Với \(0\le a\le\pi\)

\(A=\sqrt{2+\sqrt{2+2\left(2cos^2\frac{a}{2}-1\right)}}=\sqrt{2+\sqrt{4cos^2\frac{a}{2}}}\)

\(=\sqrt{2+2cos\frac{a}{2}}=\sqrt{2+2\left(2cos^2\frac{a}{4}-1\right)}\)

\(=\sqrt{4cos^2\frac{a}{4}}=2cos\frac{a}{4}\)

\(A=1-cos^2x+2cosx+1=3-\left(cosx-1\right)^2\le3\)

\(A_{max}=3\) khi \(cosx=1\)

\(B=1-sin^2x-2sin^2x-3=-1-\left(sinx+1\right)^2\le-1\)

\(B_{max}=-1\) khi \(sinx=-1\)

\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{2}-1\right)}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{4}-1\right)}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{8}-1\right)}=\sqrt{cos^2\frac{x}{8}}=cos\frac{x}{8}\)

\(B=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2cos^2\frac{a}{2}-1\right)}}}\)

\(=\sqrt{2+\sqrt{2+\sqrt{4cos^2\frac{a}{2}}}}=\sqrt{2+\sqrt{2+2cos\frac{a}{2}}}\)

\(=\sqrt{2+\sqrt{2+2\left(cos^2\frac{a}{4}-1\right)}}=\sqrt{2+\sqrt{4cos^2\frac{a}{4}}}\)

\(=\sqrt{2+2cos\frac{a}{4}}=\sqrt{2+2\left(2cos^2\frac{a}{8}-1\right)}=2cos\frac{a}{8}\)

a: \(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

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

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

b: \(=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{9-6-\sqrt{7+\sqrt{2}}}\)

\(=\sqrt{2+\sqrt{2}}\cdot\sqrt{9-7-\sqrt{2}}\)

\(=\sqrt{2}\)

a, Ta có : \(\left\{{}\begin{matrix}\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\\\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\end{matrix}\right.\)

- Thay lần lượt vào A ta được :

\(A=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)=2.2\sqrt{2}=4\sqrt{2}\)

b, \(B=\sqrt{2+\sqrt{3}}\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}=\sqrt{2+\sqrt{3}}\sqrt{4-2-\sqrt{3}}\)

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

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

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

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

a) Ta có: \(A=\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)

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

\(=2\cdot2\sqrt{2}=4\sqrt{2}\)

a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{2}+2\sqrt{3}}\)

= \(\sqrt{18-12}\)

= \(\sqrt{6}\)

b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}.\sqrt{2-2\sqrt{2-\sqrt{2}}}\)

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

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

= \(\sqrt{4-8+16\sqrt{2}-8}\)

= \(\sqrt{-12+16\sqrt{2}}\)

c) 

\(\left(\sqrt{2}-\sqrt{7}\right).\sqrt{9+2\sqrt{14}}\)

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

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

= \(\left(4-7\right).\left(\sqrt{2}+\sqrt{7}\right)\)

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

Sửa đề: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

Ta có: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

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

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

=1

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

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

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

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

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

học ghê v, đến tận đây r à :V

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

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