Đáp án:
\(\begin{array}{l}
a)x \in \emptyset \\
b)\left[ \begin{array}{l}
x = 2\\
x = - \dfrac{4}{3}
\end{array} \right.\\
c)\dfrac{{23}}{2} \ge x\\
d) - 2 < x < 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
\dfrac{{x + 1}}{{x + 2}} + \dfrac{5}{{x - 2}} = \dfrac{4}{{{x^2} - 4}} + 1\\
\to \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right) + 5\left( {x + 2} \right) - 4 - {x^2} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to {x^2} - x - 2 + 5x + 10 - {x^2} = 0\\
\to 4x + 8 = 0\\
\to x = - 2\left( l \right)\\
\to x \in \emptyset \\
b)\left| {x + 3} \right| = 2x + 1\\
\to \left[ \begin{array}{l}
x + 3 = 2x + 1\\
x + 3 = - 2x - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
3x = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{4}{3}
\end{array} \right.\\
c)\dfrac{{x + 2}}{3} \ge \dfrac{{2x - 1}}{4} - 1\\
\to \dfrac{{4\left( {x + 2} \right) - 3\left( {2x - 1} \right) + 12}}{{12}} \ge 0\\
\to 4x + 8 - 6x + 3 + 12 \ge 0\\
\to - 2x + 23 \ge 0\\
\to \dfrac{{23}}{2} \ge x\\
d)\dfrac{{2x + 1}}{{x - 1}} < 1\\
\to \dfrac{{2x + 1 - x + 1}}{{x - 1}} < 0\\
\to \dfrac{{x + 2}}{{x - 1}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x - 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 2\\
x < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 2\\
x > 1
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to - 2 < x < 1
\end{array}\)
