Giải thích các bước giải:
Ta có:
$y = \dfrac{{23 - 10x}}{{{x^2} + 2}}$
+) Tìm $Min$
Ta có:
$\begin{array}{l}
y + 1\\
= \dfrac{{23 - 10x}}{{{x^2} + 2}} + 1\\
= \dfrac{{{x^2} - 10x + 25}}{{{x^2} + 2}}\\
= \dfrac{{{{\left( {x - 5} \right)}^2}}}{{{x^2} + 2}}\\
\ge 0,\forall x\\
\Rightarrow y \ge - 1
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {x - 5} \right)^2} = 0 \Leftrightarrow x = 5$
Vậy $Miny = - 1 \Leftrightarrow x = 5$
+) Tìm $Max$
Ta có:
$\begin{array}{l}
y - \dfrac{{25}}{2}\\
= \dfrac{{23 - 10x}}{{{x^2} + 2}} - \dfrac{{25}}{2}\\
= \dfrac{{ - 25{x^2} - 20x - 4}}{{2\left( {{x^2} + 2} \right)}}\\
= \dfrac{{ - {{\left( {5x + 2} \right)}^2}}}{{2\left( {{x^2} + 2} \right)}}\\
\le 0,\forall x\\
\Rightarrow y \le \dfrac{{25}}{2}
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {5x + 2} \right)^2} = 0 \Leftrightarrow x = \dfrac{{ - 2}}{5}$
Vậy $Maxy = \dfrac{{25}}{2} \Leftrightarrow x = \dfrac{{ - 2}}{5}$
