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