Đáp án:

 17) 9-a

Giải thích các bước giải:

\(\begin{array}{l}
14)\left[ {\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right) - \sqrt x  - 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}} \right].\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \left[ {\dfrac{{x + \sqrt x  - \sqrt x  - 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}} \right].\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{x - 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}.\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{2\left( {x - 2} \right)}}{{\sqrt x \left( {x - 4} \right)}}\\
15)\left[ {\dfrac{{3\sqrt x  + 6 + \sqrt x \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}} \right].\dfrac{{\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  + 6 + x + 2\sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{1}{{\sqrt x  + 3}}\\
 = \dfrac{{x + 5\sqrt x  + 6}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{1}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{1}{{\sqrt x  + 3}}\\
 = \dfrac{1}{{\sqrt x  - 2}}\\
16)\left[ {\dfrac{{\sqrt a \left( {\sqrt a  + 1} \right) + \sqrt a \left( {\sqrt a  - 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}} \right]:\dfrac{{2\sqrt {3a} }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + a - \sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}.\dfrac{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}{{2\sqrt {3a} }}\\
 = \dfrac{{2a}}{{2\sqrt {3a} }}\\
 = \sqrt {\dfrac{a}{3}} \\
17)\left[ {\dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)}}{{\sqrt a  + 1}} + 3} \right].\left[ {3 - \dfrac{{\sqrt a \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}}} \right]\\
 = \left( {\sqrt a  + 3} \right)\left( {3 - \sqrt a } \right)\\
 = 9 - a
\end{array}\)