Question

tính A = \(\sqrt{1+2018^2+\left(\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)

Answer 1

ta xét : \(\sqrt{a^2+b^2+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}=\sqrt{\left(a+b\right)^2-2ab+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b\right)^2-2.\left(a+b\right).\frac{ab}{a+b}+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b-\frac{ab}{a+b}\right)^2}=\left|a+b-\frac{ab}{a+b}\right|\)

áp dụng vào bài toán :

\(A=\left|1+2018-\frac{2018}{2019}\right|+\frac{2018}{2019}=2019\)