Đáp án:
\[A = - \dfrac{{\sqrt 6 }}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{2\sqrt 8 - \sqrt {12} }}{{\sqrt {18} - \sqrt {48} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {30} + \sqrt {162} }}\\
= \dfrac{{2\sqrt {4.2} - \sqrt {4.3} }}{{\sqrt {9.2} - \sqrt {16.3} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {5.6} + \sqrt {6.27} }}\\
= \dfrac{{2\sqrt {{2^2}.2} - \sqrt {{2^2}.3} }}{{\sqrt {{3^2}.2} - \sqrt {{4^2}.3} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt 5 .\sqrt 6 + \sqrt 6 .\sqrt {27} }}\\
= \dfrac{{2.2\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt 6 .\left( {\sqrt 5 + \sqrt {27} } \right)}}\\
= \dfrac{{4\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{\sqrt 2 .\left( {4 - \sqrt 2 .\sqrt 3 } \right)}}{{\sqrt 3 .\left( {\sqrt 3 .\sqrt 2 - 4} \right)}} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{\sqrt 2 .\left( {4 - \sqrt 6 } \right)}}{{\sqrt 3 .\left( {\sqrt 6 - 4} \right)}} - \dfrac{1}{{\sqrt 6 }}\\
= - \dfrac{{\sqrt 2 }}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 6 }}\\
= - \dfrac{{\sqrt 2 .\sqrt 3 }}{{{{\sqrt 3 }^2}}} - \dfrac{{\sqrt 6 }}{{{{\sqrt 6 }^2}}}\\
= - \dfrac{{\sqrt 6 }}{3} - \dfrac{{\sqrt 6 }}{6}\\
= - \dfrac{{\sqrt 6 }}{2}
\end{array}\)
