Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\left( {x - 1} \right)\left( {2x + 3} \right) - 2.\left( {x + 2} \right).\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {2{x^2} + 3x - 2x - 3} \right) - 2.\left( {{x^2} + x + 2x + 2} \right) = 0\\
\Leftrightarrow \left( {2{x^2} + x - 3} \right) - 2.\left( {{x^2} + 3x + 2} \right) = 0\\
\Leftrightarrow 2{x^2} + x - 3 - 2{x^2} - 6x - 4 = 0\\
\Leftrightarrow - 5x - 7 = 0\\
\Leftrightarrow x = - \dfrac{7}{5}\\
*)\\
\left( {\dfrac{x}{2} + 3} \right).\left( {2x - 1} \right) - 3 = x.\left( {x - \dfrac{1}{2}} \right)\\
\Leftrightarrow {x^2} - \dfrac{x}{2} + 6x - 3 - 3 = {x^2} - \dfrac{1}{2}x\\
\Leftrightarrow 6x - 6 = 0\\
\Leftrightarrow x = 1
\end{array}\)
