Bài 2:
\(\dfrac{-2}{3\sqrt{11}}=\dfrac{-6\sqrt{11}}{99}\)
\(\dfrac{3}{\sqrt{7}+4}=\dfrac{3\left(\sqrt{7}-4\right)}{-9}=\dfrac{\sqrt{7}-4}{-3}\)
\(\dfrac{\sqrt{5}+3}{\sqrt{5}-3}=\dfrac{14+6\sqrt{5}}{-4}\)
\(\dfrac{31}{\sqrt{47}}=\dfrac{31\sqrt{47}}{47}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{8-2\sqrt{15}}{2}\)
\(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{3-\sqrt{3}}=3+2\sqrt{2}+\dfrac{3+\sqrt{3}}{6}\)
\(\dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}-\dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\dfrac{9-2\sqrt{14}}{9}-\dfrac{9+2\sqrt{14}}{5}\)
\(\dfrac{2}{1+\sqrt{5}}+\dfrac{2}{1-\sqrt{5}}=\dfrac{2\left(1-\sqrt{5}\right)}{6}+\dfrac{2\left(1+\sqrt{5}\right)}{-4}=\dfrac{1-\sqrt{5}}{3}+\dfrac{1+\sqrt{5}}{-2}\)
\(\dfrac{-2}{3\sqrt{11}}=\dfrac{-2\sqrt{11}}{33}\)
\(\dfrac{3}{\sqrt{7}+4}=\dfrac{3\left(\sqrt{7}-4\right)}{7-8}=12-3\sqrt{7}\)
\(\dfrac{\sqrt{5}+3}{\sqrt{5}-3}=\dfrac{\left(\sqrt{5}+3\right)^2}{5-9}=\dfrac{14+6\sqrt{5}}{-4}=-\dfrac{7+3\sqrt{5}}{2}\)
\(\dfrac{31}{\sqrt{47}}=\dfrac{31\sqrt{47}}{47}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}=\dfrac{8-2\sqrt{15}}{2}=4-\sqrt{15}\)
\(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{3-\sqrt{3}}=\dfrac{3+2\sqrt{2}}{9-8}+\dfrac{3+\sqrt{3}}{9-3}=3+2\sqrt{2}+\dfrac{3+\sqrt{3}}{6}\)
\(\dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}-\dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\dfrac{\left(\sqrt{7}-\sqrt{2}\right)^2-\left(\sqrt{7}+\sqrt{2}\right)^2}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-\sqrt{2}\right)}=\dfrac{4\sqrt{14}}{5}\)
\(\dfrac{2}{1+\sqrt{5}}+\dfrac{2}{1-\sqrt{5}}=\dfrac{2\left(1-\sqrt{5}\right)+2\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=\dfrac{4}{-4}=-1\)
Bài 2:
\(\dfrac{-2}{3\sqrt{11}}=\dfrac{-2\sqrt{11}}{33}\)
\(\dfrac{3}{\sqrt{7}+4}=\dfrac{4-\sqrt{7}}{3}\)
\(\dfrac{31}{\sqrt{47}}=\dfrac{31\sqrt{47}}{47}\)
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