• 2022-06-09
    图示应力状态,按第三强度理论校核,强度条件为( )。[img=228x172]180335df4ee991b.png[/img]
    A: τxy≤ [σ]
    B: √2τxy≤[σ]
    C: -√2τxy≤[σ]
    D: 2τxy≤[σ]
  • D

    内容

    • 0

      函数f(xy,)=xy在条件x+y=1下的极大值为()。 A: 1/4 B: 1/2 C: 1 D: 2

    • 1

      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: \({e^{xy}}({x^2}y + {y^2} + 2x)\)

    • 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: \({e^{xy}}({x}y+ {x^3} + 2y)\)

    • 3

      设\(z = z\left( {x,y} \right)\)是由方程\({z^3}{\rm{ + }}3xyz - 3\sin xy = 1\)确定的隐函数,则\( { { \partial z} \over {\partial y}}=\)( ) A: \( { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) B: \( { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\) C: \( { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) D: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)

    • 4

      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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$