Đáp án+Giải thích các bước giải:
$c/$
`(1-1/2)×(1-1/3)×(1-1/4)×...×(1-1/2019)`
`=(2-1)/2×(3-1)/3×(4-1)/4×...×(2019-1)/2019`
`=1/2×2/3×3/4×...×2018/2019`
`=1/2019`
Vậy `(1-1/2)×(1-1/3)×(1-1/4)×...×(1-1/2019)=1/2019`
$d/$
`4/(1×5)+4/(5×9)+4/(9×13)+...+4/(2017×2021)`
`=(5-1)/(1×5)+(9-5)/(5×9)+(13-9)/(9×13)+...+(2021-2017)/(2017×2021)`
`=1-1/5+1/5-1/9+1/9-1/13+...+1/2017-1/2021`
`=1-1/2021`
`=(2021-1)/2021`
`=2020/2021`
Vậy `4/(1×5)+4/(5×9)+4/(9×13)+...+4/(2017×2021)=2020/2021`
