1) \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=9+4\sqrt{5}\)

2) \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=-\sqrt{10}\)

3) \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-3\right)}{-\left(\sqrt{5}-3\right)}=-\sqrt{10}\)

4) \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}=\dfrac{\left(6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}=\dfrac{18-6\sqrt{5}-6\sqrt{5}+10}{4}=\dfrac{28-12\sqrt{5}}{4}=7-3\sqrt{5}\)

5)\(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}=\sqrt{5}+2\)

11.

\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\dfrac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\dfrac{\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+5+10\sqrt{5}}{20}+\dfrac{25+5-10\sqrt{5}}{20}\)

\(=3\)

12.

\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{2+\sqrt{3}}{4-3}\)

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)

\(=\sqrt{2}\)

11: Ta có: \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\dfrac{30+10\sqrt{5}+30-10\sqrt{5}}{20}\)

=3

12: Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)

\(=2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)

\(=\sqrt{2}\)

4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{2}\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)

\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)

=2căn 5-2-2căn 5

=-2

d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)

\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

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\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)

\(=\dfrac{5-3}{\sqrt{3}-\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\)

\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\sqrt{3}\)

\(=\sqrt{5}+\sqrt{3}+\sqrt{3}\)

\(=\sqrt{5}+2\sqrt{3}\)

\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\sqrt{5}+\dfrac{5-2\sqrt{15}+3}{5-3}\)

\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)

\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)

\(=\sqrt{5}-\sqrt{15}+4\)

#\(Toru\)