Nhan \(\sqrt{\frac{2}{2}}\) vao hai ve cua bieu thuc ta duoc

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

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

\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3}+1}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}+1}\)

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

Toi day quy dong thoi minh lam nhanh nha

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

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

TIck cho mifnh nha 

\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)

\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)

\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)

a) \(A=\frac{\sqrt{5}\sqrt{2}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}-\frac{9\left(\sqrt{10}-1\right)}{9}=\sqrt{10}-\sqrt{10}-1=1\)

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

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

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

Sửa đề câu b

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