Cho $A = \dfrac{{2x}}{{x + 3\sqrt x  + 2}} + \dfrac{{5\sqrt x  + 1}}{{x + 4\sqrt x  + 3}} + \dfrac{{\sqrt x  + 10}}{{x + 5\sqrt x  + 6}}$  với $x \ge 0$. Chọn đáp án đúng.

$A = \dfrac{{2x}}{{x + 3\sqrt x  + 2}} + \dfrac{{5\sqrt x  + 1}}{{x + 4\sqrt x  + 3}} + \dfrac{{\sqrt x  + 10}}{{x + 5\sqrt x  + 6}}$$= \dfrac{{2x}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{{5\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right)}} + \dfrac{{\sqrt x  + 10}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

$= \dfrac{{2x\left( {\sqrt x  + 3} \right) + \left( {5\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right) + \left( {\sqrt x  + 10} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

$= \dfrac{{2x\sqrt x  + 6x + 5x + 11\sqrt x  + 2 + x + 11\sqrt x  + 10}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

$= \dfrac{{2x\sqrt x  + 12x + 22\sqrt x  + 12}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

$= \dfrac{{2x\sqrt x  + 2x + 10x + 10\sqrt x  + 12\sqrt x  + 12}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

$\begin{array}{l} = \dfrac{{2x\left( {\sqrt x  + 1} \right) + 10\sqrt x \left( {\sqrt x  + 1} \right) + 12\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {2x + 10\sqrt x  + 12} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2\left( {x + 5\sqrt x + 6} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{2\left( {x + 2\sqrt x + 3\sqrt x + 6} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{2\left[ {\sqrt x \left( {\sqrt x + 2} \right) + 3\left( {\sqrt x + 2} \right)} \right]}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\= \dfrac{{2\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = 2\end{array}$

Vậy giá trị của $A$ không phụ thuộc vào biến $x.$