Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
b.4x + 12 = 2x\left( {x + 3} \right)\\
\to 4\left( {x + 3} \right) - 2x\left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {4 - 2x} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
c.2x + 2 - 10 = 7x + 2\\
\to 5x = - 10\\
\to x = - 2\\
d.6{x^2} + 3x - 4x - 2 = 0\\
\to 3x\left( {2x + 1} \right) - 2\left( {2x + 1} \right) = 0\\
\to \left( {2x + 1} \right)\left( {3x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = \frac{{ - 1}}{2}\\
x = \frac{2}{3}
\end{array} \right.\\
e.DK:x \ne \pm 2\\
\left( {x - 3} \right)\left( {x - 2} \right) + x\left( {x - 2} \right) = \left( {x - 2} \right)\left( {x + 2} \right)\\
\to \left( {x - 2} \right)\left( {x - 3 + x - x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 5\left( {TM} \right)
\end{array} \right.
\end{array}\)
