作业帮2xy对x.y求二阶偏导
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可以看成是f对于X的函数f=x^2+yx+（3-y）,当X∈《-2.2》时,函数恒大于0,判别式=y^2-4（3-y）=（y+6）（y-4）,X在-2到2时,判别式恒大于0,然后分情况讨论,当最小值X
3xy是3x原式=(x-3y)(x+2y)+3x+y+2十字相乘x-3y2×x+2y1所以原式=(x-3y+2)(x+2y+1)
对.前提是x不等于y
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²
2x+y=2xy（2x+y）/xy=22/y+(1/x)=22x+y=(2x+y)(2/y+(1/x))*(1/2)=2x/y+y/(2x)+2≥2+2=4当2x=y时,等号成立,有最小值4
【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy
=（3y^2+2xy+x^2)y''+（6yy'+2y+2xy'+2x)y再问：我也这么想的，答案看来是错了再答：恩
（xy-y^2)÷x-y/xy=y(x-y)×[xy/(x-y)]=xy²
∵(x/2+y)²-xy=x²/4+xy+y²-xy=x²/4+y²而x²/4≥0,y²≥0∴x²/4+y²≥
(x+y)(x-y)-x(y+1)-y(y+1)=(x+y)(x-y)-(x+y)(y+1)=(x+y)(x-2y-1)
(xy+1)(x+1)(y+1)+xy展开(x+1)(y+1)展开,得(xy+1)(xy+x+y+1)+xy即(xy+1)(xy+1+x+y)+xy将(xy+1)当做一个整体,展开得(xy+1)^2+
xy+y平方+x-y-2=xy+x+y^2-y-2=x(y+1)+(y+1)(y-2)=(y+1)(x+y-2)
(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/
x=y=0f(0)=2f(0)+0f(0)=0令y=0f(x)=2f(0)+x^2+3x=x^2+3x
x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²
原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x&
解(x-y)/(xy-y²)=(x-y)/y(x-y)=1/y
利用公式法逆向因式分解(x+y)(x^2-xy+y^2)=x^3+y^3(x-y)(x^2+xy+y^3)=x^3-y^3都是公式可以自己乘开来试下可以抵消的（x^2-xy+y^2)和（x^2+xy+
对所有实数x,y,都有(x-y)²≥0,即x²-2xy+y²≥0,x²+y²≥2xy,故x²+y²≥xy.