Question

 giải giúp

Answer 1

a) Ta có: \(\dfrac{1}{2+\sqrt{3}}-\dfrac{2}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}\)

\(=\dfrac{2-\sqrt{3}}{1}-\dfrac{2\left(3-\sqrt{3}\right)}{6}+\dfrac{\sqrt{12}}{6}\)

\(=\dfrac{12-6\sqrt{3}-6+2\sqrt{3}+2\sqrt{3}}{6}\)

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

b) Ta có: \(\left(3-\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}+\sqrt{3}}-3\right)\)

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

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

=5-9=-4

c) Ta có: \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)

\(=\dfrac{12\left(4+\sqrt{10}\right)}{6}-6\cdot\dfrac{\sqrt{5}}{\sqrt{2}}+\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\)

\(=2\left(4+\sqrt{10}\right)-3\sqrt{10}+\sqrt{10}\)

\(=8\)

d) Ta có: \(\sqrt[3]{\left(4+2\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}+1\right)^3}\)

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

Answer 2

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