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: $f(x,y)=\dfrac{\arcsin
x-e^{y}}{~\ln(x^2+y^2)~}$;
D: $f(x,y)=\left\{\begin{array}{cc}\dfrac{xy}{~x^2+y^2~},
&x^2+y^2\neq 0; \\0, &x^2+y^2= 0. \end{array}\right.$
A: $f(x,y)=\left|x+y\right|$;
B: $f(x,y)=\text{sgn}(x+y)$;
C: $f(x,y)=\dfrac{\arcsin
x-e^{y}}{~\ln(x^2+y^2)~}$;
D: $f(x,y)=\left\{\begin{array}{cc}\dfrac{xy}{~x^2+y^2~},
&x^2+y^2\neq 0; \\0, &x^2+y^2= 0. \end{array}\right.$
举一反三
- $(0,0)$是以下函数$f(x,y)$的定义域的内点是 A: $f(x,y)=\sqrt{x} \ln(x+y)$ B: $f(x,y)=\frac{x+y}{x^2+y^2 }$ C: $f(x,y)=\arcsin \frac{x}{y}$ D: $f(x,y)=\ln (1-x^2-y^2)$
- 函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( ) A: $\left\{(x,y)\left|~x+y\geq<br/>0\right.\right\}$; B: $\left\{(x,y)\left|~x+y\geq<br/>1~\text{或}~x+y\leq<br/>-1 \right.\right\}$; C: $\left\{(x,y)\left|~x+y\geq<br/>1\right.\right\}$; D: $\left\{(x,y)\left|~x+y\geq<br/>\dfrac{4}{~\pi^2~}\right.\right\}$．
- 已知f(x+y,xy)=x^2+y^2,则f(x,y)=（）
- 求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
- 求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]