TL:

vẫn = vô cực

-HT-

Đung thì mình kích 

\(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\)

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

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

chỗ học ko để nói xàm xí ở đây

\(\sin75^o=\cos\left(90^o-75^o\right)=\cos15^o\)

\(\cos66^o=\sin\left(90^o-66^o\right)=\sin24^o\)

\(\tan47^o=\cot\left(90^o-47^o\right)=\cot43^o\)

\(\cot65^o=\tan\left(90^o-65^o\right)=\tan25^o\)

a/ \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

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

b/ \(\sqrt{16+2\sqrt{63}}-\sqrt{16-6\sqrt{7}}\)

\(=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(3-\sqrt{7}\right)^2}\)

\(=3+\sqrt{7}-3+\sqrt{7}=2\sqrt{7}\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)

\(=\sqrt{17-3\sqrt{16.2}}+\sqrt{17-3\sqrt{16.2}}\)

\(=\sqrt{17-12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{9-2.3.2\sqrt{2}+8}+\sqrt{9-2.3.2\sqrt{2}+8}\)

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

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

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

\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)

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

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

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

Ta có: 

\(\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)

\(=1+1+\frac{2010}{2}+1+\frac{2009}{3}+...+1+\frac{1}{2011}\)

\(=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}\)

\(=2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)\)

Suy ra \(A=\frac{1}{2012}\).