x³+y³+z³+3y=xw³,x⁵+y⁵z³=xyz+x³w⁵

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

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

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-2z)(y+z-2x)(z+x-2y)

X+Y+Z=?

X+Y+Z

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

若{x+3y+10z=0 则 (x+y-z)/(x-y+z)

x+3y+10z=0就是x+3y=-10z即2x+6y=-20zA式2x-y-2z=0就是2x-y=2zB式A式-B式得到:(2x+6y)-(2x-y)=-20z-2z即7y=-22z解出y=-22z/7C式把C式代入B式,得到:2x-(-

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)

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3y+10z=5x-5y+9z-25xy+

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

x y z x+y--- = --- = ---- ----y+Z z+x x+y ,求 z 的值 .求 x+y----

x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y+z)/(2x+2y+2z)=(x+y)

已知3X=4X,5Y=6Z求X+Y:Y+Z

应该是3X=4Y,5Y=6Z吧?X+Y:Y+Z=[(4Y/3)+Y]:(Y+5Y/6)=14;11

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=18x+y+z=36①x-y=1②2x

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=kk[x+y+z]=[(y+

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,或x+y+z=0当k=1时,即(x+y-

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];如若您有不满意之处,请指出,我一定改正!希望还您一个

(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+y+z)(k-1)=0k=1或x+y+

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)k=2x+y=2z(x+y-z)/(x

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+1)x当x+y+z≠0时(x+y)(y

z/(x-y) × y/(x+y)

z/(x-y)×y/(x+y)=zy/(x-y)(x+y)=zy/(x²-y²)再问:还有两道题!麻烦你了!1.已知x-1/x=2,求x²/x四次方-x²+12.已知已知b+1/c=1,c+1/a=1

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.