Đáp án:
\(B = 2\sqrt x - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 1 + \sqrt x - 2 = 2\sqrt x - 1\\
C = \left[ {\dfrac{{\sqrt x + 2 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]:\dfrac{{2\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{2}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{2\sqrt x }}\\
= \dfrac{1}{{\sqrt x }}\\
D = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
E = \dfrac{{3\sqrt x - 3 - \sqrt x - 1 - \sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x + 1}}
\end{array}\)
