Đáp án:
d) \(\left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2{\left( {x + 2} \right)^2} - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 0\\
\to \left( {x + 2} \right)\left( {2x + 4 - {x^2} + 2x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 2\\
- {x^2} + 4x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 0\\
x = 4
\end{array} \right.\\
b)\left( {x - 1} \right)\left( {{x^2} + 5x - 2} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^2} + 5x - 2 - {x^2} - x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
4x - 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{4}
\end{array} \right.\\
c){\left( {x - 1} \right)^3} - \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^2} - 2x + 1 - x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
{x^2} - 3x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0\\
x = 3
\end{array} \right.\\
d){x^3} - 2{x^2} - 6{x^2} + 12x + 9x - 18 = 0\\
\to {x^2}\left( {x - 2} \right) - 6x\left( {x - 2} \right) + 9\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {{x^2} - 6x + 9} \right) = 0\\
\to \left( {x - 2} \right){\left( {x - 3} \right)^2} = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
e){x^4} - {x^3} + {x^3} - {x^2} + 2{x^2} - 2x + 8x - 8 = 0\\
\to {x^3}\left( {x - 1} \right) + {x^2}\left( {x - 1} \right) + 2x\left( {x - 1} \right) + 8\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} + {x^2} + 2x + 8} \right) = 0\\
\to \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} - x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\left( {do:{x^2} - x + 4 > 0\forall x} \right)
\end{array}\)
