Đáp án:
$\begin{array}{l}
1)a)\left( {5\sqrt {48} - 3\sqrt {27} + 2\sqrt 3 } \right):\sqrt 3 \\
= \left( {5.4\sqrt 3 - 3.3\sqrt 3 + 2\sqrt 3 } \right):\sqrt 3 \\
= 13\sqrt 3 :\sqrt 3 \\
= 13\\
b)\left( {\sqrt {32} - \sqrt {50} + \sqrt 8 } \right):\sqrt 2 \\
= \left( {4\sqrt 2 - 5\sqrt 2 + 2\sqrt 2 } \right):\sqrt 2 \\
= \sqrt 2 :\sqrt 2 \\
= 1\\
c)\left( {\sqrt {\dfrac{9}{2}} + \sqrt {\dfrac{1}{2}} - \sqrt 2 } \right):\sqrt 2 \\
= \left( {\dfrac{{3\sqrt 2 }}{2} + \dfrac{{\sqrt 2 }}{2} - \sqrt 2 } \right):\sqrt 2 \\
= \left( {2\sqrt 2 - \sqrt 2 } \right):\sqrt 2 \\
= \sqrt 2 :\sqrt 2 \\
= 1\\
d)\dfrac{4}{{\sqrt 3 + 1}} + \dfrac{1}{{\sqrt 3 - 2}} + \dfrac{6}{{\sqrt 3 - 3}}\\
= \dfrac{{4\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} + \dfrac{{\sqrt 3 + 2}}{{3 - 4}} + \dfrac{{6\left( {\sqrt 3 + 3} \right)}}{{3 - 9}}\\
= 2\left( {\sqrt 3 - 1} \right) - \sqrt 3 - 2 - \sqrt 3 - 3\\
= - 3\\
B2)\\
a)\sqrt {{x^2} - 2x + 1} = x - 1\left( {dkxd:x \ge 1} \right)\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} = x - 1\\
\Leftrightarrow \left| {x - 1} \right| = x - 1\\
\Leftrightarrow x - 1 \ge 0\\
\Leftrightarrow x \ge 1\\
Vậy\,x \ge 1\\
b)\sqrt {{x^2} - 6x + 10} = 1\\
\Leftrightarrow {x^2} - 6x + 10 = 1\\
\Leftrightarrow {x^2} - 6x + 9 = 0\\
\Leftrightarrow {\left( {x - 3} \right)^2} = 0\\
\Leftrightarrow x = 3\\
Vậy\,x = 3
\end{array}$
