Giá trị của biểu thức 11.2+12.3+13.4+14.5+...+12018.2019\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{2018.2019}}1.21+2.31+3.41+4.51+...+2018.20191 là
20182019\dfrac{{2018}}{{2019}}20192018
20192018\dfrac{{2019}}{{2018}}20182019
111
12019\dfrac{1}{{2019}}20191
Đáp án: <p><span data-type="math-inline" data-latex="\dfrac{{2018}}{{2019}}">$\dfrac{{2018}}{{2019}}$</span></p>
Phương pháp giải:
Sử dụng tính chất:
Với số tự nhiên n≠0n \ne 0n=0 ta có 1n(n+1)=1n−1n+1\dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{1}{n} - \dfrac{1}{{n + 1}}n(n+1)1=n1−n+11
Lời giải chi tiết:
11.2+12.3+13.4+14.5+...+12018.2019\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{2018.2019}}1.21+2.31+3.41+4.51+...+2018.20191
=1−12+12−13+13−14+14−15+...−12018+12018−12019= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... - \dfrac{1}{{2018}} + \dfrac{1}{{2018}} - \dfrac{1}{{2019}}=1−21+21−31+31−41+41−51+...−20181+20181−20191
=1−12019= 1 - \dfrac{1}{{2019}}=1−20191
=20182019= \dfrac{{2018}}{{2019}}=20192018 .
