Z=x^2 y^2-2xy,求dZ
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=[(X+Z)+(X-Y)]/[X(X-Y)+Z(X-Y)]-[(X+Y)+(X+Z)]/[X(X+Y)+Z(X+Y)]=[(X+Z)+(X-Y)]/[(X+Z)(X-Y)]-[(X+Y)+(X+Z)
实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.
dz/dx=y(yf1'+2f2')dz/dy=f(xy,2x+y)++y(xf1'+f2')da/dxdy=(yf1'+2f2')+y【f1'+y(xf1'+f2')+2(xf1'+f2')】=2y
瀑布汗....(X^2+Y^2)/(X^2+Y^2)=1E(1),=1
把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.
第一步,找|x|+|y|
z²-4z+4=xy-9又x=6-y,代入得z²-4z+4=(6-y)y-9(z-2)²=-(y-3)²(z-2)²+(y-3)²=0所以(
两端对x求偏导得:-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得:-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,
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-------
你只要X看成是是常数求导就行了,答案就不给你了,自己动手丰衣足食
δz/δx=y^2*f1+(2y-1)*f2δz/δy=2xy*f1+x^2y*2*f2再问:f1和f2是什么?再答:f1表示z对x求导,也可写成fx,(x为下标,在右下角,我不好打,不好意思!)这只
将x+y=2两边分别平方,得x2+2xy+y2=4(1)把方程xy-z2=1两边都乘以2得2xy-2z2=2(2)(1)-(2)得:x2+y2+2z2=2(3)由x+y=2得2x+2y=4(4)(3)
(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&
∵x-y=4,y-z=2,∴x-z=6∴x^2y+y^2z+z^2x-(xy^2+yz^2+zx^2)=(x^2y-xy^2)+(y^2z-yz^2)+(z^2x-zx^2)=xy(x-y)+yz(y
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
X=0.00934063110.84511098*i-43.00467-0.84511098*i-43.00467Y=86.0186811.690222*i-0.0093406311-1.690222
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
3x-y=-2zx+2y=-3z那么:x=-z,y=-z(3x^-xy+2y^)/(2x^+4xy+y^)=(3z^2-z^2+2z^2)/(2z^2+4z^2+z^2)=4z^2/7z^2=4/7
∵x-y=4,y-z=2,∴x-z=6∴x^2y+y^2z+z^2x-(xy^2+yz^2+zx^2)=(x^2y-xy^2)+(y^2z-yz^2)+(z^2x-zx^2)=xy(x-y)+yz(y