Đáp án:
\(\begin{array}{l}
B4:\\
a)\left[ \begin{array}{l}
x = - 3\\
x = 4
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a)\left( {x - 3} \right)\left( {x + 3} \right) - \left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {x - 3 - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 3 = 0\\
x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 4
\end{array} \right.\\
b)5x\left( {4x - 3} \right) = \left( {3 - 4x} \right)\left( {3 + 4x} \right)\\
\to 5x\left( {4x - 3} \right) = - \left( {4x - 3} \right)\left( {3 + 4x} \right)\\
\to 5x\left( {4x - 3} \right) + \left( {4x - 3} \right)\left( {3 + 4x} \right) = 0\\
\to \left( {4x - 3} \right)\left( {5x + 3 + 4x} \right) = 0\\
\to \left[ \begin{array}{l}
4x - 3 = 0\\
9x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{4}\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
