Giải thích các bước giải:
$\begin{array}{l} B2)a)3x\left( {2x + 1} \right) - 3{x^2} - 6 = 0\\ \Leftrightarrow 6{x^2} + 3x - 3{x^2} - 6 = 0\\ \Leftrightarrow 3{x^2} + 3x - 6 = 0\\ \Leftrightarrow {x^2} + x - 2 = 0\\ \Leftrightarrow {x^2} - x + 2x - 2 = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 0\\ x + 2 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 2 \end{array} \right.\\ \text{Vậy}\,x = 1;x = - 2\\ b)\left( {x - 1} \right)\left( {3 - 2x} \right) + 2x\left( {x - 3} \right) + 10 = 0\\ \Leftrightarrow 3x - 2{x^2} - 3 + 2x + 2{x^2} - 6x + 10 = 0\\ \Leftrightarrow - x + 7 = 0\\ \Leftrightarrow x = 7\\ \text{Vậy}\,x = 7\\ c){x^2} - 4x = 0\\ \Leftrightarrow x\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 4 \end{array} \right.\\ \text{Vậy}\,x = 0;x = 4\\ d){\left( {x - 1} \right)^2} = {\left( {2x + 3} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 2x + 3\\ x - 1 = - 2x - 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 4\\ 3x = - 2 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 4\\ x = - \frac{2}{3} \end{array} \right.\\ \text{Vậy}\,x = - 4;x = - \frac{2}{3}\\ e){x^2} - 3x + 2 = 0\\ \Leftrightarrow {x^2} - x - 2x + 2 = 0\\ \Leftrightarrow x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = 2 \end{array} \right.\\ \text{Vậy}\,x = 1;x = 2 \end{array}$
