Bài tập 5.48 trang 208 SBT Toán 11

a)

\(\begin{array}{l}
f\prime \left( x \right) = 3 - \frac{{60}}{{{x^2}}} + \frac{{192}}{{{x^4}}} = \frac{{3{x^4} - 60{x^2} + 192}}{{{x^4}}}\\
 \Rightarrow f\prime \left( x \right) = 0 \Rightarrow 3{x^4} - 60{x^2} + 192 = 0 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 16\\
{x^2} = 4
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x =  \pm 4\\
x =  \pm 2
\end{array} \right.
\end{array}\)

b)
\(\begin{array}{l}
f\prime (x) = \cos 3x - \sin x - \sqrt 3 (\cos x - \sin 3x)\\
 \Rightarrow f\prime (x) = 0 \Leftrightarrow (\cos 3x + \sqrt 3 \sin 3x) - (\sin x + \sqrt 3 \cos x) = 0\\
 \Leftrightarrow \left( {\frac{{\cos 3x}}{2} + \sin 3x.\frac{{\sqrt 3 }}{2}} \right) - \left( {\cos x.\frac{{\sqrt 3 }}{2} + \sin x.\frac{1}{2}} \right) = 0\\
 \Leftrightarrow \cos \left( {3x - \frac{\pi }{3}} \right) - \cos \left( {x - \frac{\pi }{6}} \right) = 0\\
 \Leftrightarrow \cos \left( {3x - \frac{\pi }{3}} \right) = \cos \left( {x - \frac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - \frac{\pi }{3} = x - \frac{\pi }{6} + k2\pi \\
3x - \frac{\pi }{3} =  - x + \frac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{\pi }{8} + k\pi 2
\end{array} \right.(k \in Z)
\end{array}\)