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

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

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

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

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

\(=\sqrt{2}\)

\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)

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

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

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

\(\text{a) }\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{8+1+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\\ =\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{13+30\sqrt{2}+30}\\ =\sqrt{43+30\sqrt{2}}\\ =\sqrt{25+18+30\sqrt{2}}\\ =\sqrt{\left(5+\sqrt{18}\right)^2}\\ =5+3\sqrt{2}\)

\(\text{b) }\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\\ =\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\\ =\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\\ =\sqrt{m-1}+1+\sqrt{m-1}-1\\ =2\sqrt{m-1}\)

b, \(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\) + \(\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)

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

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

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

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

= -2

b)\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)

\(\Leftrightarrow\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)

\(\Leftrightarrow\sqrt{m-1}+1+\sqrt{m-1}-1\Leftrightarrow2\sqrt{m-1}\)

Câu 1 phá từng lớp ra :VD\(9+4\sqrt{2}\) =\((\sqrt{2}+2)^2\)

Câu 2:m+2\(\sqrt{m-1}\) =m-1+1+2\(\sqrt{m-1}\) =\((\sqrt{m-1} -1)^2 \)

a)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(\Leftrightarrow\sqrt{13+30\sqrt{2}+30}\)

\(\Leftrightarrow\sqrt{43+30\sqrt{2}}\)

\(\Leftrightarrow\sqrt{\left(3\sqrt{2}+5\right)^2}\)

\(\Leftrightarrow3\sqrt{2}+5\)