Đáp án:
$\begin{array}{l}
f){\left( {x + 2} \right)^2} - x + 4 = 0\\
\Leftrightarrow {x^2} + 4x + 4 - x + 4 = 0\\
\Leftrightarrow {x^2} + 3x + 8 = 0\\
\Leftrightarrow {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{23}}{4} = 0\\
\Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{23}}{4} = 0\left( {vn} \right)\\
Vậy\,x \in \emptyset \\
g){\left( {{x^2} - 2} \right)^2} + 4{\left( {x - 1} \right)^2} - 4\left( {{x^2} - 2} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow {\left( {{x^2} - 2 - 2\left( {x - 1} \right)} \right)^2} = 0\\
\Leftrightarrow {x^2} - 2 - 2x + 2 = 0\\
\Leftrightarrow {x^2} - 2x = 0\\
\Leftrightarrow x\left( {x - 2} \right) = 0\\
\Leftrightarrow x = 0;x = 2\\
Vậy\,x = 0;x = 2
\end{array}$
