Đáp án:
$\begin{array}{l}
1a)\sqrt {48} - \sqrt {363} + \sqrt {147} - \dfrac{1}{4}\sqrt {192} \\
= 4\sqrt 3 - 11\sqrt 3 + 7\sqrt 3 - \dfrac{1}{4}.8\sqrt 3 \\
= 4\sqrt 3 - 11\sqrt 3 + 7\sqrt 3 - 2\sqrt 3 \\
= - 2\sqrt 3 \\
b)\left( {\sqrt {14} - \sqrt {10} } \right).\sqrt {6 + \sqrt {35} } \\
= \left( {\sqrt 7 - \sqrt 5 } \right).\sqrt 2 .\sqrt {6 + \sqrt {35} } \\
= \left( {\sqrt 7 - \sqrt 5 } \right).\sqrt {12 + 2.\sqrt 5 .\sqrt 7 } \\
= \left( {\sqrt 7 - \sqrt 5 } \right).\sqrt {{{\left( {\sqrt 7 + \sqrt 5 } \right)}^2}} \\
= \left( {\sqrt 7 - \sqrt 5 } \right).\left( {\sqrt 7 + \sqrt 5 } \right)\\
= 7 - 5\\
= 2\\
c)\dfrac{2}{{3 - \sqrt 7 }}.\sqrt {\dfrac{{6\sqrt 2 - 2\sqrt {14} }}{{3\sqrt 2 + \sqrt {14} }}} \\
= \dfrac{2}{{3 - \sqrt 7 }}.\sqrt {\dfrac{{2\sqrt 2 \left( {3 - \sqrt 7 } \right)}}{{\sqrt 2 \left( {3 + \sqrt 7 } \right)}}} \\
= \dfrac{2}{{3 - \sqrt 7 }}.\dfrac{{\sqrt 2 .\sqrt {3 - \sqrt 7 } }}{{\sqrt {3 + \sqrt 7 } }}\\
= \dfrac{{2\sqrt 2 }}{{\sqrt {\left( {3 - \sqrt 7 } \right).\left( {3 + \sqrt 7 } \right)} }}\\
= \dfrac{{2\sqrt 2 }}{{\sqrt {9 - 7} }}\\
= \dfrac{{2\sqrt 2 }}{{\sqrt 2 }}\\
= 2\\
B2)\\
Dkxd:x \ge - 2\\
\sqrt {{x^2} - 4x + 4} = x + 2\\
\Leftrightarrow {x^2} - 4x + 4 = {x^2} + 4x + 4\\
\Leftrightarrow 8x = 0\\
\Leftrightarrow x = 0\left( {tm} \right)\\
Vậy\,x = 0\\
B3)\\
a = \dfrac{{3 - m}}{5} > 0\\
\Leftrightarrow 3 - m > 0\\
\Leftrightarrow m < 3\\
Vậy\,m < 3
\end{array}$
