x-y-z=-1 3x 5y-7z=11 4x-y 2z=-1

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/30 15:39:56
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

已知x+y-z/z=x-y+z/y=-x+y+z/x,且xyz不等于0,求分式[(x+y)(x+z)(y+z)]/xyz

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z

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

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

{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

三元一次 5X-3Y+4Z=13 2X+7Y-3Z=19 3X+2Y-Z=18

5X-3Y+4Z=13(1)2X+7Y-3Z=19(2)3X+2Y-Z=18(3)(1)+(3)*417X+5Y=85(4)(3)*3-(2)7X-Y=35(5)(4)+(5)*552X=260X=5

若z分之x+y+z=y分之x-y+z=x分之-x+y+z,求xyz分之(x+y)(y+z)(z+x)

设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,

解下列方程组(1)3x-y+z=3 2x+y-3z=11 x+y+z=12(2)5x-4y+4z=13 2x+7y-3z

第一题3x-y+z=3①;2x+y-3z=11②;x+y+z=12③;①减③2X-2Y=-9④;3倍的③即3X+3Y+3Z=36⑤;⑤加②5X+4Y=47⑥;2倍的④4X-4Y=-18⑦;⑥式加⑦9X

解方程 3x+2y+z=13 x+5y+2z=7 2x+3y-z=12

x=4y=1z=-1需要详细步骤么?再问:要再答:3x+2y+z=13(1)x+5y+2z=7(2)2x+3y-z=12(3)(1)+(3)=5x+5y=25(2)+2(3)=x+5y+2z=4x+6

y+z÷x=Z+X÷y=X+Y÷z,X+Y+Z不等0求X+Y-Z÷X+Y+z值

∵y+z÷x=Z+X÷y=X+Y÷z容易发现x,y,z位置互换也成立∴式子与x,y,z值无关∴x=y=z∴(X+Y-Z)÷(X+Y+z)=x/3x=1/3明教为您解答,请点击[满意答案];如若您有不满

3x+2y=z=13 x+y+2z=7 2x+2y-z=12三元一次方程组的解

3x+2y+z=13①x+y+2z=7②2x+2y-z=12③①-3*②得-y-5z=-8④2*①-3*③得-2y+5z=-10⑤2*④-⑤得-15z=-6z=2/5代入z到④,得y=6代入y和z到①

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz

设:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y=k{x+y-z=kz(1){y+z-x=kx(2){z+x-y=ky(3)(1)+(2)+(3)得:(x+y+z)=k(x+y+z)(x

x分之y+z=y分之z+x=z分之x+y(x+y+z不等于0),求x+y+z分之x+y-z

令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0

2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X

2x+y+z=10(1)x+2y+z=-6(2)x+y+z=8(3)(2)-(3):y=-14(4)(1)-(2):x-y=16(5)把(4)代入(5):x+14=16x=2(6)把(4)和(6)代入

已知:(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)

设x+y-z/z=x-y+z/y=y+z-x/x=k有x+y-z=kzx-y+z=kyy+z-x=kx三式相加得x+y+z=k(x+y+z)k=1得x+y=(k+1)zx+z=(k+1)yy+z=(k

5x+2y=5,y-z=-7,4z+3x=13

5x+2y=5,(1){y-z=-7,(2)4z+3x=13(3)解方程组:(1)-(2)*2得:5x+2z=19(4)(4)*2-(3)得:7x=25x=25/7把x=25/7代入(1)得y=-45

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.

{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2

这个题目没有问题么,我是说最后一个式子确定是z+5y+8z=-2?如果没有问题的话:x+y+z=1;①x+3y+7z=-1;②z+5y+8z=-2③①-②2Y+6Z=-2Y=(-2-6Z)/2=-1-

X+Y+Z=?

X+Y+Z

f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)

f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)