Question

Rút gọn biểu thức: \(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

Answer 1

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

Answer 2

Ta có:

\(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{2+\sqrt{2}-\sqrt{3}+2-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=1+\frac{\sqrt{2}(2+\sqrt{2}-\sqrt{3})}{2+\sqrt{2}-\sqrt{3}}\)

\(=1+\sqrt{2}\)

Vậy \(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}=1+\sqrt{2}\)

Answer 3

Ta có

\(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}=\frac{2+\sqrt{2}-\sqrt{3}+2-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=1+\frac{\sqrt{2}(2+\sqrt{2}-\sqrt{3})}{2+\sqrt{2}-\sqrt{3}}=1+\sqrt{2}\)

\(\text{Vậy }\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}=1+\sqrt{2}\)