Question

\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}\)

Answer 1

\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}\)

=\(\dfrac{1\left(\sqrt{3}+\sqrt{2}+1\right)}{\left(\sqrt{3}+\sqrt{2}-1\right)\left(\sqrt{3}+\sqrt{2}+1\right)}\)

=\(\dfrac{\sqrt{3}+\sqrt{2}-1}{4+2\sqrt{6}}\)

Answer 2

làm tiếp bạn nhé nhỡ giửi:

= \(\dfrac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(4-2\sqrt{6}\right)}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}\)

=-\(\dfrac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(4-2\sqrt{6}\right)}{8}\)

 

Answer 3

\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}-1}{4+2\sqrt{6}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(4-2\sqrt{6}\right)}{-8}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{2}-1\right)\cdot\left(\sqrt{6}-2\right)}{4}\)