Đáp án:
g. \( - 4 < x < 2\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ {2;3} \right\}\\
A = \dfrac{{ - 2{x^2} + 4x - 1 + \left( {x + 3} \right)\left( {x - 3} \right) + \left( {2x + 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
= \dfrac{{ - 2{x^2} + 4x - 1 + {x^2} - 9 + 2{x^2} - 3x - 2}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
= \dfrac{{{x^2} + x - 12}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
= \dfrac{{\left( {x - 3} \right)\left( {x + 4} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
= \dfrac{{x + 4}}{{x - 2}}\\
f.A\left( {x - 2} \right) = \left| {x - 4} \right|\\
\to \left| {x - 4} \right| = \dfrac{{x + 4}}{{x - 2}}.\left( {x - 2} \right)\\
\to \left| {x - 4} \right| = x + 4\\
\to \left[ \begin{array}{l}
x - 4 = x + 4\\
x - 4 = - x - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 4 = 4\left( l \right)\\
2x = 0
\end{array} \right.\\
\to x = 0\\
g.A < 0\\
\to \dfrac{{x + 4}}{{x - 2}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 4 > 0\\
x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 4 < 0\\
x - 2 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 4\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 4\\
x > 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to - 4 < x < 2
\end{array}\)
