Đáp án: B+D+C
Giải thích các bước giải:
$\begin{array}{l}
1)\dfrac{{a + 2\sqrt {ab} + b}}{{\sqrt a + \sqrt b }} - \dfrac{{a - b}}{{\sqrt a - \sqrt b }}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a - \sqrt b }}\\
= \sqrt a + \sqrt b - \left( {\sqrt a + \sqrt b } \right)\\
= 0\\
\Leftrightarrow B\\
2)\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
\Leftrightarrow D\\
3)\dfrac{{\sqrt b {{\left( {\sqrt a - \sqrt b } \right)}^2} - \left( {a\sqrt b - 3b\sqrt a } \right)}}{{{b^2} + b\sqrt {ab} }}\\
= \dfrac{{\sqrt b \left( {a - 2\sqrt {ab} + b} \right) - a\sqrt b + 3b\sqrt a }}{{{b^2} + b\sqrt {ab} }}\\
= \dfrac{{a\sqrt b - 2b\sqrt a + b\sqrt b - a\sqrt b + 3b\sqrt a }}{{b\sqrt b \left( {\sqrt b + \sqrt a } \right)}}\\
= \dfrac{{b\sqrt a + b\sqrt b }}{{b\sqrt b \left( {\sqrt b + \sqrt a } \right)}}\\
= \dfrac{{b\left( {\sqrt b + \sqrt a } \right)}}{{b\sqrt b \left( {\sqrt b + \sqrt a } \right)}}\\
= \dfrac{1}{{\sqrt b }}\\
\Leftrightarrow C
\end{array}$
