Đáp án:

Giải thích các bước giải:

\(\begin{array}{l}
a)\quad x^4 +4x^3 -2x^2 -12x + 3 =0\\
\Leftrightarrow (x^4+ 4x^3+4x^2) - 6x^2 - 12x + 3 =0\\
\Leftrightarrow (x^2 + 2x)^2 - 6(x^2 + 2x) + 3 =0\\
Đặt\,\,t = x^2 + 2x\qquad (t \geq -1)\\
\text{Phương trình trở thành:}\\
\quad t^2 - 6t + 3 =0\\
\Leftrightarrow t^2 - 6t + 9 = 6\\
\Leftrightarrow (t -3)^2 =6\\
\Leftrightarrow \left[\begin{array}{l}t - 3 = \sqrt6\\t - 3 = -\sqrt6\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}t =3+ \sqrt6\\t=3 -\sqrt6\end{array}\right.\quad (nhận)\\
+)\quad Với\,t = 3 + \sqrt6\,\,ta\,\,được:\\
\quad x^2 + 2x = 3 + \sqrt6\\
\Leftrightarrow (x + 1)^2 = 4 + \sqrt6\\
\Leftrightarrow \left[\begin{array}{l}x+1 = \sqrt{4+ \sqrt6}\\x+1 = -\sqrt{4+ \sqrt6}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x=-1+ \sqrt{4+ \sqrt6}\\x=-1-\sqrt{4+ \sqrt6}\end{array}\right.\\
+)\quad Với\,t = 3 - \sqrt6\,\,ta\,\,được:\\
\quad x^2 + 2x = 4 - \sqrt6\\
\Leftrightarrow (x + 1)^2 = 4 - \sqrt6\\
\Leftrightarrow \left[\begin{array}{l}x+1 = \sqrt{4- \sqrt6}\\x+1 = -\sqrt{4- \sqrt6}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x=-1+ \sqrt{4- \sqrt6}\\x=-1-\sqrt{4- \sqrt6}\end{array}\right.\\
Vậy\,\,x = -1 \pm \sqrt{4 + \sqrt6}\,\,hoặc\,\,x = - 1 \pm \sqrt{4 - \sqrt6}\\
b)\quad x^2 - 2x + 2 - 3|x-1| =0\\
\Leftrightarrow (x^2 - 2x + 1) - 3|x-1| + 1 =0\\
\Leftrightarrow (|x-1|)^2 - 3|x-1| + 1 =0\\
\Leftrightarrow \left(|x-1| -\dfrac32\right)^2 = \dfrac54\\
\Leftrightarrow \left[\begin{array}{l}|x-1| - \dfrac32 = \dfrac{\sqrt{5}}{2}\\|x-1| - \dfrac32 =- \dfrac{\sqrt{5}}{2}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}|x-1| =\dfrac32+ \dfrac{\sqrt{5}}{2}\\|x-1| = \dfrac32 - \dfrac{\sqrt{5}}{2}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x-1 =\dfrac32+ \dfrac{\sqrt{5}}{2}\\x-1 =-\dfrac32- \dfrac{\sqrt{5}}{2}\\x-1 = \dfrac32 - \dfrac{\sqrt{5}}{2}\\x-1 = -\dfrac32 + \dfrac{\sqrt{5}}{2}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x =\dfrac52+ \dfrac{\sqrt{5}}{2}\\x =-\dfrac12- \dfrac{\sqrt{5}}{2}\\x = \dfrac52 - \dfrac{\sqrt{5}}{2}\\x = -\dfrac12 + \dfrac{\sqrt{5}}{2}\end{array}\right.\\
Vậy\,\,x = \dfrac52 \pm \dfrac{\sqrt{5}}{2}\,\,hoặc\,\,x = -\dfrac12\pm \dfrac{\sqrt5}{2}\\
c)\quad x^4 -4x^2 + 5|x^2 - 2| =-8\\
\Leftrightarrow x^4 - 4x^2 + 4 + 5|x^2 - 2| + 4 =0\\
\Leftrightarrow (|x^2 - 2|)^2 + 5|x^2 - 2| + 4 =0\\
\Leftrightarrow (|x^2 - 2| +1)(|x^2 -2| +4) =0\\
\Leftrightarrow \left[\begin{array}{l}|x^2 -2| = -1\quad \text{(vô nghiệm)}\\|x^2 - 2|= -4\quad \text{(vô nghiệm)}\end{array}\right.\\
\text{Vậy phương trình đã cho vô nghiệm}\\
d)\quad x^4 - 6x^2 +4|x^2 - 3| = -4\\
\Leftrightarrow x^4 - 6x^2 + 9 + 4|x^2 - 3| -5 =0\\
\Leftrightarrow (|x^2 - 3|)^2 + 4|x^2 - 3| - 5 =0\\
\Leftrightarrow (|x^2 - 3|)^2 + 4|x^2 - 3| + 4 = 9\\
\Leftrightarrow (|x^2 - 3| + 2)^2 = 9\\
\Leftrightarrow \left[\begin{array}{l}|x^2 - 3| +2 = 3\\|x^2 - 3| + 2 = -3\quad \text{(vô nghiệm)}\end{array}\right.\\
\Leftrightarrow |x^2 - 3| = 1\\
\Leftrightarrow \left[\begin{array}{l}x^2 - 3 = 1\\x^2 - 3 = -1\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x^2 =4\\x^2 =2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x=\pm 2\\=\pm \sqrt2\end{array}\right.\\
Vậy\,\,x = \pm 2\,\,hoặc\,\,x = \pm \sqrt2
\end{array}\)