Đáp án:
c. \(x = - \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left[ \begin{array}{l}
4x - 3 = 0\\
x - \dfrac{3}{2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = \dfrac{3}{2}
\end{array} \right.\\
b.2x - 5 = 0\left( {do:{x^2} + 4 > 0\forall x \in R} \right)\\
\to x = \dfrac{5}{2}\\
c.DK:x \ne \left\{ { - 2; - 1} \right\}\\
\dfrac{{3\left( {x + 2} \right) - 2\left( {x + 1} \right) - 4x - 5}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = 0\\
\to 3x + 6 - 2x - 2 - 4x - 5 = 0\\
\to - 3x - 1 = 0\\
\to x = - \dfrac{1}{3}\left( {TM} \right)
\end{array}\)
