Đáp án:

$\begin{array}{l}
a){x^2} - 2.\left| x \right| - 3 = 0\\
 \Rightarrow {\left( {\left| x \right|} \right)^2} - 2\left| x \right| - 3 = 0\\
 \Rightarrow {\left( {\left| x \right|} \right)^2} - 3\left| x \right| + \left| x \right| - 3 = 0\\
 \Rightarrow \left( {\left| x \right| - 3} \right)\left( {\left| x \right| + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\left| x \right| = 3\\
\left| x \right| =  - 1\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow x =  \pm 3\\
b){x^2} - 2x + 3 - 3\left| {x - 1} \right| = 0\\
 \Rightarrow {\left( {x - 1} \right)^2} - 3\left| {x - 1} \right| + 2 = 0\\
 \Rightarrow {\left( {\left| {x - 1} \right|} \right)^2} - 3\left| {x - 1} \right| + 2 = 0\\
 \Rightarrow \left( {\left| {x - 1} \right| - 1} \right)\left( {\left| {x - 1} \right| - 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\left| {x - 1} \right| = 1\\
\left| {x - 1} \right| = 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 =  - 1\\
x - 1 = 2\\
x - 1 =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0\\
x = 3\\
x =  - 1
\end{array} \right.
\end{array}$

Đáp án + giải thích các bước giải:

`x^2-2|x|-3=0`

`->x^2+|x|-3|x|-3=0`

`->|x|(|x|+1)-3(|x|+1)=0`

`->(|x|+1)(|x|-3)=0`

`->`\(\left[ \begin{array}{l}|x|+1=0\\|x|-3=0\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}|x|=-1\\|x|=3\end{array} \right.\) 

`->x=3`

b) `x^2-2x+3-3|x-1|=0`

Với `x-1>=0` hay `x>=1`, có:

`x^2-2x+3-3(x-1)=0`

`->x^2-2x+3-3x+3=0`

`->x^2-5x+6=0`

`->x^2-2x-3x+6=0`

`->x(x-2)-3(x-2)=0`

`->(x-2)(x-3)=0`

`->`\(\left[ \begin{array}{l}x=2\\x=3\end{array}(TM) \right.\) 

Với `x-1<0` hay `x<1`, có:

`x^2-2x+3-3(1-x)=0`

`->x^2-2x+3-3+3x=0`

`->x^2+x=0`

`->x(x+1)=0`

`->`\(\left[ \begin{array}{l}x=0\\x=-1\end{array}(TM) \right.\)