Đáp án:

\(\begin{array}{l}
d,\\
\left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{3}\\
{y_{\max }} = 1
\end{array} \right.\\
g,\\
\left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{4}\\
{y_{\max }} = \dfrac{3}{4}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
d,\\
y = \dfrac{1}{{2 - \cos x}}\\
 - 1 \le \cos x \le 1\\
 \Leftrightarrow  - 1 \le  - \cos x \le 1\\
 \Leftrightarrow  - 1 + 2 \le  - \cos x + 2 \le 1 + 2\\
 \Leftrightarrow 1 \le 2 - \cos x \le 3\\
 \Leftrightarrow \dfrac{1}{3} \le \dfrac{1}{{2 - \cos x}} \le \dfrac{1}{1}\\
 \Leftrightarrow \dfrac{1}{3} \le y \le 1\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{3} \Leftrightarrow \cos x =  - 1 \Leftrightarrow x = \pi  + k2\pi \\
{y_{\max }} = 1 \Leftrightarrow \cos x = 1 \Leftrightarrow x = k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{3}\\
{y_{\max }} = 1
\end{array} \right.\\
g,\\
y = \dfrac{{1 + 2{{\sin }^2}x}}{4}\\
 - 1 \le \sin x \le 1\\
 \Leftrightarrow 0 \le {\sin ^2}x \le 1\\
 \Leftrightarrow 0 \le 2{\sin ^2}x \le 2\\
 \Leftrightarrow 1 \le 1 + 2{\sin ^2}x \le 3\\
 \Leftrightarrow \dfrac{1}{4} \le \dfrac{{1 + 2{{\sin }^2}x}}{4} \le \dfrac{3}{4}\\
 \Leftrightarrow \dfrac{1}{4} \le y \le \dfrac{3}{4}\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{4} \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = \dfrac{3}{4} \Leftrightarrow {\sin ^2}x = 1 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = \dfrac{1}{4}\\
{y_{\max }} = \dfrac{3}{4}
\end{array} \right.
\end{array}\)