\(\begin{array}{l}\dfrac{1}{{\sqrt 3 - \sqrt 2 }} \\= \dfrac{{\sqrt 3 + \sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}\\
= \dfrac{{\sqrt 3 + \sqrt 2 }}{{3 - 2}} = \sqrt 3 + \sqrt 2
\end{array}\)
So sánh \(\sqrt 3 + \sqrt 2 \) và \(\sqrt 5 + 1\)Xét \(A = \sqrt 3 + \sqrt 2 >0\)\({A^2} = {(\sqrt 3 + \sqrt 2 )^2} = 5 + 2\sqrt 6 \) \({A^2} - 5 = 2\sqrt 6 \)Xét \(B = \sqrt 5 + 1>0\)\({B^2} = {(\sqrt 5 + 1)^2} = 6 + 2\sqrt 5 \)\({B^2} - 5 = 1 + 2\sqrt 5 \)Ta so sánh: \(2\sqrt 6 \) và \(1 + 2\sqrt 5 \)\({(2\sqrt 6 )^2} = 24=21+3\)\({(1 + 2\sqrt 5 )^2} = 21 + 4\sqrt 5 \)
Do \(3 < 4\sqrt 5 \Leftrightarrow 24 < 21 + 4\sqrt 5 \)
Vậy \(\begin{array}{l}{A^2} - 5 < {B^2} - 5\\\Leftrightarrow {A^2} < {B^2}\\ \Rightarrow A<B\end{array}\)Hay \(\dfrac{1}{{\sqrt 3 - \sqrt 2 }} < \sqrt 5 + 1\).
