Question

Tính   \(\sqrt[4]{17+12\sqrt{2}}-\sqrt{2}\)

 

Answer 1

A = \(\sqrt[4]{3^2+2.3.\left(2\sqrt{2}\right)+\left(2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}\)

A = \(\sqrt{3+2\sqrt{2}}-\sqrt{2}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}.1+1}-\sqrt{2}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{2}\)

A = \(\sqrt{2}+1-\sqrt{2}=1\)

Answer 2

\(\sqrt[4]{\left(3+2\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{3+2\sqrt{2}}-\sqrt{2}=\sqrt{\left(1+\sqrt{2}\right)^2}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

Answer 3

\(\sqrt[4]{17+12\sqrt{2}}-\sqrt{2}=\sqrt[4]{9+2.3\sqrt{8}+8}-\sqrt{2}=\sqrt[4]{\left(3+\sqrt{2}\right)^2}-\sqrt{2}\)

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