4x-9z=17 3x y 15z=18 x 2y 3z=2

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

（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

一3x+4z=7,(1)2x+3y+z=9,(2)5x-9y+7z=8,(3)(2)*3得6x+9y+3z=27,(4)(3)+(4)得11x+10z=35,(5)(1)*5得15x+20z=35,(

把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.

4x-9z=17…………………A3x+y+15z=18……………Bx+2y+3z=2………………CB*2-C：5x+27z=34……D联立A与D就可推得：x=5,y=1/3,z=-22x+4y+3z=

1、x+4y-9/2z=0即2x+8y-9z=0(1)2x-3y+2z=0(2)(1)-(2)11y-11z=0y=z(1)×2+(2)×94x+16y-18z+18x-27y+18z=022x-11

z²-4z+4=xy-9又x=6-y,代入得z²-4z+4=（6-y）y-9(z-2)²=-(y-3)²(z-2)²+(y-3)²=0所以(

（1）x-z=4,①x+y-z=-1,②2y-z=1③②-①得：y=-5代入③：-10-z=1-z=11z=-11代入①：x+11=4x=-7所以,x=-7,y=-5,z=-11

x+y-z=6,①x-3y+2z=1,②3x+2y-z=-7③①-②得4y-3z=5④①*3-③得y-2z=25⑤④-⑤*4得5z=-95z=-19代入⑤得y+38=25y=-13代入①得x-13+1

1:2x+3y+z=9和5x-9y+7z=8利用这2个式子消除y…前面那个式子左右都乘以3等于6x+9x+3z=27加上后面的式子消除y得11x+10z=35再和3x+4z=7这是二元一次!前面那个乘

1\a4x+9y=12,b3y-2z=1,c2x+6z=3a-3b4x+6z=9与C联立x=3z=-1/2带入ay=0x=3y=0z=-1/22\a3x-y+2z=3,b2x+y-3z=11,cx+y

X=5Y=-2Z=1/3

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,(2

4x-9z=171式3x+y+15z=182式x+2y+3z=23式2式×2-3式得到：5x+27z=344式1式×5-2式×4得到：20x-45z-20x-108z=85-136-153z=-51z

4X-9Z=17（1）3X+Y+15Z=18（2）X+2Y+3Z=2（3）(2)*2－(3)5x+27z=34（4）(1)*3+(4)17x=85x=5代入(1)得z=1/3两者共代入(3)得y=-2

【参考答案】x=5,y=-2,z=1/34x-9z=17(1)3x+y+15z=18（2）x+2y+3z=2(3)(2)x2得6x+2y+30z=36(4)(4)-(3)得5x+27z=34(5)(1

2）2式乘以2减去3式得5X+27Z=34记为4式,1式乘以3与4式相加得17X=85,所以X=5,代入1式得Z=1./3,一起代入3式得Y=-23）2式乘以2加上1式得8X+13Z=31,记为4式,

3.4X-9.8=1.4X-93.4X-1.4X=-9+9.82X=0.8X=0.4Z+2/7=4/9Z-3/74/9Z-Z=3/7+2/7-5/9Z=5/7Z=-9/7

1.设X,Y,Z成等差数列,代数式（X-Z）*（X-Z）+4(X-Y)(Z-Y)=(-2d)^2-4d*d=02.设数列｛An｝的通项公式为An=4n+3求证：｛An｝为等差数列.An=4n+3An+

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.