Đáp án:
$\begin{array}{l}
Dkxd:x \ge 1\\
\sqrt {{x^2} - 1} = x + 1 + \sqrt {x + 1} \\
\Rightarrow \sqrt {\left( {x + 1} \right).\left( {x - 1} \right)} = {\left( {\sqrt {x + 1} } \right)^2} + \sqrt {x + 1} \\
\Rightarrow \sqrt {x + 1} .\sqrt {x - 1} - {\left( {\sqrt {x + 1} } \right)^2} - \sqrt {x + 1} = 0\\
\Rightarrow \sqrt {x + 1} .\left( {\sqrt {x - 1} - \sqrt {x + 1} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\left( {ktm} \right)\\
\sqrt {x - 1} - \sqrt {x + 1} - 1 = 0
\end{array} \right.\\
\Rightarrow \sqrt {x - 1} = \sqrt {x + 1} + 1\\
\Rightarrow x - 1 = x + 1 + 2\sqrt {x + 1} + 1\\
\Rightarrow 2\sqrt {x + 1} = - 3\left( {ktm} \right)
\end{array}$
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