Đáp án:
a) \(\dfrac{{{x^2} + 8}}{{{x^2} - 4}}\)
b) \(\left[ \begin{array}{l}
x = \pm 4\\
x = \pm 1\\
x = 0
\end{array} \right.\)
c) \( - 2 < x < 2\)
d) Không tồn tại x để A đạt GTNN
e) \(x = 1,716188659\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
A = \dfrac{{x\left( {x - 2} \right) + 2\left( {x + 2} \right) + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 2x + 2x + 4 + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 8}}{{{x^2} - 4}}\\
b)A = \dfrac{{{x^2} - 4 + 12}}{{{x^2} - 4}}\\
= 1 + \dfrac{{12}}{{{x^2} - 4}}\\
A \in Z \to \dfrac{{12}}{{{x^2} - 4}} \in Z\\
\to {x^2} - 4 \in U\left( {12} \right)\\
\to \left[ \begin{array}{l}
{x^2} - 4 = 12\\
{x^2} - 4 = 6\\
{x^2} - 4 = 4\\
{x^2} - 4 = 3\\
{x^2} - 4 = 2\\
{x^2} - 4 = 1\\
{x^2} - 4 = - 1\\
{x^2} - 4 = - 2\\
{x^2} - 4 = - 3\\
{x^2} - 4 = - 4\\
{x^2} - 4 = - 6\left( l \right)\\
{x^2} - 4 = - 12\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = \pm 4\\
x = \pm \sqrt {10} \left( l \right)\\
x = \pm \sqrt 8 \left( l \right)\\
x = \pm \sqrt 7 \left( l \right)\\
x = \pm \sqrt 6 \left( l \right)\\
x = \pm \sqrt 5 \left( l \right)\\
x = \pm \sqrt 3 \left( l \right)\\
x = \pm \sqrt 2 \left( l \right)\\
x = \pm 1\\
x = 0
\end{array} \right.\\
c)A \le 1\\
\to 1 + \dfrac{{12}}{{{x^2} - 4}} \le 1\\
\to \dfrac{{12}}{{{x^2} - 4}} \le 0\\
\to {x^2} - 4 < 0\\
\to - 2 < x < 2\\
d)Do:{x^2} \ge 0\forall x\\
\to {x^2} - 4 \ge - 4\\
\to \dfrac{{12}}{{{x^2} - 4}} \le - 3\\
\to 1 + \dfrac{{12}}{{{x^2} - 4}} \le - 2\\
\to Max = - 2\\
\to x = 0
\end{array}\)
⇒ Không tồn tại x để A đạt GTNN
\(\begin{array}{l}
e)\dfrac{{{x^2} + 8}}{{{x^2} - 4}}.\left( {x - 2} \right) = {x^2}\\
\to \dfrac{{{x^2} + 8}}{{x + 2}} = {x^2}\\
\to {x^2} + 8 = {x^3} + 2{x^2}\\
\to {x^3} + {x^2} - 8 = 0\\
\to x = 1,716188659
\end{array}\)
