2x y-z w=1,3x-2y z-3w=4,x 4y-3z 5w=-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
x-3=y-2x-y=1y-2=z-1y-z=1x-3=z-1z-x=-2x^2+y^2+z^2-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)=x+y-2zx-3=z-1y-2=z-12
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
由xy/(x+y)=1,yz/(y+z)=2,zx/(z+x)=3,得:(x+y)/xy=1,(y+z)/yz=1/2,(z+x)/zx=1/3,(取倒数)所以1/x+1/y=1,(1)1/y+1/z
(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^
本题解法较多,以下举二个解法:解法一:设α、β、γ>0,则依基本不等式得α²x²+y²≥2αxy,β²y²+z²≥2βyz,γ²z
x²-xy+xz-yz=x(x-y)+z(x-y)=(x+z)(x-y)若仍有疑问,欢迎追问!
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
x+y分之xy=1,y+z分之yz=2,z+x分之zx=3每个等式左右均取倒数,所以:1/x+1/y=11/y+1/z=1/21/z+1/x=1/3设:1/x=a1/y=b1/z=ca+b=1----
y=-12;一共是三个方程,因为xy/(x+y)=3推出(x+y)/(xy)=1/3-------方程1;同理:(y+z)/(yz)=1/2-------方程2;(x+z)/(xz)=1-------
左式可化为[(xy)^3+(xz)^3+(yz)^3]/xyz+6xyz;然后[(xy)^3+(xz)^3+(yz)^3]/xyz>=3xyz(这一步是将分子利用(a+b+c)>=3*(abc)^(1
设α、β、γ>0,则有α^2x^2+y^2≥2αxy,β^2y^2+z^2≥2βyzγ^2z^2+w^2≥2rzw.∴xy+2yz+zw≤(α/2)x^2+(1/2α+β)y^2+(1/β+γ/2)z
题目是这样吧1=xy/(x+y),2=yz/(y+z),3=xz/(x+z)倒数法,写成每个式子的倒数;1=1/x+1/y,(1)1/2=1/y+1/z,(2)1/3=1/x+1/z(3)三式相加,得
1/Y+1/X=1(1)1/Z+1/Y=2(2)1/X+1/Z=3(3)(1)+(2)+(3):1/X+1/Y+1/Z=3(4)(4)-(1):1/Z=2Z=1/2(4)-(2):1/X=1X=1题目
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)
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:
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
xy:yz:zx=3:2:1xy:yz=3:2则x:z=3:2同理y:z=3:1=6:2故(x+y):z=(3+6):2=9:2
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