Giải phương trình \(f'\left( x \right) = 0\), biết rằng:
a) \({f\left( x \right) = 3x + \frac{{60}}{x} - \frac{{64}}{{{x^3}}} + 5}\)
b) \({f\left( x \right) = \frac{{\sin 3x}}{3} + \cos x - \sqrt 3 \left( {\sin x + \frac{{\cos 3x}}{3}} \right)}\)
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)
\(\begin{array}{l}
f\prime (x) = \cos 3x - \sin x - \sqrt 3 (\cos x - \sin 3x)\\
\Rightarrow f\prime (x) = 0 \Leftrightarrow (\cos 3x + \sqrt 3 \sin 3x) - (\sin x + \sqrt 3 \cos x) = 0\\
\Leftrightarrow \left( {\frac{{\cos 3x}}{2} + \sin 3x.\frac{{\sqrt 3 }}{2}} \right) - \left( {\cos x.\frac{{\sqrt 3 }}{2} + \sin x.\frac{1}{2}} \right) = 0\\
\Leftrightarrow \cos \left( {3x - \frac{\pi }{3}} \right) - \cos \left( {x - \frac{\pi }{6}} \right) = 0\\
\Leftrightarrow \cos \left( {3x - \frac{\pi }{3}} \right) = \cos \left( {x - \frac{\pi }{6}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \frac{\pi }{3} = x - \frac{\pi }{6} + k2\pi \\
3x - \frac{\pi }{3} = - x + \frac{\pi }{6} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{\pi }{8} + k\pi 2
\end{array} \right.(k \in Z)
\end{array}\)
-- Mod Toán 11