(x y-2z)(x-2y-z)多项式

=[(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)

∑是循环和例如∑a=a+b+c∑a^2=a^2+b^2+c^2∑(z-y)(x-y)/(x+y-2z)(y+z-2x)＝∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z

1=(x-z-2ab)/xy2=(a²-2ab+b²)/a-b=(a-b)²/a-b=a-

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.

δz/δx=1/(xy+x/y)*(y+1/y)=(y²+1)/(xy²+x)=1/xδ^2z/δxδy=δ(δz/δx)/δy=0

x=6-3y               &nbs

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,

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

如果可以用排序不等式证明的话x^2+y^2+z^2>=x^1.5y^0.5+y^1.5z^0.5+z^1.5x^0.5=2xxy/2(xy)^0.5+2yyz/2(yz)^0.5+2zzx/2(zx)

由题式可以看出当x=y或y=z或x=z时式子为0所以肯定有因式(x-y)(y-z)(z-x)展开后x最高项为-x^2y与x^2z而原式中x最高次项为x^3y和-x^3z所以还差x的1次项因式,所以实际

(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)&

对称性不妨设：x≥y≥za=|x-y|=x-y,b=|y-z|=y-z,c=|z-x|=x-z有：a、b、c≥0;c=a+b则：c≥a、b≥0A的最大值=c已知得出：16=a^2+b^2+c^2=2c

原式=[(x--y)+(x--z)]/(x--y)(x--z)+[(y--x)+(y--z)]/(y--x)(y--z)+[(z--x)+(z--y)]/(z--x)(z--y)=1/(x--z)+1

-2x²y（3xy²z-2y²z)=-6x³y³z+4x²y³z(ab²c)²÷（ab³c²

第二个分母写错了?(y-x)(z-x)/(x-2y+z)/(x+y-2z)+(z-y)(x-y)/(x+y-2z)/(y+z-2x)+(x-z)(y-z)/(y+z-2x)/(x-2y+z)=1

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)

把x=y+根号2代入得2y^2+2根号2y+2根号2*z^2+1=02[y+(根号2)/2]^2+2根号2*Z^2=0∴y+(根号2)/2=02根号2*z^2=0∴y=-(根号2)/2z=0x=(根号

（x^2-yz）/[x^2-(y+z)x+yz]+（y^2-zx）/[y^2-(z+x)y+zx]+（z^2-xy）/[z^2-(x+y)z+xy]=（yz-x^2）/（x-y）(z-x)+（zx-y

-2x²y(3xy²z-2y平方z)=-6x³y³z+4x²y³z