Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} - 2x + \left| {x - 1} \right| - 1 = 0\\
\Leftrightarrow \left( {{x^2} - 2x + 1} \right) + \left| {x - 1} \right| - 2 = 0\\
\Leftrightarrow {\left| {x - 1} \right|^2} + \left| {x - 1} \right| - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {x - 1} \right| = 1\\
\left| {x - 1} \right| = - 2
\end{array} \right. \Rightarrow \left| {x - 1} \right| = 1 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
b,\\
{x^2} - 2x - 5\left| {x - 1} \right| + 7 = 0\\
\Leftrightarrow \left( {{x^2} - 2x + 1} \right) - 5\left| {x - 1} \right| + 6 = 0\\
\Leftrightarrow {\left| {x - 1} \right|^2} - 5\left| {x - 1} \right| + 6 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {x - 1} \right| = 2\\
\left| {x - 1} \right| = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = 4\\
x = - 2
\end{array} \right.\\
c,{T^2}b\\
d,\\
{x^2} + 4x + 3\left| {x + 2} \right| = 0\\
\Leftrightarrow \left( {{x^2} + 4x + 4} \right) + 3\left| {x + 2} \right| - 4 = 0\\
\Leftrightarrow {\left| {x + 2} \right|^2} + 3\left| {x + 2} \right| - 4 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {x + 2} \right| = 1\\
\left| {x + 2} \right| = - 4
\end{array} \right. \Rightarrow \left| {x + 2} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
