A: τxy≤ [σ]
B: √2τxy≤[σ]
C: -√2τxy≤[σ]
D: 2τxy≤[σ]
举一反三
- 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(X,Y)的联合密度为[img=246x61]17de89815546a18.png[/img],则EY=_____, E(XY)=_____. A: EY=4/5, E(XY)=7/5 B: EY=7/5, E(XY)=2/5 C: EY=3/5, E(XY)=1/2 D: EY=2/3, E(XY)=2/5
- 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 )
- 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: \({e^{xy}}(y+ xy + {y^2})\)
- 在下图中,已知斜截面上无应力,该x、y面上的应力分量满足关系 ( )[img=363x330]1803be4a6959759.png[/img] A: σx>σy, τyx>τxy B: σx<σy, τxy=τxy C: σx>σy, τyx=τxy D: σx<σy, τyx>τxy
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$