Đáp án:
\[x = \dfrac{{\sqrt {105} }}{6}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt {{x^2} + 2x + 9} = 1 + \sqrt {{x^2} - 2x + 9} \\
\Leftrightarrow {\sqrt {{x^2} + 2x + 9} ^2} = {\left( {1 + \sqrt {{x^2} - 2x + 9} } \right)^2}\\
\Leftrightarrow {x^2} + 2x + 9 = 1 + 2.1.\sqrt {{x^2} - 2x + 9} + \left( {{x^2} - 2x + 9} \right)\\
\Leftrightarrow {x^2} + 2x + 9 = {x^2} - 2x + 10 + 2\sqrt {{x^2} - 2x + 9} \\
\Leftrightarrow 4x - 1 = 2\sqrt {{x^2} - 2x + 9} \\
\Leftrightarrow \left\{ \begin{array}{l}
4x - 1 \ge 0\\
{\left( {4x - 1} \right)^2} = 4.\left( {{x^2} - 2x + 9} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{4}\\
16{x^2} - 8x + 1 = 4{x^2} - 8x + 36
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{4}\\
12{x^2} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{4}\\
{x^2} = \dfrac{{35}}{{12}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{4}\\
\left[ \begin{array}{l}
x = \dfrac{{\sqrt {105} }}{6}\\
x = - \dfrac{{\sqrt {105} }}{6}
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{{\sqrt {105} }}{6}
\end{array}\)
Vậy \(x = \dfrac{{\sqrt {105} }}{6}\)
