Question

[Nguyễn Duy Khang]

Thực hiện phép tính ( rút gọn biểu thức ) 

a) \(\dfrac{\sqrt{2}}{2\sqrt{2}-3}\)+\(\dfrac{1}{3+2\sqrt{2}}\)                                                 b) \(\dfrac{1}{\sqrt{10}+\sqrt{6}}\)+\(\dfrac{1}{\sqrt{6}-\sqrt{10}}\)

c) \(\dfrac{-2}{3\sqrt{8}}\)+\(\dfrac{1}{3-2\sqrt{2}}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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

\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)

b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)

\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)

c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)

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

\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)