Đáp án:
f. \(\left[ \begin{array}{l}
x = 6\\
x = - \dfrac{4}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
c.{\left( {3x + 1} \right)^2} = 16{\left( {x + 1} \right)^2}\\
\to \left| {3x + 1} \right| = 4\left| {x + 1} \right|\\
\to \left[ \begin{array}{l}
- 3x - 1 = 4x + 4\\
3x + 1 = 4x + 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
7x = - 5\\
x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{5}{7}\\
x = - 3
\end{array} \right.\\
d.x\left( {9{x^2} - 12x + 4} \right) = 0\\
\to x{\left( {3x - 2} \right)^2} = 0\\
\to \left[ \begin{array}{l}
x = 0\\
3x - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{2}{3}
\end{array} \right.\\
e.3{x^2} - 5x - 8 = 0\\
\to 3{x^2} + 3x - 8x - 8 = 0\\
\to 3x\left( {x + 1} \right) - 8\left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {3x - 8} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = \dfrac{8}{3}
\end{array} \right.\\
f.5{x^2} - 26x - 24 = 0\\
\to 5{x^2} - 30x + 4x - 24 = 0\\
\to 5x\left( {x - 6} \right) + 4\left( {x - 6} \right) = 0\\
\to \left( {x - 6} \right)\left( {5x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 6\\
x = - \dfrac{4}{5}
\end{array} \right.
\end{array}\)
