z=6-2x²-y²和z=x² 2y²

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x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y

(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+y-z)(x-y+z)=

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

2x 6y z=7和3x 9y z=8求x y z=几解二愿以此方程,我知道x y

2x+6y+z=73x+9y+z=8两式相减x+3y=1x=1-3y两式相加5x+15y+2z=155-15y+15y+2z=15z=5x+y+z=1-3y+y+5=6-2y得不到x+y+z=5的结论

已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值

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

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

已知方程组{4x-3y-6z=0 x+2y-7z=0 且z≠0,则(x+y+3z)÷(4x-y-5z)=?结果和思路!

亲,题目看似很麻烦,仔细想还是有思路的.解:由4x-3y-6z=0,(1式)x+2y-7z=0(2式)(2式)*4得:4x+8y-28z=0(3式)(3式)-(1式)得:4x+8y-28z-(4x-3

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

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

x,y,z为实数且(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-

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-

已知,对于有理数x有|x-3|+|x+2|+|z+2|+|y+3|=13-|y-1|-|z+6|,求x+y+z的最大值和

这道题看似复杂,但是按步骤解就不难了.对于有绝对值的方程就一定要讨论了.x,y,z都有三种可能,x《-2,-2《x《3,x》3,y《-3,-3《y《1,y>1,z

解方程组{x(x+y+z)=6,y(x+y+z)=12,z(x+y+z)=18

x(x+y+z)=6(1)y(x+y+z)=12(2)z(x+y+z)=18(3)(1)/(2)x/y=1/2y=2x(1)/(3)x/z=1/3z=3xx(x+2x+3x)=66x^2=6x=1y=

若x-y=6,xy=-8,求代数式(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)的值

(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)&

X+Y+Z=0,X-Y-Z=0,Y=2,求x和z

0和-2再答:0和-2再问:说清是X等于几和z等干几

解方程组:5x+6y+2z=80 和 4x-3y+z=16 和 3x-2y+6z=92

5x+6y+2z=80和4x-3y+z=16两个方程化简求得13x+4z=1125x+6y+2z=80和3x-2y+6z=92两个方程化简求得14x+20z=356然后两个化简求得的方程求得x=4z=

X+Y+Z=?

X+Y+Z

分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)

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

已知x,y,z满足方程组{x+y-z=6,y+z-x=2,z+x-y=0,求x,

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

x/2=y/3=z/5 x+3y-z/x-3y+z

设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是:(x+3y-z)/(x-3y+z)?若是,如下:(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3

(x+y-z)^2-(x-y+z)^2=?

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