Tìm x 1+1/3+1/6+1/10+….+2/x(x+1)=4040/2021 23/09/2021 Bởi Kennedy Tìm x 1+1/3+1/6+1/10+….+2/x(x+1)=4040/2021
`1 + 1/3 + 1/6 + 1/10 +…+ 2/(x(x+1)) = 4040/2021` `⇔ 2 . (1/2 + 1/6 + 1/12 + 1/20 +…+ 1/(x(x+1))) = 4040/2021` `⇔ 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +…+ 1/(x(x+1)) = 2020/2021` `⇔ 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/x – 1/(x+1) = 2020/2021` `⇔ 1 – 1/(x+1) = 2020/2021` `⇔ 1/(x+1) = 1/2021` `⇔ x + 1 = 2021 ⇔ x = 2020` Bình luận
Đáp án:
Giải thích các bước giải:
Đây bn
`1 + 1/3 + 1/6 + 1/10 +…+ 2/(x(x+1)) = 4040/2021`
`⇔ 2 . (1/2 + 1/6 + 1/12 + 1/20 +…+ 1/(x(x+1))) = 4040/2021`
`⇔ 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +…+ 1/(x(x+1)) = 2020/2021`
`⇔ 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/x – 1/(x+1) = 2020/2021`
`⇔ 1 – 1/(x+1) = 2020/2021`
`⇔ 1/(x+1) = 1/2021`
`⇔ x + 1 = 2021 ⇔ x = 2020`