Đáp án:
d)x=-4
Giải thích các bước giải:
\(\begin{array}{l}
a){\left( {3x + 1} \right)^2} = 4{\left( {x - 1} \right)^2}\\
\to \left| {3x + 1} \right| = 2\left| {x - 1} \right|\\
\to \left[ \begin{array}{l}
3x + 1 = 2x - 2\left( {DK:x \ge 1} \right)\\
3x + 1 = - 2x + 2\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 3\left( l \right)\\
x = \dfrac{1}{5}\left( {TM} \right)
\end{array} \right.\\
b){x^3} - 4{x^2} + 2x - 8 - {x^3} - 6{x^2} - 12x - 8 = - 16\\
\to - 10{x^2} - 10x = 0\\
\to {x^2} + x = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.\\
c)7{x^3} + 7{x^2} - 4{x^2} - 4x + x + 1 = 0\\
\to 7{x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) + \left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {7{x^2} - 4x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
7{x^2} - 4x + 1 = 0\left( {vô nghiệm} \right)
\end{array} \right.\\
\to x = - 1\\
d){x^3} + 3{x^2} + 3x + 1 = - 27\\
\to {\left( {x + 1} \right)^3} = - {3^3}\\
\to x + 1 = - 3\\
\to x = - 4
\end{array}\)
