代数式yz(xz 2)-2y(3xz^2 z x) 5xyz^2
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xy/x+y=-2,取倒数得1/x+1/y=-1/2①yz/y+z=3/4取倒数得1/y+1/z=4/3②zx/z+x=-3/4取倒数得1/x+1/z=-4/3③①+②+③得2(1/x+1/y+1/z
a,b互为倒数,ab=1|x|=3,所以x=3,x=-32y^m+3次方与-1/3y^5-n次方是同类项m+3=5-n,即m+n=2当x=3,原式=3-(-1+2+1)x9+2^2010-1=2^20
首先,显然x,y,z均不为0.然后分开看xy:yz=3:2,两边除以y,得x:z=3:2yz:zx=2:1,除以z,得y:x=2:1,两边同时乘以3,得x:y=3:6所以:x:y:z=3:6:2,不能
(x+3)(y+3)(x+3)=(xy+3x+3y+9)(z+3)=xyz+3xy+3xz+9x+3yz+9y+9z+27=xyz+3(xy+yz+zx)+9(x+y+z)+27=9m+3n+p+27
f(x,y,z)=yz+xz使得,y^2+z^2=1,yz=3令F(x,y,z)=yz+xz+a(y²+z²-1)+b(yz-3)Fx=z=0Fy=z+2ay+bz=0Fz=y+x
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz>3(xy+yz+zx)所以只要求证x^2+y^2+z^2>xy+yz+zx2(x^2+y^2+z^2)>2(xy+yz+zx)(x^
应该是设X/2=Y/1=Z/3=K则X=2KY=KZ=3K则有xy+xz+yz=992K^2+6K^2+3K^2=99==>K^2=9所以4x^2-2xz+3yz-9y^2=2X(2X-Z)+3Y(Z
xy+yz+xz=1/2x(y+z)+1/2y(x+z)+1/2z(x+y)=(1/2x)*(1/2yz)+1/2y*(1/3zx)+1/2z*(xy)=11/12xyz应该知道答案了吧
xy+2z=xy+2(2-x-y)=(x-2)(y-2)同理,yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).原式=z−2+x−2+y−2(x−2)(y−2)(z−2)=(x+y+
=(x+y+z)^2+yz(y+z+x)=(x+y+z)(x+y+z+yz)
答:x-y=a,y-z=8两式相加:x-z=a+8x^2+y^2+z^2-xy-yz-xz=(1/2)*(2x^2+2y^2+2z^2-2xy-2yz-2xz)=(1/2)*[(x-y)^2+(y-z
4x-3y-3z=0,x-3y-z=0(xyz≠0)4x-3y=3zx-3y=zx=2z/3,y=-z/9(xy+2yz)/(x²+y²-z²)=(-2z²/2
令a=xx=6a.y=4a.z=3a(2x^2-yz+z^2)/(x^2-2xy+z^2)=(72a^2-12a^2+9a^2)/(36a^2-48a^2+9a^2)=69a^2/(-3a^2)=-2
你的题有问题,总的思路设x=2k,y=3k,z=4k,代入原式可求
设a=x-1,b=y-1,c=z-1,于是a+b+c=0xyz-xy-xz-yz=(a+1)(b+1)(c+1)-(a+1)(b+1)-(b+1)(c+1)-(c+1)(a+1)=abc-2利用a^3
答案是:(2*X)/((X-Z)*(X+Z))再问:解题过程给我写下1再答:=(2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)
(x+2y-z)^2+(z-x)^2=0所以x+2y-z=0,z-x=0x=z所以2y=0,y=0代入xz^2+yz-5√(xz^2+yz+9)+3=0x^3-5√(x^3+9)+3=0(x^3+9)
解题思路:本题的关键是将三个方程两边取倒数,化简后分别将方程等号左边和右边相加,得到1/x+1/y+1/z的值,最后将要求的分式化简,把1/x+1/y+1/z的值带入即可。解题过程:
设x/3=y/4=z/5=k3k*4k+4k*5k+3k*5k=94k的平方=22X²+12y²+9z²=18k²+192k²+255k²=
①x:y:z因为xy:yz:zx=3:2:1所以xy:yz=3:2所以x:z=3:2同理yz:zx=2:1所以y:x=2:1=6:3所以x:y:z=3:6:2②x/yz:y/zx=x^2:y^2=(x