Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = - 2\\
x = \dfrac{1}{3}\\
x = - \dfrac{7}{3}
\end{array} \right.\]
Giải thích các bước giải:
\(\begin{array}{l}
\left( {3x - 2} \right){\left( {x + 1} \right)^2}.\left( {3x + 8} \right) = - 16\\
\Leftrightarrow \left[ {\left( {3x - 2} \right)\left( {3x + 8} \right)} \right].{\left( {x + 1} \right)^2} = - 16\\
\Leftrightarrow \left( {9{x^2} + 18x - 16} \right).\left( {{x^2} + 2x + 1} \right) = - 16\\
\Leftrightarrow \left( {9.\left( {{x^2} + 2x + 1} \right) - 25} \right).\left( {{x^2} + 2x + 1} \right) = - 16\,\,\,\,\,\,\,\,\left( 1 \right)\\
t = {x^2} + 2x + 1 = {\left( {x + 1} \right)^2} \ge 0,\,\,\,\forall x\\
\left( 1 \right) \Leftrightarrow \left( {9t - 25} \right).t = - 16\\
\Leftrightarrow 9{t^2} - 25t + 16 = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {9t - 16} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = \dfrac{{16}}{9}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{\left( {x + 1} \right)^2} = 1\\
{\left( {x + 1} \right)^2} = \dfrac{{16}}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 1\\
x + 1 = - 1\\
x + 1 = \dfrac{4}{3}\\
x + 1 = - \dfrac{4}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2\\
x = \dfrac{1}{3}\\
x = - \dfrac{7}{3}
\end{array} \right.
\end{array}\)
Em xem lại đề câu b nhé, đề bị thiếu rồi.
