Đáp án:
a) \(\left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {4x - 9} \right| - 2x = - 3\\
\to \left[ \begin{array}{l}
4x - 9 = - 3 + 2x\\
4x - 9 = 3 - 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 6\\
6x = 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\left( {TM} \right)\\
b){x^2} - 3\left| {x - 1} \right| - 1 = 0\\
\to {x^2} - 1 = 3\left| {x - 1} \right|\\
\to \left[ \begin{array}{l}
3x - 3 = {x^2} - 1\left( {DK:x \ge 1} \right)\\
3x - 3 = - {x^2} + 1\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 3x + 2 = 0\\
{x^2} + 3x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1\\
x = - 4
\end{array} \right.\\
c)\sqrt {{x^2} + x + 2} = x + 1\\
\to {x^2} + x + 2 = {x^2} + 2x + 1\left( {DK:x \ge - 1} \right)\\
\to x = 1
\end{array}\)
