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NT
7 tháng 9 2021 lúc 14:29

Bài 2: 

a: \(\dfrac{\sqrt{5}}{\sqrt{3}}=\dfrac{\sqrt{15}}{3}\)

b: \(\dfrac{13}{5-\sqrt{3}}=\dfrac{65+13\sqrt{3}}{22}\)

c: \(\dfrac{2\sqrt{2}}{\sqrt{7}+\sqrt{3}}=\dfrac{2\sqrt{14}-2\sqrt{6}}{4}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)

d: \(\dfrac{5-\sqrt{3}}{\sqrt{6}-2\sqrt{2}}=\dfrac{-3\sqrt{6}-7\sqrt{2}}{2}\)

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NT
7 tháng 9 2021 lúc 14:26

Bài 3:

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

\(=\sqrt{3}+1-\sqrt{3}+1\)

=2

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

\(=\dfrac{3+\sqrt{3}}{6}+\dfrac{3-\sqrt{2}}{7}\)

\(=\dfrac{21+7\sqrt{3}+18-6\sqrt{2}}{42}\)

\(=\dfrac{39+7\sqrt{3}-6\sqrt{2}}{42}\)

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