sửa đề: \(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\)

\(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\\ =\left(\dfrac{\sqrt{2}+\sqrt{3}+2\sqrt{2}-2\sqrt{3}}{2-3}\right).\sqrt{3}.\sqrt{7-4\sqrt{3}}\\ =\dfrac{3\sqrt{2}-\sqrt{3}}{-1}.\sqrt{3}.\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left(-3\sqrt{2}+\sqrt{3}\right).\sqrt{3}.\left(2-\sqrt{3}\right)\\ =\left(\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}-3\right)\\ =6-3\sqrt{3}-6\sqrt{6}+9\sqrt{2}\)

tới đó thì ko bt còn hay hết

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

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

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

chịu,,, chắc toàn dấu cộng chứ tự nhiên có dấu trừ sao làm

nếu là dấu cộng cx khó đấy

\(a,\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)

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

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{4-6}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{2}.\sqrt{3}}.\dfrac{4\sqrt{3}}{-2}-\dfrac{1}{\sqrt{2}}\)

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

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

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

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

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

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

.

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

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

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

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

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

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

   = \(-\sqrt{3}\)

a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)

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

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

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

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

d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)

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

e, Ghi đúng đề.

\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

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

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

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

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

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

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

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

\(=2\sqrt{2}-\frac{6\sqrt{2}}{6}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)