2x+3y+z=38,3x+4y+2z=56,4x+5y+z=66
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1.x+y+z2.-x-3y-1
(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
(1)x+y+z=23x-y-4z=52x+3y-2z=0消去y4x-3z=711x-14z=15解是x=25/7,y=4,z=17/7(2)2x+3y+z=383x-y-4z=52x+3y-2z=0
方程(1)+(2)得:3x-z=9④,方程(2)+(3)得:2x-z=7⑤,④-⑤得:x=2,把它代入⑤得:z=-3,把它代入(1)得:y=-3,∴原方程的解为x=2y=−3z=−3.
x-y+4z=10(1)x+3y+2z=2(2)x+2y+3z=11(3)(2)-(1):4y-2z=-8,即2y-z=-4(4)(3)-(1):3y-z=1(5)(5)-(4):y=5代入(5):z
x+2y+3z=10,(1)x-y+4z=10,(2)x+3y+2z=2(3)(1)-(2)得:3y-z=0z=3y(4)(3)-(2)得:4y-2z=-8(5)(4)代入(5)得:-2y=-8y=4
1.x+y=16①y+z=12②z+x=10③①-②x-z=4④③+④2x=14x=7⑤⑤代入①y=9⑥⑥代入②z=3x=7,y=9,z=3(2)3x-y+z=4①2x+3y-z=12②x+y+z=6
X+Y+Z=4,2*(X+Y+Z)+X=10,可以解出X=2.套入第二个和第一个.4+3Y-Z=66+2Y+2Z=10那么3Y=Z+2,2Y+2Z=4.Y=1,X=1X+Y+Z=4=2+1=12x+3
楼主好,4x-9z=17.A3x+y+15z=18.Bx+2y+3z=2.C3C->3x+6y+9z=6.DA+D得:7x+6y+0Z=23.E5C->5x+10y+15z=10.FF-B得:2x+9
【解】视z为常数,由已知两方程,可解得x=3zy=2z将其代入待求值式中,得3x*x+2y*y+5z*z/5x*x+y*y-9z*z=[3(3z)^2+2(2z)^2+5z^2]/[5(3z)^2+(
x+2y-7z=a3x-4y+6z=b4x-2y-z=c则a+b=ca^3+b^-c^3=(a+b)^-3a^2b-3ab^2-c^3=-3ab(a+b)=-3abc=-3(x+2y-7z)(3x-4
(x+2y+3z)-(3x+2y-z)+(2x-3y+4z)=x+2y+3z-3x-2y+z+2x-3y+4z=-3y+8z
即y=3x/2z=2x所以原式=(2x+3x/2-2x)/(3x-3x+2x)=(3x/2)/(2x)=3/4
①x+2y-4z=0②3x+y-z=0①-2②x-6x-4z+2z=05x=2z代入①z=5x/2x+2y-10x=02y=9xy=9x/2x:y:z=1:9/2:5/2=2:9:5
2x+y+3z=38①3x+2y+4z=56②4x+y+5z=66③③-①得:2x+2z=28,即x+z=14④,①×2-②得:x+2z=20⑤,由④和⑤组成方程组:x+z=14x+2z=20,解得:
=-(3x-2y-3z)[(x+y+z)^2-(x+y+z)(4x-y-2z)+(4x-y-2z)^2]+(3x-2y-3z)^3然后提取公因式就好了
x+2y+3z=1 ①2x+3y+z=2 &nb
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-2y-[-4x+(z+3y)]=3x-2y-(-4x+z+3y)=3x-2y+4x-z-3y=7x-5y-z