函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( )
A: $\left\{(x,y)\left|~x+y\geq
0\right.\right\}$;
B: $\left\{(x,y)\left|~x+y\geq
1~\text{或}~x+y\leq
-1 \right.\right\}$;
C: $\left\{(x,y)\left|~x+y\geq
1\right.\right\}$;
D: $\left\{(x,y)\left|~x+y\geq
\dfrac{4}{~\pi^2~}\right.\right\}$.
A: $\left\{(x,y)\left|~x+y\geq
0\right.\right\}$;
B: $\left\{(x,y)\left|~x+y\geq
1~\text{或}~x+y\leq
-1 \right.\right\}$;
C: $\left\{(x,y)\left|~x+y\geq
1\right.\right\}$;
D: $\left\{(x,y)\left|~x+y\geq
\dfrac{4}{~\pi^2~}\right.\right\}$.
举一反三
- 下列函数是多元初等函数的是( ) A: $f(x,y)=\left|x+y\right|$; B: $f(x,y)=\text{sgn}(x+y)$; C: $f(x,y)=\dfrac{\arcsin<br/>x-e^{y}}{~\ln(x^2+y^2)~}$; D: $f(x,y)=\left\{\begin{array}{cc}\dfrac{xy}{~x^2+y^2~},<br/>&x^2+y^2\neq 0; \\0, &x^2+y^2= 0. \end{array}\right.$
- \(\left\{ {\left( {x,y} \right)\left| {2 \le {x^2} + {y^2} \le 4} \right.} \right\}\)是闭区域.
- 若\({y_1}\left( x \right), {y_2}\left( x \right)\)都是\(y' + P\left( x \right)y = Q\left( x \right)\)的特解,且 \({y_1}\left( x \right), {y_2}\left( x \right)\) 线性无关,则通解可表为\(y\left( x \right) = {y_1}\left( x \right) + C\left[ { { y_1}\left( x \right) - {y_2}\left( x \right)} \right]\)。
- 曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为( ) A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
- 函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)