Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)

Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

\(\dfrac{1}{33}>\dfrac{1}{40}\)

\(\dfrac{1}{34}>\dfrac{1}{40}\)

\(\dfrac{1}{35}>\dfrac{1}{40}\)

\(\dfrac{1}{36}>\dfrac{1}{40}\)

\(\dfrac{1}{37}>\dfrac{1}{40}\)

\(\dfrac{1}{38}>\dfrac{1}{40}\)

\(\dfrac{1}{39}>\dfrac{1}{40}\)

\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)

Vậy \(S>\dfrac{1}{4}\)

tham khảo:

99/100

Các bạn giúp mình nhé

\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)

\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)

\(=13\left(1+...+3^7\right)⋮13\)

\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)

\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)

\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)

\(S=4\left(3^2+3^4+3^6+3^8\right)\)

\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)

\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)

\(S=4x\left(1+3^2+...+3^8\right)\)

Vì 4 chia hết cho 4 nên S chia hết cho 4

đề bài sai

Cách đơn giản nhất và lố nhất : 

Cộng tất cả vào rùi tìm S = bao nhiêu 

Rồi so sánh thôi , đã chứng tỏ

\(\dfrac{2}{30}\)

`(1 xx 2 xx 3 xx 4)/(3 xx 4 xx 5 xx 6)`

`= (1 xx 2)/(5 xx 6)`

`= 2/12`

`= 1/6`

Bổ sung:

`(1 xx 2)/(5 xx 6)`

`= 2/30`

`= 1/15`