Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {\dfrac{{x + 4\sqrt x + 4}}{{x + \sqrt x - 2}} + \dfrac{{x + \sqrt x }}{{1 - x}}} \right):\left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{{1 - \sqrt x }}} \right)\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right)\\
= \left( {\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}} + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}} \right):\dfrac{{\left( {1 - \sqrt x } \right) - \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \left( {\dfrac{{\sqrt x + 2}}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{1 - \sqrt x }}} \right):\dfrac{{1 - 2\sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}\\
= \dfrac{{ - \left( {\sqrt x + 2} \right) + \sqrt x }}{{1 - \sqrt x }}.\dfrac{{\sqrt x \left( {1 - \sqrt x } \right)}}{{1 - 2\sqrt x }}\\
= \dfrac{{ - 2}}{{1 - \sqrt x }}.\dfrac{{\sqrt x \left( {1 - \sqrt x } \right)}}{{1 - 2\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{2\sqrt x - 1}}\\
*)\\
A \ge \dfrac{{1 + \sqrt {2018} }}{{\sqrt {2018} }}\\
\Leftrightarrow \dfrac{{2\sqrt x }}{{2\sqrt x - 1}} \ge \dfrac{{1 + \sqrt {2018} }}{{\sqrt {2018} }}\\
\Leftrightarrow 2\sqrt {2018} .\sqrt x \ge \left( {1 + \sqrt {2018} } \right)\left( {2\sqrt x - 1} \right)\\
\Leftrightarrow 2\sqrt {2018} .\sqrt x \ge 2\sqrt x - 1 + 2\sqrt {2018} .\sqrt x - \sqrt {2018} \\
\Leftrightarrow 0 \ge 2\sqrt x - 1 - \sqrt {2018} \\
\Leftrightarrow 2\sqrt x \le 1 + \sqrt {2018} \\
\Leftrightarrow \sqrt x \le \dfrac{{1 + \sqrt {2018} }}{2}\\
\Rightarrow 0 \le x \le {\left( {\dfrac{{1 + \sqrt {2018} }}{2}} \right)^2}\\
x \in Z,x \ne 1 \Rightarrow x \in \left\{ {0;2;3;4;5;....527} \right\}
\end{array}\)
Vậy có 527 số nguyên x thỏa mãn.
