Đáp án:
\[{E_{\min }} = 4 \Leftrightarrow x = 4\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
E = \dfrac{{x + \sqrt x }}{{x - 2\sqrt x + 1}}:\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} - \dfrac{1}{{1 - \sqrt x }} + \dfrac{{2 - x}}{{x - \sqrt x }}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{{2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \sqrt x + \left( {2 - x} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{\left( {x - 1} \right) + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{{{\sqrt x }^2}}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {{{\sqrt x }^2} - 1} \right) + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x - 1}}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x - 1}}\\
= \left[ {\left( {\sqrt x - 1} \right) + \dfrac{1}{{\sqrt x - 1}}} \right] + 2\\
\ge 2.\sqrt {\left( {\sqrt x - 1} \right).\dfrac{1}{{\sqrt x - 1}}} + 2\,\,\,\,\,\,\,\,\,\,\,\left( {x > 1 \Rightarrow \sqrt x - 1 > 0} \right)\\
= 2 + 2 = 4\\
\Rightarrow {E_{\min }} = 4 \Leftrightarrow \sqrt x - 1 = \dfrac{1}{{\sqrt x - 1}} \Leftrightarrow \sqrt x - 1 = 1 \Leftrightarrow x = 4
\end{array}\)
Vậy \({E_{\min }} = 4 \Leftrightarrow x = 4\)
