Rút gọn:. A= \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Cho A=\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{120}+\sqrt{121}}\)
B=\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{35}}\)
C/m A<B
\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)
\(=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}\)
=\(=\frac{12-6}{6}=\frac{6}{6}=1\)
Vậy A = \(\sqrt{2}\)
b, Có bạn làm câu này rồi, bạn thử tham khảo nhé: https://olm.vn/hoi-dap/question/971018.html
cảm ơn bn nhé