Giải bài 5.48 trang 207 - SBT Đại số và Giải tích lớp 11

a)

$\begin{aligned} & f'\left( x \right)=3-\dfrac{60}{{{x}^{2}}}+\dfrac{192}{{{x}^{4}}}=\dfrac{3{{x}^{4}}-60{{x}^{2}}+192}{{{x}^{4}}} \\ & \Rightarrow f'\left( x \right)=0 \\ & \Rightarrow 3{{x}^{4}}-60{{x}^{2}}+192=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{x}^{2}}=16 \\ & {{x}^{2}}=4 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\pm 4 \\ & x=\pm 2 \\ \end{aligned} \right. \\ \end{aligned}$

b)
$\begin{aligned} & f'\left( x \right)=\cos 3x-\sin x-\sqrt{3}\left( \cos x-\sin 3x \right) \\ & \Rightarrow f'\left( x \right)=0 \\ & \Leftrightarrow \left( \cos 3x+\sqrt{3}\sin 3x \right)-\left( \sin x+\sqrt{3}\cos x \right)=0 \\ & \Leftrightarrow \left( \dfrac{\cos 3x}{2}+\sin 3x.\dfrac{\sqrt{3}}{2} \right)-\left( \cos x.\dfrac{\sqrt{3}}{2}+\sin x.\dfrac{1}{2} \right)=0 \\ & \Leftrightarrow \cos \left( 3x-\dfrac{\pi }{3} \right)-\cos \left( x-\dfrac{\pi }{6} \right)=0 \\ & \Leftrightarrow \cos \left( 3x-\dfrac{\pi }{3} \right)=\cos \left( x-\dfrac{\pi }{6} \right) \\ & \Leftrightarrow \left[ \begin{aligned} & 3x-\dfrac{\pi }{3}=x-\dfrac{\pi }{6}+k2\pi \\ & 3x-\dfrac{\pi }{3}=-x+\dfrac{\pi }{6}+k2\pi \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{\pi }{12}+k\pi \\ & x=\dfrac{\pi }{8}+\dfrac{k\pi }{2} \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$