Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{ - 19 + \sqrt {253} }}{9}\\
x = \dfrac{{ - 19 - \sqrt {253} }}{9}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{\left( {3 + 2x} \right)^3} - \left( {6x - 1} \right)\left( {6x + 1} \right) = {\left( {2x - 1} \right)^3} + {\left( {x + 4} \right)^2} - {x^3} + \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\
\to 27 + 27.2x + 3.3.4{x^2} + 8{x^3} - 36{x^2} + 1 = 8{x^3} - 3.4{x^2} + 3.2x.1 - 1 + {x^2} + 8x + 16\\
- {x^3} + {x^3} + {x^2} + x + {x^2} + x + 1\\
\to 27 + 54x + 36{x^2} - 36{x^2} + 1 = - 12{x^2} + 6x - 1 + {x^2} + 8x + 16 + 2{x^2} + 2x + 1\\
\to 9{x^2} + 38x + 12 = 0\\
\to {\left( {3x} \right)^2} + 2.3x.\dfrac{{19}}{3} + {\left( {\dfrac{{19}}{3}} \right)^2} - \dfrac{{253}}{9} = 0\\
\to {\left( {3x + \dfrac{{19}}{3}} \right)^2} = \dfrac{{253}}{9}\\
\to \left[ \begin{array}{l}
3x + \dfrac{{19}}{3} = \sqrt {\dfrac{{253}}{9}} \\
3x + \dfrac{{19}}{3} = - \sqrt {\dfrac{{253}}{9}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 19 + \sqrt {253} }}{9}\\
x = \dfrac{{ - 19 - \sqrt {253} }}{9}
\end{array} \right.
\end{array}\)