Đáp án:
$\begin{array}{l}
1)y = \sin \dfrac{x}{2}.\cos 2x\\
\Leftrightarrow y' = \dfrac{1}{2}.\cos \dfrac{x}{2}.\cos 2x + \sin \dfrac{x}{2}.2.\left( { - \sin 2x} \right)\\
= \dfrac{1}{2}.\cos \dfrac{x}{2}.\cos 2x - 2\sin \dfrac{x}{2}.\sin 2x\\
2)Dkxd:x\# 5\\
f\left( x \right) = \dfrac{{{x^2} - 10x + 26}}{{x - 5}} = \dfrac{{{{\left( {x - 5} \right)}^2} + 1}}{{x - 5}}\\
= x - 5 + \dfrac{1}{{x - 5}}\\
\Leftrightarrow f'\left( x \right) = 1 - \dfrac{1}{{{{\left( {x - 5} \right)}^2}}} \le 0\\
\Leftrightarrow \dfrac{1}{{{{\left( {x - 5} \right)}^2}}} \ge 1\\
\Leftrightarrow {\left( {x - 5} \right)^2} \le 1\\
\Leftrightarrow - 1 \le x - 5 \le 1\\
\Leftrightarrow 4 \le x \le 6\\
Vậy\,4 \le x \le 6;x\# 5
\end{array}$
