Đáp án:
$\begin{array}{l}
1)B1)a)\sqrt {\dfrac{1}{7}} + \dfrac{1}{7}\sqrt {28} - \sqrt {{{\left( {\sqrt 7 - 3} \right)}^2}} \\
= \dfrac{{\sqrt 7 }}{7} + \dfrac{1}{7}.2\sqrt 7 - \left( {3 - \sqrt 7 } \right)\\
= \dfrac{{3\sqrt 7 }}{7} - 3 + \sqrt 7 \\
= \dfrac{{10\sqrt 7 - 21}}{7}\\
b)\dfrac{5}{{\sqrt 5 - 2}} - \dfrac{5}{{\sqrt 5 + 2}}\\
= \dfrac{{5\left( {\sqrt 5 + 2} \right) - 5\left( {\sqrt 5 - 2} \right)}}{{\left( {\sqrt 5 - 2} \right)\left( {\sqrt 5 + 2} \right)}}\\
= \dfrac{{5\sqrt 5 + 10 - 5\sqrt 5 + 10}}{{5 - 4}}\\
= 20\\
B2)\\
a)3 - \sqrt {49 - 14x + {x^2}} = 1\\
\Leftrightarrow 3 - \sqrt {{{\left( {7 - x} \right)}^2}} = 1\\
\Leftrightarrow \sqrt {{{\left( {7 - x} \right)}^2}} = 2\\
\Leftrightarrow \left[ \begin{array}{l}
7 - x = 2\\
7 - x = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 9
\end{array} \right.\\
Vậy\,x = 5;x = 9\\
b)Dkxd:x \ge 3\\
\dfrac{1}{3}\sqrt {9x - 27} + \sqrt {\dfrac{{x - 3}}{4}} + 2 = 0\\
\Leftrightarrow \dfrac{1}{3}.3\sqrt {x - 3} + \dfrac{1}{2}\sqrt {x - 3} + 2 = 0\\
\Leftrightarrow \sqrt {x - 3} + \dfrac{1}{2}\sqrt {x - 3} = - 2\\
\Leftrightarrow \dfrac{3}{2}\sqrt {x - 3} = - 2\\
\Leftrightarrow \sqrt {x - 3} = - \dfrac{4}{3}\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$
