Đáp án:
a) \(\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){\left( {{x^2} - 2} \right)^2} - 2.\left( {{x^2} - 2} \right).2\left( {x - 1} \right) + 4{\left( {x - 1} \right)^2} = 0\\
\to {\left( {{x^2} - 2 - 2\left( {x - 1} \right)} \right)^2} = 0\\
\to {x^2} - 2 - 2x + 2 = 0\\
\to {x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
b)4{x^2} = 81\\
\to {x^2} = \dfrac{{81}}{4}\\
\to \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x = - \dfrac{9}{2}
\end{array} \right.\\
c){x^2} + 4x + 4 - x + 4 = 0\\
\to {x^2} + 3x + 8 = 0\\
\to {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{23}}{4} = 0\\
\to {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{23}}{4} = 0\left( {KTM} \right)\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{23}}{4} > 0\forall m\\
\to x \in \emptyset \\
d){\left( {x + 2} \right)^2} = {\left( {2x - 1} \right)^2}\\
\to \left[ \begin{array}{l}
x + 2 = 2x - 1\\
x + 2 = - 2x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
