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: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
*下列 反应的 △ r H m θ (298K) 就是 △ f H m θ (CO 2 , g,298K) 。 (A) CO(g)+1/2O 2 (g)=CO 2 (g) (B) C( 金刚石 )+O 2 (g)=CO 2 (g) (C) C( 石墨 )+O 2 (g)=CO 2 (g) (D) CO 2 (g) = C( 石墨 )+O 2 (g)
*下列 反应的 △ r H m θ (298K) 就是 △ f H m θ (CO 2 , g,298K) 。 (A) CO(g)+1/2O 2 (g)=CO 2 (g) (B) C( 金刚石 )+O 2 (g)=CO 2 (g) (C) C( 石墨 )+O 2 (g)=CO 2 (g) (D) CO 2 (g) = C( 石墨 )+O 2 (g)
有关常规细胞培养箱气体比例的描述,下列正确的是() A: 1%CO<sub>2</sub>,99%空气的CO<sub>2</sub>培养箱 B: 2%CO<sub>2</sub>,98%空气的CO<sub>2</sub>培养箱 C: 3%CO<sub>2</sub>,97%空气的CO<sub>2</sub>培养箱 D: 4%CO<sub>2</sub>,96%空气的CO<sub>2</sub>培养箱 E: 5%CO<sub>2</sub>,95%空气的CO<sub>2</sub>培养箱
有关常规细胞培养箱气体比例的描述,下列正确的是() A: 1%CO<sub>2</sub>,99%空气的CO<sub>2</sub>培养箱 B: 2%CO<sub>2</sub>,98%空气的CO<sub>2</sub>培养箱 C: 3%CO<sub>2</sub>,97%空气的CO<sub>2</sub>培养箱 D: 4%CO<sub>2</sub>,96%空气的CO<sub>2</sub>培养箱 E: 5%CO<sub>2</sub>,95%空气的CO<sub>2</sub>培养箱
1)z^2=z拔(2)z^2+|z|=0
1)z^2=z拔(2)z^2+|z|=0
【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
\( xoz \) 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: \( {z^2} + {x^2} = {(y - 2)^2} \)
代偿性代谢性酸中毒时() A: CO<sub>2</sub>CP↓pH↓ B: CO<sub>2</sub>CP↓pH不变 C: CO<sub>2</sub>CP↓pH↑ D: CO<sub>2</sub>CP↑pH↑ E: CO<sub>2</sub>CP↑pH不变
代偿性代谢性酸中毒时() A: CO<sub>2</sub>CP↓pH↓ B: CO<sub>2</sub>CP↓pH不变 C: CO<sub>2</sub>CP↓pH↑ D: CO<sub>2</sub>CP↑pH↑ E: CO<sub>2</sub>CP↑pH不变
下列各组化合物中,对热稳定性判断正确的是( ) A: H<sub>2</sub>CO<sub>3</sub>>Ca(HCO<sub>3</sub>)<sub>2</sub> B: Na<sub>z</sub>CO<sub>3</sub>>PbCO<sub>3</sub> C: (NH<sub>4</sub>)<sub>2</sub>CO<sub>3</sub>>K<sub>2</sub>CO<sub>3</sub> D: Na<sub>2</sub>SO<sub>3</sub>>Na<sub>2</sub>SO<sub>4</sub>
下列各组化合物中,对热稳定性判断正确的是( ) A: H<sub>2</sub>CO<sub>3</sub>>Ca(HCO<sub>3</sub>)<sub>2</sub> B: Na<sub>z</sub>CO<sub>3</sub>>PbCO<sub>3</sub> C: (NH<sub>4</sub>)<sub>2</sub>CO<sub>3</sub>>K<sub>2</sub>CO<sub>3</sub> D: Na<sub>2</sub>SO<sub>3</sub>>Na<sub>2</sub>SO<sub>4</sub>
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,