Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
\sqrt {5{x^2} + 10x + 1} = 7 - 2x - {x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {5{x^2} + 10x + 1 \ge 0} \right)\\
\Leftrightarrow {x^2} + 2x - 7 + \sqrt {5{x^2} + 10x + 1} = 0\\
\Leftrightarrow 5{x^2} + 10x - 35 + 5\sqrt {5{x^2} + 10x + 1} = 0\\
\Leftrightarrow \left( {5{x^2} + 10x + 1} \right) + 5\sqrt {5{x^2} + 10x + 1} - 36 = 0\\
\Leftrightarrow \left( {\sqrt {5{x^2} + 10x + 1} + 9} \right)\left( {\sqrt {5{x^2} + 10x + 1} - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {5{x^2} + 10x + 1} = - 9\\
\sqrt {5{x^2} + 10x + 1} = 4
\end{array} \right.\\
\Leftrightarrow \sqrt {5{x^2} + 10x + 1} = 4\\
\Leftrightarrow 5{x^2} + 10x + 1 = 16\\
\Leftrightarrow 5{x^2} + 10x - 15 = 0\\
\Leftrightarrow {x^2} + 2x - 3 = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1
\end{array} \right.\\
3,\\
5\sqrt x + \dfrac{5}{{2\sqrt x }} = 2x + \dfrac{1}{{2x}} + 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x > 0} \right)\\
\Leftrightarrow 5.\left( {\sqrt x + \dfrac{1}{{2\sqrt x }}} \right) = 2.\left( {x + \dfrac{1}{{4x}}} \right) + 4\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
t = \sqrt x + \dfrac{1}{{2\sqrt x }}\,\,\,\left( {t > 0} \right)\\
\Rightarrow {t^2} = x + 2.\sqrt x .\dfrac{1}{{2\sqrt x }} + \dfrac{1}{{4x}}\\
\Rightarrow x + \dfrac{1}{{4x}} = {t^2} - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{t^2} \ge 1 \Rightarrow t \ge 1} \right)\\
\left( 1 \right) \Leftrightarrow 5t = 2.\left( {{t^2} - 1} \right) + 4\\
\Leftrightarrow 2{t^2} - 5t + 2 = 0\\
\Leftrightarrow \left( {2t - 1} \right)\left( {t - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = \dfrac{1}{2}\\
t = 2
\end{array} \right.\\
t \ge 1 \Rightarrow t = 2 \Leftrightarrow \sqrt x + \dfrac{1}{{2\sqrt x }} = 2\\
\Leftrightarrow \dfrac{{2x + 1}}{{2\sqrt x }} = 2\\
\Leftrightarrow 2x - 4\sqrt x + 1 = 0\\
\Leftrightarrow \sqrt x = \dfrac{{2 \pm \sqrt 2 }}{2}\\
\Leftrightarrow x = \dfrac{{3 \pm 2\sqrt 2 }}{2}
\end{array}\)
