Ta có :\(\left(\frac{1}{\sqrt{7-2\sqrt{10}}}-\frac{\sqrt{2}}{\sqrt{10}+2}+1\right):\left(\sqrt{2}+1\right)^2\)
=\(\left(\frac{1}{\sqrt{2-2\sqrt{2}\sqrt{5}+5}}-\frac{\sqrt{2}}{\sqrt{10}+2}+1\right):\left(\sqrt{2}+1\right)^2\)
=\(\left(\frac{1}{\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}}-\frac{\sqrt{2}}{\sqrt{10}+2}+1\right):\left(\sqrt{2}+1\right)^2\)
= \(\left(\frac{1}{|\sqrt{2}-\sqrt{5}|}-\frac{\sqrt{2}}{\sqrt{10}+2}+1\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{10}+2}+1\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\left(\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}-\frac{\sqrt{2}\left(\sqrt{10}-2\right)}{\left(\sqrt{10}+2\right)\left(\sqrt{10}-2\right)}+1\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\left(\frac{\sqrt{5}+\sqrt{2}}{5-2}-\frac{\sqrt{20}-2\sqrt{2}}{10-2^2}+1\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\left(\frac{\sqrt{5}+\sqrt{2}}{3}-\frac{\sqrt{20}-2\sqrt{2}}{6}+1\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\left(\frac{2\left(\sqrt{5}+\sqrt{2}\right)}{6}-\frac{\sqrt{20}-2\sqrt{2}}{6}+\frac{6}{6}\right)\left(\frac{1}{\sqrt{2}+1}\right)^2\)
= \(\frac{2\left(\sqrt{5}+\sqrt{2}\right)-\left(\sqrt{20}-2\sqrt{2}\right)+6}{6}.\frac{1^2}{\left(\sqrt{2}+1\right)^2}\)
= \(\frac{2\sqrt{5}+2\sqrt{2}-\sqrt{20}+2\sqrt{2}+6}{6}.\frac{1}{2+2\sqrt{2}+1}\)
= \(\frac{2\sqrt{5}+2\sqrt{2}-2\sqrt{5}+2\sqrt{2}+6}{6}.\frac{1}{2+2\sqrt{2}+1}\)
= \(\frac{4\sqrt{2}+6}{6}.\frac{1}{2+2\sqrt{2}+1}=\frac{4\sqrt{2}+6}{6\left(2+2\sqrt{2}+1\right)}\)
= \(\frac{4\sqrt{2}+6}{12+12\sqrt{2}+6}=\frac{4\sqrt{2}+6}{12\sqrt{2}+18}=\frac{4\left(\sqrt{2}+1,5\right)}{12\left(\sqrt{2}+1.5\right)}\)
= \(\frac{4}{12}=\frac{1}{3}\)
