sin(-<span>$$\frac{9\pi}{4}$$</span>)的值是（）选项： A:<p>-<span class=

$$ \frac{x-1}{4}=\frac{y+2}{-2}=\frac{z+3}{2} $$

$$ \frac{x-1}{4}=\frac{y-2}{2}=\frac{z-3}{2}$$

">1.$$过点(-1,2,3)且平行于\frac{x-1}{4}=\frac{y}{-2}=\frac{z+1}{4}的直线是$$ 选项： A:

$$\frac{x+1}{4}=\frac{y+2}{2}=\frac{z+3}{2}$$

$$\frac{x+1}{4}=\frac{y-2}{-2}=\frac{z-3}{2}$$

$$ \frac{x-1}{4}=\frac{y+2}{-2}=\frac{z+3}{2} $$

$$ \frac{x-1}{4}=\frac{y-2}{2}=\frac{z-3}{2}$$

$$\frac{x+1}{4}=\frac{y+2}{2}=\frac{z+3}{2}$$

$$ \frac{e^{\frac{1}{x}} \cdot \cos \left(e^{\frac{1}{x}}\right)}{x^{2}}$$

$$ \frac{e^{\frac{1}{x}} \cdot \cos \left(e^{\frac{1}{x}}\right)}{-x^{2}}$$

$$ \frac{ \cos \left(e^{\frac{1}{x}}\right)}{-x^{2}}$$

">2.$$ 设 y=\sin \left(e^{\frac{1}{x}}\right), 则 \frac{d y}{d x}=$$ 选项： A:

$$ \frac{ \cos \left(e^{\frac{1}{x}}\right)}{x^{2}}$$

$$ \frac{e^{\frac{1}{x}} \cdot \cos \left(e^{\frac{1}{x}}\right)}{x^{2}}$$

$$ \frac{e^{\frac{1}{x}} \cdot \cos \left(e^{\frac{1}{x}}\right)}{-x^{2}}$$

$$ \frac{ \cos \left(e^{\frac{1}{x}}\right)}{-x^{2}}$$

$$ \frac{ \cos \left(e^{\frac{1}{x}}\right)}{x^{2}}$$

$$-tan^{-1}0.25\omega-tan^{-1}\frac{0.4\omega}{1-0.25\omega^{2}}$$

$$-tan^{-1}0.25\omega+tan^{-1}\frac{0.4\omega}{1-0.25\omega^{2}}$$

$$-tan^{-1}0.25\omega+tan^{-1}\frac{0.4\omega}{1+0.25\omega^{2}}$$

">3.设某系统的开环传递函数为$$G(s)=\frac{10}{(0.25s+1)(0.25s^{2}+0.4s+1)}$$，则其相频特性$$\varphi(\omega)$$=（）。选项： A:

$$-tan^{-1}0.25\omega-tan^{-1}\frac{0.4\omega}{1+0.25\omega^{2}}$$

$$-tan^{-1}0.25\omega-tan^{-1}\frac{0.4\omega}{1-0.25\omega^{2}}$$

$$-tan^{-1}0.25\omega+tan^{-1}\frac{0.4\omega}{1-0.25\omega^{2}}$$

$$-tan^{-1}0.25\omega+tan^{-1}\frac{0.4\omega}{1+0.25\omega^{2}}$$