\[\begin{array}{l}
\sqrt {{x^2} + 1} + \sqrt {{x^2} + 3x + 2} = \sqrt {{x^2} - x} + \left| {x - 1} \right|\\
DK:\left\{ \begin{array}{l}
{x^2} + 3x + 2 \ge 0\\
{x^2} - x \ge 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \le - 2
\end{array} \right.\\
TH1\,\,\,\,\,\,\,\,x \ge 1\\
PT \leftrightarrow \sqrt {{x^2} + 1} + \sqrt {{x^2} + 3x + 2} = \sqrt {{x^2} - x} + x - 1\\
\leftrightarrow 2{x^2} + 3x + 3 + 2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} + 3x + 2} \right)} = 2{x^2} - 3x + 1 + 2\sqrt {\left( {{x^2} - x} \right)} \left( {x - 1} \right)\\
\leftrightarrow 6x + 2 = 2\sqrt {\left( {{x^2} - x} \right)} \left( {x - 1} \right) - 2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} + 3x + 2} \right)} \,\,\,\,\\
+ 2\sqrt {\left( {{x^2} - x} \right)} \left( {x - 1} \right) - 2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} + 3x + 2} \right)} \,\,\,\, < 0\left( {moi\,\,x} \right)\\
\to PT:VN\\
TH2\,\,\,\,\,\,x \le - 2\\
PT \leftrightarrow \sqrt {{x^2} + 1} + \sqrt {{x^2} + 3x + 2} = \sqrt {{x^2} - x} + 1 - x\\
\leftrightarrow 6x + 2 = 2\sqrt {\left( {{x^2} - x} \right)} \left( {1 - x} \right) - 2\sqrt {\left( {{x^2} + 1} \right)\left( {{x^2} + 3x + 2} \right)} \\
\leftrightarrow 3x + 1 = \sqrt {x\left( {x - 1} \right){{\left( {x + 1} \right)}^2}} - \sqrt {\left( {{x^2} + 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)} \\
.........
\end{array}\]
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