Đáp án:
a) \(\dfrac{{{x^2}}}{{x - 2}}\)
b) A=-1
c) \(x < 2;x \ne \left\{ { - 2;0} \right\}\)
d) \(\left[ \begin{array}{l}
x = 6\\
x = 4\\
x = 3\\
x = 1
\end{array} \right.\)
e) \(Min = 6\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}:\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) + x + 6 - {x^2}}}{{x\left( {x - 2} \right)}}\\
= \dfrac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}.\dfrac{{x\left( {x - 2} \right)}}{{{x^2} - 4 - {x^2} + x + 6}}\\
= \dfrac{{x\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}.\dfrac{{x\left( {x - 2} \right)}}{{x + 2}}\\
= \dfrac{{{x^2}}}{{x - 2}}\\
b)\left| {2x + 1} \right| = 3\\
\to \left[ \begin{array}{l}
2x + 1 = 3\\
2x + 1 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2\left( l \right)
\end{array} \right.\\
Thay:x = 1\\
\to A = \dfrac{{{1^2}}}{{1 - 2}} = - 1\\
c)A < 0\\
\to \dfrac{{{x^2}}}{{x - 2}} < 0\\
\to x - 2 < 0\left( {do:{x^2} > 0\forall x \ne 0} \right)\\
\to x < 2;x \ne \left\{ { - 2;0} \right\}\\
d)A = \dfrac{{{x^2} - 4 + 4}}{{x - 2}} = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 4}}{{x - 2}}\\
= x + 2 + \dfrac{4}{{x - 2}}\\
A \in Z \to \dfrac{4}{{x - 2}} \in Z\\
\to x - 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 4\\
x - 2 = - 4\\
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = - 2\left( l \right)\\
x = 4\\
x = 0\left( l \right)\\
x = 3\\
x = 1
\end{array} \right.\\
e)A = x + 2 + \dfrac{4}{{x - 2}}\\
= \left( {x - 2} \right) + \dfrac{4}{{x - 2}} + 4\\
Do:x > 2\\
BDT:Co - si:\left( {x - 2} \right) + \dfrac{4}{{x - 2}} \ge 2\sqrt {\left( {x - 2} \right).\dfrac{4}{{x - 2}}} = 2\\
\to \left( {x - 2} \right) + \dfrac{4}{{x - 2}} + 4 \ge 6\\
\to Min = 6\\
\Leftrightarrow \left( {x - 2} \right) = \dfrac{4}{{x - 2}}\\
\to {\left( {x - 2} \right)^2} = 4\\
\to \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
