分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2

设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial y}}=\)（ ）。 A: \({e^{xy}}({x}y^2 + {x^3} + 2y)\) B: \({e^{xy}}({x^2}y + {x^3} + 2y)\) C: \({e^{xy}}({x}y^2 + {x^3} + 2x)\) D: \({e^{xy}}({x}y+ {x^3} + 2y)\)

设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}({x^2}y + {y^3} + 2x)\) B: \({e^{xy}}({x}y^2 + {y^3} + 2x)\) C: \({e^{xy}}({x}y + {y^3} + 2x)\) D: \({e^{xy}}({x^2}y + {y^2} + 2x)\)

9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定，则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$（ ） A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

设\(z = u{e^v}\)，\(u = x + y\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}(1 + xy + {y^2})\) B: \({e^{xy}}(1 + xy + {y^3})\) C: \({e^{xy}}(x+ xy + {y^2})\) D: \({e^{xy}}(y+ xy + {y^2})\)

【填空题】已知正数 x , y 满足 x 2 +2 xy -3=0, 则 2 x + y 的最小值是 ______.

下列函数中，( )不是方程\( xy' + y - x^2 = 0 \)的解。 A: \( y = { { {x^2}} \over 3} + {1 \over x} \) B: \( y = { { {x^2}} \over 3} \) C: \( y = { { {x^2}} \over 3} + 2 \) D: \( y = { { {x^2}} \over 3} - {1 \over x} \)

已知｜x｜=3，｜y｜=2，xy＜0，则x+y=________．

应力圆的半径是( )。 A: (σx +σy)/2 B: (σx -σy)/2 C: τxy D: sqrt( [(σx -σy)/2]^2 + τxy^2 )

由方程\({z^3} - 3xyz = {a^3}\)所确定的隐函数\(z= f(x,y)\)的偏导数\( { { \partial z} \over {\partial x}} = \) A: \( { { yz} \over { { z^2} - xy}}\) B: \(- { { yz} \over { { z^2} - xy}}\) C: \( { { yz} \over { { z^2} +xy}}\) D: \(- { { yz} \over { { z^2}+xy}}\)