y)^2]+[(

x+2y-z=6①x-y+2z=3②，①×2+②，得x+y=5，则y=5-x③，①+2×②，得x+z=4，则z=4-x④，把③④代入x2+y2+z2得，x2+（5-x）2+（4-x）2=3x2-18x

（1）2x+3y-4z=-5（2）x+y+z=6（两边同时×33x+3y+3z=18（与（1）相减得（5）（3）x-y+3z=10（与（2）相加得（4））（4）2x+4z=16（5）x+7z=23（两

{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

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

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

2x-y+2z=-17①3x+y-3z=-4②x+y+z=6③③*2：2x+2y+2z=12④④-①：3y=29y=29/3带入③：x+z=-11/3⑤带入②：3x-3z=-41/3即x-z=-41/

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

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

X+Y+Z=4,2*(X+Y+Z)+X=10,可以解出X=2.套入第二个和第一个.4+3Y-Z=66+2Y＋2Z＝10那么3Y=Z+2,2Y+2Z=4.Y=1,X=1X+Y+Z=4=2+1=12x+3

设a=x-y,b=y-z,-a-b=z-x(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方b^2+a^2+(-a-b)^2=(-a-b-

x+y+z=41式x+y+2z=52式3x+y-z=63式2-1式z=13-2式2x-3z=14式z=1代入4式x=2再代入1式y=1∴x=2,y=1,z=1请点击下面的【选为满意回答】按钮,再问：�

x+y+z=6(1)2x-y+z=3(2)3x+9y+z=24(3)(1)-(2)得：2y-x=3(4)(3)-(1)得：2x+8y=18即x+4y=9(5)(4)+(5)得：6y=12y=2代入(4

(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&

2x+y+z=2(1)x+2y+z=4(2)x+y+2z=6(3)(1)+(2)+(3)4x+4y+4z=12x+y+z=3(4)(1)-(4),x=-1(2)-(4),y=1(3)-(4),z=3

=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)

X+Y-Z=6.aY+Z-X=2.bZ+X-Y=0.ca,b,c三式相加X+Y+Z=8.dd式-a式2Z=2Z=1d式-b式2X=6X=3d式-C式2Y=8Y=4

括号是什么意思?只有一半

根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开：得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)合并同类项