Đáp án:
$\begin{array}{l}
a)\sqrt {7\dfrac{7}{{48}}} = \sqrt {\dfrac{{7.48 + 7}}{{48}}} = \sqrt {\dfrac{{7.49}}{{3.16}}} \\
= \dfrac{{\sqrt 7 .7}}{{\sqrt 3 .4}} = \dfrac{{7\sqrt 7 .\sqrt 3 }}{{3.4}}\\
= \dfrac{{7\sqrt {21} }}{{12}}\\
b)A = \sqrt {4 - \sqrt 7 } - \sqrt {4 + \sqrt 7 } \left( {A < 0} \right)\\
\Rightarrow {A^2} = 4 - \sqrt 7 - 2\sqrt {4 - \sqrt 7 } .\sqrt {4 + \sqrt 7 } + 4 + \sqrt 7 \\
= 8 - 2\sqrt {16 - 7} \\
= 8 - 2\sqrt 9 \\
= 8 - 2.3\\
= 2\\
\Rightarrow A = - \sqrt 2 \\
c)\\
B = \dfrac{1}{{1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \dfrac{1}{{\sqrt {99} + \sqrt {100} }}\\
= \dfrac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \dfrac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + ... + \dfrac{{\sqrt {100} - \sqrt {99} }}{{100 - 99}}\\
= \sqrt 2 - 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + .. + \sqrt {100} - \sqrt {99} \\
= \sqrt {100} - 1\\
= 10 - 1\\
= 9
\end{array}$
