Đáp án:
x=3
Giải thích các bước giải:
\(\begin{array}{l}
DK:\,x \ge 2\\
Dat\,\sqrt {x + 1} = a;\sqrt {x - 2} = b\,\,\left( {a,b \ge 0} \right)\\
\Rightarrow {a^2} - {b^2} = 3\\
Ta\,co\,hpt\\
\left\{ \begin{array}{l}
\left( {a - b} \right)\left( {1 + ab} \right) = 3\\
{a^2} - {b^2} = 3
\end{array} \right.\\
\Rightarrow \left( {a - b} \right)\left( {a + b} \right) - \left( {a - b} \right)\left( {1 + ab} \right) = 0\\
\Leftrightarrow \left( {a - b} \right)\left( {a + b - ab - 1} \right) = 0\\
\Leftrightarrow \left( {a - b} \right)\left( {a - 1 - b\left( {a - 1} \right)} \right) = 0\\
\Leftrightarrow \left( {a - b} \right)\left( {a - 1} \right)\left( {b - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = b\\
a = 1\\
b = 1
\end{array} \right.\\
TH1:\,a = b \Leftrightarrow \sqrt {x + 1} = \sqrt {x - 2} \left( {VN} \right)\\
TH2:\,a = 1 \Rightarrow {b^2} = - 2\left( {VN} \right)\\
TH3:\,b = 1 \Rightarrow a = 2\\
Suy\,ra\,\left\{ \begin{array}{l}
\sqrt {x + 1} = 2\\
\sqrt {x - 2} = 1
\end{array} \right. \Leftrightarrow x = 3\left( {tm} \right)
\end{array}\)
