X+Y+Z=9

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

X+Y+Z

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)得x=1,代入（1）得z=3.故得：x

2x+y+z=71x+2y+z=82x+y+2z=931+2+3,得x+y+z=641-4,得2x+y+z-x-y-z=7-6x=12-4,得x+2y+z-x-y-z=8-6y=23-4,得x+y+2z-x-y-z=9-6z=3所以,原方程

z=2x+3y-11然后代入式得到5x+5y=20可得到x+y=4得到z=-2,然后代入1和3式然后1式乘以2,3式乘以2,可得到y=1,然后代入任意一式得到x值.再问：过程再答：你敢不敢给我给分啊？？

实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.

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)

（一）x(x+y+z)=4-yz.===>x²+(y+z)x+yz=4.===>(x+y)(x+z)=4①.同理,将后面两个方程变形可得(x+y)(y+z)=9,②(x+z)(y+z)=25.③,将三个方程相乘可得(x+y)(y+

思路：(x-2y+z)/9=(2x+y+3z)/10=-(3x+2y-4z)/3=1即(x-2y+z)/9=1,(2x+y+3z)/10=1,-(3x+2y-4z)/3=1即(x-2y+z)=9,(2x+y+3z)=10,(3x+2y-4z

2x+y-3z=1,①x-2y+z=6,②3x-y+2z=9③①+③得：5x-z=10④①×2+②得：5x-5z=8⑤④-⑤得：4z=2∴z=1/2x=21/10=2.1y=-1.7

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)

(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+

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

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+

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

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)

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

x/10=y/8=z/9=(x+y+z)/(8+9+10)=(y+z)/(8+9)所以：(x+y+z)/(8+9+10)=(y+z)/(8+9)所以：(x+y+z)/(y+z)=(8+9+10)/(8+9)=27/17

设x/10=y/8=z/9=KX=10K,Y=8K,Z=9K（X+2Y+3Z）/（Y-5Z）=（10K+16K+27K）/（8K-45K）=-53/37