\(=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

\(\sqrt{27}-3\sqrt{48}+2\sqrt{108}-\sqrt{2-\sqrt{3}}^2=3\sqrt{3}-12\sqrt{3}+12\sqrt{3}-2+\sqrt{3}=3\sqrt{3}-2+\sqrt{3}=4\sqrt{3}-2=2\left(2\sqrt{3}-1\right)\)

Ta có: \(\sqrt{27}-3\sqrt{48}+2\sqrt{108}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

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

\(2\sqrt{12}-3\sqrt{48}+2\sqrt{75}\)

\(=2\sqrt{2^2\cdot3}-3\sqrt{2^4\cdot3}+2\sqrt{5^2\cdot3}\)

\(=2\cdot2\sqrt{3}-3\cdot2^2\sqrt{3}+2\cdot5\sqrt{3}\)

\(=4\sqrt{3}-3\cdot4\sqrt{3}+10\sqrt{3}\)

\(=4\sqrt{3}-12\sqrt{3}+10\sqrt{3}\)

\(=\left(4-12+10\right)\sqrt{3}\)

\(=2\sqrt{3}\)

a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)

\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)

\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)

\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)

\(=-8\sqrt{3}\)

b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)

\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)

Ta có: \(\sqrt{12}+\sqrt{27}-\sqrt{3}\)

= \(\sqrt{4}.\sqrt{3}+\sqrt{9}.\sqrt{3}-\sqrt{3}\)

= \(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\)

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

Ý bạn là thế này : \(\sqrt{27+12\sqrt{2}}\)Không rút gọn được bạn nhé ^^

Gợi ý cho bạn : \(\sqrt{17+12\sqrt{2}}=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)

Mk ko hiểu bn ghi đề bài gì cả sao lại căn bậc hai của 27+ 12 căn bậc hai của 2 hai căn gần nhau à

\(\sqrt{12}+\sqrt{27}-\sqrt{3}=\sqrt{3}.\left(2+3-1\right)=4\sqrt{3}\)

\(\sqrt{12}\) + \(\sqrt{27}\) - \(\sqrt{3}\)

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

= \(\sqrt{3}\) . ( 2 + 3 - 1 )

= \(\sqrt{3}\) . 4

= \(\sqrt{48}\)

Cái cuối là căn bậc mấy vậy bạn?

\(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\)

= \(\sqrt{2.2^2}-2\sqrt{4^2.2}+3\sqrt{5^2.2}\)

= \(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\)

= \(9\sqrt{2}\)

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

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

= 0

\(\left(\sqrt{12}+\sqrt{27}-\sqrt{108}\right).2\sqrt{3}\)

= \(\left(2\sqrt{3}+3\sqrt{3}-6\sqrt{3}\right).2\sqrt{3}\)

= \(-\sqrt{3}.2\sqrt{3}\)

= \(-6\)