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a. \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
b. \(\dfrac{26}{5-2\sqrt{3}}=\dfrac{26\left(5+2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}=\dfrac{26\left(5+2\sqrt{3}\right)}{13}=2\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
c. \(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}=\dfrac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\dfrac{3\sqrt{10}}{6}=\dfrac{\sqrt{10}}{2}\)
d. \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\dfrac{23\sqrt{6}}{46}=\dfrac{\sqrt{6}}{2}\)
\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)
\(=\frac{11-\sqrt{10}}{6}\)
\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)
\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..
\(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}=\sqrt{6}+\sqrt{2}+\sqrt{5}\)
cảm ơn pn nhiều nha .