Đáp án:
e) \(x = \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge - 2\\
\sqrt {2 + x} \left( { - x + 2x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
2 + x = 0\\
- x + 2x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\left( {TM} \right)\\
x = 1 + \sqrt 2 \left( {TM} \right)\\
x = 1 - \sqrt 2 \left( {TM} \right)
\end{array} \right.\\
b)DK:{x^2} - 3 \ge 0 \to \left[ \begin{array}{l}
x \ge \sqrt 3 \\
x \le - \sqrt 3
\end{array} \right.\\
\sqrt {{x^2} - 3} = 1 \to {x^2} - 3 = 1\\
\to {x^2} = 4\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\left( {TM} \right)\\
c)DK:x \ge - \dfrac{5}{2}\\
\sqrt {3{x^2} - 4x - 4} = \sqrt {2x + 5} \\
\to 3{x^2} - 4x - 4 = 2x + 5\\
\to 3{x^2} - 6x - 9 = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\left( {TM} \right)\\
d)\left| {2x - 4} \right| = 3 - 5x\\
\to \left[ \begin{array}{l}
2x - 4 = 3 - 5x\left( {DK:x \ge 2} \right)\\
2x - 4 = - 3 + 5x\left( {DK:x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
7x = 7\\
3x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - \dfrac{1}{3}\left( {TM} \right)
\end{array} \right.\\
e)DK:x \ne \pm 1\\
\dfrac{{2x}}{{{x^2} - 1}} - 2 = \dfrac{1}{{x + 1}}\\
\to \dfrac{{2x - 2{x^2} + 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{x + 1}}\\
\to \dfrac{{2x - 2{x^2} + 2 - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to - 2{x^2} + x + 3 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\left( {TM} \right)\\
x = - 1\left( l \right)
\end{array} \right.
\end{array}\)
