Đáp án:
$\begin{array}{l}
a)3x\left( {x + 1} \right) - \left( {x - 2} \right)\left( {3x + 1} \right) = 12\\
\Leftrightarrow 3{x^2} + 3x - 3{x^2} + 5x + 2 - 12 = 0\\
\Leftrightarrow 8x - 10 = 0\\
\Leftrightarrow x = \dfrac{5}{4}\\
Vậy\,x = \dfrac{5}{4}\\
b)\left( {3{x^2} - x + 1} \right)\left( {x - 1} \right) + {x^2}\left( {4 - 3x} \right) = \dfrac{5}{2}\\
\Leftrightarrow 3{x^3} - 3{x^2} - {x^2} + x + x - 1 + 4{x^2} - 3{x^3} = \dfrac{5}{2}\\
\Leftrightarrow 2x = \dfrac{5}{2} + 1\\
\Leftrightarrow 2x = \dfrac{7}{2}\\
\Leftrightarrow x = \dfrac{7}{4}\\
Vậy\,x = \dfrac{7}{4}\\
c)\left( {x + 3} \right)\left( {5x - 1} \right) = 5\left( {x + 1} \right)\left( {x - 2} \right)\\
\Leftrightarrow 5{x^2} + 14x - 3 = 5{x^2} - 5x - 10\\
\Leftrightarrow 19x = - 7\\
\Leftrightarrow x = \dfrac{{ - 7}}{{19}}\\
Vậy\,x = \dfrac{{ - 7}}{{19}}\\
d)\left( {2x - 3} \right)\left( {x + 1} \right) = 2x\left( {x - 1} \right)\\
\Leftrightarrow 2{x^2} - x - 3 = 2{x^2} - 2x\\
\Leftrightarrow x = 3\\
Vậy\,x = 3\\
e)\dfrac{1}{4}{x^2} - \left( {\dfrac{1}{2}x - 4} \right).\dfrac{1}{2}x = - 14\\
\Leftrightarrow \dfrac{1}{4}{x^2} - \dfrac{1}{4}{x^2} + 2x = - 14\\
\Leftrightarrow 2x = - 14\\
\Leftrightarrow x = - 7\\
Vậy\,x = - 7\\
f)3.\left( {1 - 4x} \right)\left( {x - 1} \right) + 4\left( {3x - 2} \right)\left( {x + 3} \right) = - 27\\
\Leftrightarrow 3.\left( { - 4{x^2} + 5x - 1} \right) + 4.\left( {3{x^2} + 7x - 6} \right) = - 27\\
\Leftrightarrow - 12{x^2} + 15x - 3 + 12{x^2} + 28x - 24 = - 27\\
\Leftrightarrow 43x = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$
