4x+9y=8 3y-2z=2 5x+3z=7
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[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
(1)因为2x+3y-4z=7(1)3x-2y+z=3(2)x+5y+9z=20(3)(1)+(2)+(3)得(2x+3y-4z)+(3x-2y+z)+(x+5y+9z)=7+3+20整理得:6x+6
(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-3
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
x+y+z=41)2x+y+3z=92)x+3y+2z=143)2)-1):x+2z=54)1)*3-3):2x+z=-25)4)*2-5):3z=12,得:z=4代入4):x=5-2z=-3代入1)
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
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\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
(2)×2+(3)得x+2y=53(4)(1)+(4)得2x=44∴x=22把x=22代入(1)得22-2y=-9∴y=31/2把x=22代入(3)得2z+22=47∴z=25/2∴x=22y=31/
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
不妨方程x-2y=-9标注为(1),方程3x-y+z=4标注为(2),方程y-z=3标注为(3),则有(2)-(1)*3=(3x-y+z)-(x-2y)*3=4-(-9)x3,即为5y+z=31,标注
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
1运行结果为:1,32分析x=y--
x-2y=-9...(1)y-z=3.(2)2z+x=47.(3)(1)+(2)*2得到:x-2z=-3.(4)(3)+(4)得到:x=22;代入(4)得到:z=25/2;进而可得到:y=31/2;所
【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
根据第三个方程式得X=9-4Y①根据第一个方程式得X+Y=-Z②乘以3得3X+3Y=-3Z③第二个方程减③得X-Y+1=O④①代入④得9-4Y-Y+1=0解得Y=2X=1Z=-3
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.
3x+4y=152y-z=2x+2z=9y=(2+z)/2x=9-2z3(9-2z)+4+2z=15z=4y=3x=1x+y+z=8(1)2x+3y-z=6(2)3x-y+z=8(3)(1)+(2)3