Đáp án:
2019
Giải thích các bước giải:
\[\begin{array}{l}
B = \frac{{x\sqrt x - 1}}{{x + \sqrt x + 1}} - \frac{{x - \sqrt x }}{{\sqrt x }} + 2019\,\,\,\,\left( {x > 0} \right)\\
= \frac{{x\sqrt x - 1}}{{x + \sqrt x + 1}} - \frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x }} + 2019\,\\
= \frac{{x\sqrt x - 1}}{{x + \sqrt x + 1}} - \sqrt x + 1 + 2019\\
= \frac{{x\sqrt x - 1 + \left( { - \sqrt x + 2020} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \frac{{x\sqrt x - 1 - x\sqrt x - x - \sqrt x + 2020x + 2020\sqrt x + 2020}}{{x + \sqrt x + 1}}\\
= \frac{{2019x + 2019\sqrt x + 2019}}{{x + \sqrt x + 1}}\\
= \frac{{2019.\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}} = 2019
\end{array}\]
