Question

Bài 1. Tính P = \(\dfrac{1}{\dfrac{2015}{\dfrac{5}{2015}}}+\dfrac{1}{\dfrac{2016}{\dfrac{5}{2016}}}-\dfrac{1}{\dfrac{2017}{\dfrac{5}{2017}}}\)(chỗ trống mình không viết là dấu "+" và dấu "-" đấy

Answer 1

P = \(\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{\dfrac{5}{2015}+\dfrac{5}{2016}-\dfrac{5}{2017}}=\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{5\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}\right)}=\dfrac{1}{5}\)

Answer 2

\(P=\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{\dfrac{5}{2015}+\dfrac{5}{2016}-\dfrac{5}{2017}}=\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{5\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}\right)}=\dfrac{1}{5}\)

Answer 3

\(P=\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{\dfrac{5}{2015}+\dfrac{5}{2016}-\dfrac{5}{2017}}=\dfrac{\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}}{5.\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}\right)}=\dfrac{1}{5}\)