y-3x-2z=1 x 2y 3z=9 5x-7y z=34

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y-3x-2z=1 x 2y 3z=9 5x-7y z=34
三元一次方程组数学题x+2y+2z=33x+y-2z=72x+3y-2z=10x-y=2z-x=3y+z=-1x-y-z

1.x=1,y=2,z=-12.x=-1,y=-3,z=23.a=-5/2,b=7/2,c=2其他的我也不说了,慢慢想吧~

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

{2x-2y+z=0,2x+y-z+1,x+3y-2z=1

①2x-2y+z=0②2x+y-z=1③x+3y-2z=1你可以2式-3式,得出4式x-2y+z=01式=4式,得出x=o,代入1式,得出2y=z继续代入2式或者3式,得出Y=-1,Z=-2

2x-2y+z=0 2x+y-z=1 x+3y-2z=1

①+②4x-y=12②-③3x-y=1有上得x=0y=-1代入②得z=-2

x+y+z=2 4x+2y+z=4 2x+3y+z=1

x+y+z=2(1)4x+2y+z=4(2)2x+3y+z=1(3)(2)-(1)3x+y=2(4)(2)-(3)2x+y=3(5)(4)-(5)所以x=-1y=3-2x=5z=2-x-y=-2

x-y-z=-1 3x+5y+7z=11 4x-y+2z=-1 分别求出x=?y=?z=?

x-y-z=-1(1)3x+5y+7z=11(2)4x-y+2z=-1(3)(1)*2+(3)得6x-3y=-32x-y=-1(4)所以2x-y=4x-y+2z=-1x+z=0代入(2)有5y+4z=

已知x:y:z=1:2:3,x+y+z=24,求x,y,z

解∵x:y:z=1:2:3∴x=k,y=2k,z=3k∵x+y+z=24∴k+2k+3k=24即6k=24∴k=4∴x=4.y=8,z=12

3x+2y+z=14 x+y+2z+-3 2x+3y-z=1

(1)十(3)消去了Z得x十y=3代入(2)得知z的值,(2)有问题.

x+y+z=2 x-3y+2z=1 2x+2y+z+5 要具体步骤..

x+y+z=2(1)x-3y+2z=1(2)2x+2y+z=5(3)(1)×2-(2)得:x+5y=3(4)(3)-(1)得:x+y=3(5)(4)-(5)得:y=0代入(5)得:x=3代入(1)得:

3道高数题,1,函数F(x,y,z)=(e^x) * y * (z^2) ,其中z=z(x,y)是由x+y+z+xyz=

1、隐函数对x求导得1+az/ax+yz+xy*az/ax=0,故az/ax=-(1+yz)/(1+xy);F对x求导得aF/ax=e^x*y*z^2+e^x*y*2z*az/ax;当x=0,y=1时

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

如果,根号x-3+| y-2 |+z^2=2z-1 求 (x+z)^y

根号x-3+|y-2|+z^2=2z-1根号x-3+|y-2|+(z^2-2z+1)=0根号x-3+|y-2|+(z-1)^2=0由于数值开根号,绝对值和平方数均为大于等于0的数则上式要成立只有X-3

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+2y=3 x+y+z=36 2x+y+z=15 2y=3z x-y=1 x+2y+z x-z=-1 2x+z-y=1

x+2y=32y=3zx-y=-1x+2y=3①2y=3z②x-y=-1③①-③得3y=4,得y=4/3代入③,得x=y-1=1/3代入②,得z=2/3y=8/9x+y+z=36x-y=12x+z-y

{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-

解方程组:x-2y+z=-1,x+y+z=2,x+2y+3z=-1

x-2y+z=-1①x+y+z=2②x+2y+3z=-1③①+③2x+4z=-2x+2z=-1④①+②×23x+3z=-1+4x+z=1⑤由⑤得x=1-z代入④1-z+2z=-1z=-2∴x=1-(-

[3x+2y+z=14,x+y+z=10,2x+3y-z=1]

3x+2y+z=14.(1)x+y+z=10.(2)2x+3y-z=1.(3)解(1)-(2)得2x+y=4.(4)(2)+(3)得3x+4y=11.(5)4*(4)-(5)得5x=5x=1把x=1代

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