Đáp án:
`2019/1010`.
Giải thích các bước giải:
Đặt `A=1+1/(1+2)=1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+2019)`
`=>`$A=1+\dfrac{1}{3}+\dfrac16+\dfrac1{10}\ +\,.\!.\!.+\ \dfrac{1}{\dfrac{(2019+1).2019}{2}}$
`=>`$A=1+\dfrac{1}{3}+\dfrac16+\dfrac1{10}\ +\,.\!.\!.+\ \dfrac{1}{\dfrac{2019.2020}{2}}$
`=>A/2=1/2+1/6+1/12+1/20+...+1/2019.2020`
`=>A/2=1/1.2+1/2.3+1/3.4+1/4.5+...+1/2019.2020`
`=>A/2=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2019-1/2020`
`=>A/2=1-1/2020`
`=>A/2=2019/2020`
`=>A=2019/2020*2`
`=>A=2019/1010`
Vậy biểu thức có giá trị là `2019/1010`.
