Đáp án:
$\begin{array}{l}
a){x^4} - 4{x^2} + 4x - 1 = 0\\
= > {x^4} - \left( {4{x^2} - 4x + 1} \right) = 0\\
= > {x^4} - {\left( {2x - 1} \right)^2} = 0\\
= > \left( {{x^2} - 2x + 1} \right)\left( {{x^2} + 2x - 1} \right) = 0\\
= > {\left( {x - 1} \right)^2}.\left( {{x^2} + 2x - 1} \right) = 0\\
= > \left[ \begin{array}{l}
x = 1\\
{x^2} + 2x - 1 = 0
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = 1\\
{\left( {x + 1} \right)^2} = 2
\end{array} \right.\\
= > \left[ \begin{array}{l}
x = 1\\
x = - 1 \pm \sqrt 2
\end{array} \right.\\
Vậy\,x = 1;x = - 1 \pm \sqrt 2 \\
b)2{x^3} - {x^2} - 4x + 3 = 0\\
= > 2{x^3} - 2{x^2} + {x^2} - x - 3x + 3 = 0\\
= > 2{x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) - 3\left( {x - 1} \right) = 0\\
= > \left( {x - 1} \right)\left( {2{x^2} + x - 3} \right) = 0\\
= > \left( {x - 1} \right)\left( {2x + 3} \right)\left( {x - 1} \right) = 0\\
= > {\left( {x - 1} \right)^2}\left( {2x + 3} \right) = 0\\
= > x = 1;x = - \frac{3}{2}\\
Vậy\,x = 1;x = - \frac{3}{2}\\
c){\left( {{x^2} + 2} \right)^2} + 2\left( {{x^2} + 2} \right)\left( {2x + 1} \right) = 3{\left( {2x + 1} \right)^2}\\
Dat:{x^2} + 2 = a;2x + 1 = b\\
= > {a^2} + 2ab = 3{b^2}\\
= > {a^2} + 2ab - 3{b^2} = 0\\
= > {a^2} + 3ab - ab - 3{b^2} = 0\\
= > \left( {a + 3b} \right)\left( {a - b} \right) = 0\\
TH1:a + 3b = 0\\
= > {x^2} + 2 + 6x + 3 = 0\\
= > {x^2} + 6x + 5 = 0\\
= > \left( {x + 1} \right)\left( {x + 5} \right) = 0\\
= > x = - 1;x = - 5\\
TH2:a - b = 0\\
= > {x^2} + 2 - 2x - 1 = 0\\
= > {x^2} - 2x + 1 = 0\\
= > x = 1\\
Vậy\,x = - 1;x = - 5;x = 1
\end{array}$
