Đáp án:
$\begin{array}{l}
d)\left( {\sqrt 6 + \sqrt 2 } \right)\left( {4 - 2\sqrt 3 } \right)\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).\sqrt 2 .{\left( {\sqrt 3 - 1} \right)^2}.\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 3 + 1} \right).{\left( {\sqrt 3 - 1} \right)^2}.\sqrt {4 + 2\sqrt 3 } \\
= {\left( {\sqrt 3 - 1} \right)^2}.\left( {\sqrt 3 + 1} \right).\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= {\left( {\sqrt 3 - 1} \right)^2}.\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)\\
= {\left( {3 - 1} \right)^2}\\
= 4\\
B2)a)\\
\left( {\dfrac{{a\sqrt a + b\sqrt b }}{{\sqrt a + \sqrt b }} - \sqrt {ab} } \right){\left( {\dfrac{{\sqrt a + \sqrt b }}{{a - 2}}} \right)^2}\\
= \left( {\dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {a - \sqrt {ab} + b} \right)}}{{\sqrt a + \sqrt b }} - \sqrt {ab} } \right){\left( {\dfrac{{\sqrt a + \sqrt b }}{{a - 2}}} \right)^2}\\
= \left( {a - 2\sqrt {ab} + b} \right).{\left( {\dfrac{{\sqrt a + \sqrt b }}{{a - 2}}} \right)^2}\\
= {\left( {\sqrt a - \sqrt b } \right)^2}.\dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}}\\
= \dfrac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}}\\
b)4 - x - \dfrac{{\sqrt {{x^2} - 6x + 9} }}{{3 - x}}\\
= 4 - x - \dfrac{{\sqrt {{{\left( {x - 3} \right)}^2}} }}{{3 - x}}\\
= 4 - x - \dfrac{{\left| {x - 3} \right|}}{{3 - x}}\\
= \left[ \begin{array}{l}
4 - x - \dfrac{{x - 3}}{{3 - x}} = 4 - x + 1 = 5 - x\left( {khi:x > 3} \right)\\
4 - x - \dfrac{{3 - x}}{{3 - x}} = 3 - x\left( {khi:x < 3} \right)
\end{array} \right.
\end{array}$
