Thực hiện phép tính sau :
\((\dfrac{1}{\sqrt{5}-\sqrt{2}} - \dfrac{1}{\sqrt{5}+\sqrt{2}} +1). \dfrac{1}{(\sqrt{2}+1)^2}\)
= \({(}\) \(\dfrac{\sqrt{5}+\sqrt{2}}{(\sqrt{5}-\sqrt{2}).(\sqrt{5}+\sqrt{2})}\) \(- \) \(\dfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2}).(\sqrt{5}-\sqrt{2})} \) + \(1\) \({{)}}\) \(. \dfrac{1}{\sqrt{2^2}+2.\sqrt{2}.1+1^2}\)
=( \(\dfrac{{(\sqrt{5}+\sqrt{2})-(\sqrt{5}-\sqrt{2})}}{{(\sqrt{5}+\sqrt{2}).(\sqrt{5}-\sqrt{2})} }\) + 1 ) \(. \dfrac{1}{3+2\sqrt{2}}\)
= ( \(\dfrac{{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}}}{\sqrt{5^2}-\sqrt{2^2}} \) + 1 ) \(. \dfrac{1}{3+2\sqrt{2}} \)
= \((\dfrac{2\sqrt{2}}{5-2} +1) \) \(. \dfrac{1}{3+2\sqrt{2}} \)
= \(( \dfrac{2\sqrt{2}}{3}+1) \) \(. \dfrac{1}{3+2\sqrt{2}} \)
= \(\dfrac{3+2\sqrt{2}}{3}\) \(. \dfrac{1}{3+2\sqrt{2}} \)
= \(\dfrac{1}{3} \) ( vì rút gọn chéo \(3+2\sqrt{2} \) nha bn )
\(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right).\frac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\frac{\sqrt{5}+2}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{5}-2}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}+1\right).\frac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\frac{\sqrt{5}+2-\sqrt{5}+2}{5-2}\right).\frac{1}{3+2\sqrt{2}}\)
\(=\frac{4}{3}.\frac{1}{3+2\sqrt{2}}=\frac{4}{9+6\sqrt{2}}\)
