Đáp án+Giải thích các bước giải:

$a)|4x^2 − 1| − 3x|2x − 1| = 0\\ \Leftrightarrow |(2x − 1)(2x + 1)| − 3x|2x − 1| = 0 (1)$

BXD:

\begin{array}{|c|cc|} \hline x&-\infty&&-\dfrac{1}{2}&&\dfrac{1}{2}&&+\infty\\\hline 2x + 1&&-&0&+&|&+& \\\hline 2x − 1&&-&|&-&0&+& \\\hline (2x − 1)(2x + 1) &&+&0&-&0&+&\\\hline \end{array}

$\circledast x <\dfrac{-1}{2}\\ (1) \Leftrightarrow (2x − 1)(2x + 1) + 3x(2x − 1) = 0\\ \Leftrightarrow 4x^2-1 + 6x^2 − 3x = 0\\ \Leftrightarrow  10x^2 − 3x -1= 0\\ \Leftrightarrow \left[\begin{array}{l} x=-\dfrac{1}{5}(L) \\ x=\dfrac{1}{2} (L)\end{array} \right.\\ \circledast \dfrac{-1}{2} \le x \le \dfrac{1}{2}\\ (1) \Leftrightarrow -(2x − 1)(2x + 1) + 3x(2x − 1) = 0\\ \Leftrightarrow 1-4x^2 + 6x^2 −3x = 0\\ \Leftrightarrow  2x^2 −3x+1 = 0\\ \Leftrightarrow  \left[\begin{array}{l} x=\dfrac{1}{2} \\ x=1 (L)\end{array} \right.\\ \circledast x>\dfrac{1}{2}\\ (1) \Leftrightarrow (2x − 1)(2x + 1) - 3x(2x − 1) = 0\\ \Leftrightarrow 4x^2-1 -6x^2+3x= 0\\ \Leftrightarrow -2x^2+3x-1= 0\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{1}{2} (L)\\ x=1 \end{array} \right.$

Vậy phương trình có nghiệm $x=\dfrac{1}{2};x=1$

$b)|x^2-2x|+4=2x\\ \Leftrightarrow |x^2-2x|-(2x-4)=0\\ \Leftrightarrow |x(x-2)|-(x-2)=0(2)$

BXD:

\begin{array}{|c|cc|} \hline x&-\infty&&0&&2&&+\infty\\\hline x&&-&0&+&|&+& \\\hline x − 2&&-&|&-&0&+& \\\hline x(x-2) &&+&0&-&0&+&\\\hline \end{array}

$\circledast \left[\begin{array}{l} x<0 \\ x >2\end{array} \right.\\ (2) \Leftrightarrow x(x-2)-(x-2)=0\\ \Leftrightarrow (x-1)(x-2)=0\\ \Leftrightarrow x=1(L);x=2(L)\\ \circledast 0 \le x \le 2\\ (2) \Leftrightarrow -x(x-2)-(x-2)=0\\ \Leftrightarrow -(x+1)(x-2)=0\\ \Leftrightarrow x=-1(L);x=2$

Vậy phương trình có nghiệm $x=2.$