Question

* rút gọn biểu thức:

a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)

b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)

c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)

d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)

e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)

Answer 1

a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)

= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)

= \(\sqrt{15}.0\)

= 0

b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)

= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)

= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)

= \(\sqrt{5}.\sqrt{5}\)

= 5

c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)

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

d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)

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

= \(2\sqrt{5}.4\)

= \(8\sqrt{5}\)

e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)

= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)

= \(\sqrt{3}.\frac{20}{3}\)

= \(\frac{20\sqrt{3}}{3}\)

Answer 2

a, $\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}$

b, $\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}$

c, $\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}$

d, ${\left(2+\sqrt{5}\right)}^2-{\left(2+\sqrt{5}\right)}^2$

e, $\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}$