Các câu hỏi tương tự
Rút gọn \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2018}+\sqrt{2019}}{1+\sqrt{2018}+\sqrt{2019}}\)
\(A=\frac{1}{\sqrt{1.2018}}+\frac{1}{\sqrt{2.2017}}+....+\frac{1}{\sqrt{k.\left(2018-k+1\right)}}+....+\frac{1}{\sqrt{2018.1}}\)
So sánh \(A\) với \(2.\frac{2018}{2019}\)
Tính:
A= \(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+ \(\frac{1}{3\sqrt{2}+2\sqrt{3}}\)+....+ \(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Tính \(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
rút gon:\(\frac{1+2019\sqrt{2018}-2018\sqrt{2019}}{\sqrt{2018}+\sqrt{2019}+\sqrt{2018.2019}}\)
tính \(\sqrt{1+2018^2+\frac{2018}{2019}^2}+\frac{2018}{2019}\)
Tính:
\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
Chứng minh :
A = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
là 1 số hữu tỉ .
tính
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{2018}-\sqrt{2019}}\)