z=sin(xy^2),求Zxy
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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
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
XY/X+Y=-2,-->(x+y)/(xy)=-1/2,-->1/x+1/y=-1/2YZ/Y+Z=4/3,-->(y+z)/(yz)=3/4,-->1/y+1/z=3/4&
本题考查最值不等式:a+b≥2√ab当且仅当a=b时,取等号x√yz+y√zx+z√xy≤x(y+z)/2+y(z+x)/2+z(x+y)/2当且仅当y=z,z=x,x=y,即:x=y=z时,取等号,
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
设2/x=3/y=4/z=1/k则x=2k,y=3k,z=4k将xyz代入x2+y2+z2/xy+yz+zx原式=(4k^2+9k^2+16k^2)/(6k^2+12k^2+8k^2)=29k^2/2
xy/yz=x/z=1/2,∴z=2x,代入zx=3,解得x=±√(3/2),z=±√6,y=±√(2/3);∵xy=1yz=2zx=3全为正数,∴xyz也全为正数;∴x=√(3/2),z=√6,y=
答:Zx=2xf1'+yf2'Zxy=2x²f12''+f2'+xyf22''
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-------
(x+y+z)^2=4x^2+y^2+z^2+2xy+2xz+2yz=4x^2+y^2+z^2+2(-5)=4x^2+y^2+z^2=14
x+y+z=5,xy+yz+zx=9所以(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25所以x^2+y^2+z^2=25-2×9=25-18=7
-4再问:请问第三步是怎么算出-1/4的可以写一下过程么再答:1/x+1/y+1/y+1/z+1/z+1/x=2(1/x+1/y+1/z)=-1/2∴1/x+1/y+1/z=-1/2/2=-1/4
第一题题目(求z-zy+x-3的值)修改为求(z-2y+x-3)的值已知-4(xy-zx-y²+yz)=-z²+2zx-x²,左边括号里的1,3项提个y出来等于y(x-y
令x/3=y/1=z/2=kx=3ky=kz=2kxy+yz+zx=993k*k+k*2k+3k*2k=993k^2+2k^2+6k^2=9911k^2=99k^2=9k=±3x^2=9k^2=9*9
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
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+y+z)²=x²+y²+z²+2xy+2yz+2xz所以可得:xy+yz+xz=[(x+y+z)²-(x²+y²+z
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