Đáp án:
\(\dfrac{{x + \sqrt x }}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne \left\{ {\dfrac{1}{4};1} \right\}\\
D = \left[ {\dfrac{{2x\sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \left[ {\dfrac{{2x\sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x }}{{\sqrt x - 1}}} \right].\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{2x\sqrt x + x - \sqrt x - \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{2x\sqrt x + x - \sqrt x - x\sqrt x - x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{x\sqrt x - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{x\sqrt x - 2\sqrt x }}{{\left( {x + \sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{x\sqrt x - 2\sqrt x + \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {x + \sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{x\sqrt x - 2\sqrt x + x\sqrt x + x + \sqrt x }}{{\left( {x + \sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{2x\sqrt x + x - \sqrt x }}{{\left( {x + \sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {x + \sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}} = \dfrac{{x + \sqrt x }}{{x + \sqrt x + 1}}
\end{array}\)
