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

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

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

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

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

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

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

\(=2\sqrt{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\sqrt{3}\cdot\sqrt{27}\)

\(=\sqrt{3}\cdot\sqrt{3^3}\)

\(=\sqrt{3\cdot3^3}\)

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

\(=\sqrt{9^2}\)

\(=9\)

Với \(a,b>0;a\ne b\)ta có: 

 \(\left(\sqrt{a}-\sqrt{b}\right)^2>0\Leftrightarrow a-2\sqrt{ab}+b>0\Leftrightarrow2\left(a+b\right)>\left(\sqrt{a}+\sqrt{b}\right)^2\)

\(\Leftrightarrow\sqrt{a}+\sqrt{b}< \sqrt{2\left(a+b\right)}\)

Áp dụng ta được: 

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< \sqrt{2\left(2+6\right)}+\sqrt{2\left(12+20\right)}\)

\(=\sqrt{16}+\sqrt{64}=4+8=12\)

Ta có đpcm.