Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a\left( {x - 1} \right)\left( {6x - 3 - 5x - 40} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 43
\end{array} \right.\\
b.\left( {3x + 1} \right)\left( {3x - 1 - 4x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - \frac{1}{3}\\
x = - 2
\end{array} \right.\\
c.\left( {x + 7} \right)\left( {3x - 1 - 7 + x} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 7\\
4x = 8
\end{array} \right. \to \left[ \begin{array}{l}
x = - 7\\
x = 2
\end{array} \right.\\
d.4{x^2} + 4x + 1 = {x^2} - 2x + 1\\
\to 3{x^2} + 6x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 3
\end{array} \right.\\
e.\left[ \begin{array}{l}
x = 0\\
{x^2} - 5x + 6 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
\left( {x - 3} \right)\left( {x - 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3\\
x = 2
\end{array} \right.\\
g.\left( {x - 4} \right)\left( {x + 4} \right)\left( {2x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 4\\
x = - \frac{3}{2}
\end{array} \right.
\end{array}\)
