• 2022-06-08 问题

    设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\),则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)

    设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\),则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)

  • 2022-06-06 问题

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

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

  • 2022-06-17 问题

    设z=1+i(i是虚数单位),则2z2+z=______.

    设z=1+i(i是虚数单位),则2z2+z=______.

  • 2022-05-31 问题

    1)z^2=z拔(2)z^2+|z|=0

    1)z^2=z拔(2)z^2+|z|=0

  • 2021-04-14 问题

    【简答题】设 z 1 =4 + 3i , z 2 =2 - 3i ,计算 z 1 · z 2

    【简答题】设 z 1 =4 + 3i , z 2 =2 - 3i ,计算 z 1 · z 2

  • 2022-06-16 问题

    \( xoz \) 坐标面上的直线\( x = z - 2 \)绕\( z \)轴旋转而成的圆锥面的方程为( ) A: \( {x^2} - {y^2} = {(z - 2)^2} \) B: \( {x^2} + {y^2} = {(z - 2)^2} \) C: \( {z^2} + {y^2} = {(x - 2)^2} \) D: \( {z^2} + {x^2} = {(y - 2)^2} \)

    \( xoz \) 坐标面上的直线\( x = z - 2 \)绕\( z \)轴旋转而成的圆锥面的方程为( ) A: \( {x^2} - {y^2} = {(z - 2)^2} \) B: \( {x^2} + {y^2} = {(z - 2)^2} \) C: \( {z^2} + {y^2} = {(x - 2)^2} \) D: \( {z^2} + {x^2} = {(y - 2)^2} \)

  • 2022-06-18 问题

    f(z)=e^(z^2)*sin(z^2),求f(z)展成Z的幂级数,

    f(z)=e^(z^2)*sin(z^2),求f(z)展成Z的幂级数,

  • 2021-04-14 问题

    若复数z 1=1+i ,z 2=3-i ,则z 1·z 2等于

    若复数z 1=1+i ,z 2=3-i ,则z 1·z 2等于

  • 2022-06-18 问题

    z=0为f(z)=z^2 (e^(z^2 )-1)的 级零点,

    z=0为f(z)=z^2 (e^(z^2 )-1)的 级零点,

  • 2022-06-01 问题

    若复数z满足:z+.z=2,z?.z=2,则|z-.z|=______.

    若复数z满足:z+.z=2,z?.z=2,则|z-.z|=______.

  • 1 2 3 4 5 6 7 8 9 10