Đáp án:
d. \(\left[ \begin{array}{l}
x = 3\\
x = - 3\\
x = 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^3} - 4x - 14x\left( {x - 2} \right) = 0\\
\to x\left( {{x^2} - 4} \right) - 14x\left( {x - 2} \right) = 0\\
\to x\left( {x - 2} \right)\left( {x + 2} \right) - 14x\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {x\left( {x + 2} \right) - 14x} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 2x - 14x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
{x^2} - 12x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x\left( {x - 12} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 0\\
x = 12
\end{array} \right.\\
b.{x^3} - 7x - 6 = 0\\
\to {x^3} - 3{x^2} + 3{x^2} - 9x + 2x - 6 = 0\\
\to {x^2}\left( {x - 3} \right) + 3x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\\
\to \left( {x - 3} \right)\left( {{x^2} + 3x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
{x^2} + x + 2x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
\left( {x + 1} \right)\left( {x + 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = - 2
\end{array} \right.\\
c.{\left( {x + 3} \right)^2} = 9{\left( {2x - 1} \right)^2}\\
\to \left| {x + 3} \right| = 3\left| {2x - 1} \right|\\
\to \left[ \begin{array}{l}
x + 3 = 6x - 3\\
- x - 3 = 6x - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = 6\\
7x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{6}{5}\\
x = 0
\end{array} \right.\\
d.{x^3} - 9x - 5{x^2} + 45 = 0\\
\to x\left( {{x^2} - 9} \right) - 5\left( {{x^2} - 9} \right) = 0\\
\to \left( {{x^2} - 9} \right)\left( {x - 5} \right) = 0\\
\to \left( {x - 3} \right)\left( {x + 3} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 3\\
x = 5
\end{array} \right.
\end{array}\)
