Đáp án:
x=1
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 1\\
\sqrt {\left( {x - 1} \right)\left( {x + 2} \right)} + \sqrt {\left( {x - 1} \right)\left( {x + 1} \right)} = 2\sqrt {x\left( {x - 1} \right)} \\
\to \sqrt {x - 1} \left( {\sqrt {x + 2} + \sqrt {x + 1} - 2\sqrt x } \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
\sqrt {x + 2} + \sqrt {x + 1} - 2\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( {TM} \right)\\
\sqrt {x + 2} + \sqrt {x + 1} = 2\sqrt x
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x + 2 + 2\sqrt {{x^2} + 3x + 2} + x + 1 = 4x
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
2x - 3 = 2\sqrt {{x^2} + 3x + 2} \left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to 4{x^2} - 12x + 9 = 4{x^2} + 12x + 8\\
\to 24x = 1\\
\to x = \dfrac{1}{{24}}\left( l \right)\\
KL:x = 1
\end{array}\)
