Trả lời

\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)

\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)

\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}\)

\(\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{2}\)

Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\left(đpcm\right)\)

1/2 lớn hơn

vì phân số 1/2 có mẫu số nhỏ hơn các phân số kia nên phân số 1/2 sẽ lớn hơn các phân số kia

không biết ?

Ta thấy:  \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}\)

               \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\)

               \(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}\)

               \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)

Do đó:   \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}\cdot3+\frac{1}{45}\cdot3=\frac{1}{2}\)

\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}<\frac{1}{2}\)

Ta có: Gọi dãy số cần chứng minh là A

\(A<\frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)\)

\(A<\frac{1}{3}+\frac{3}{30}+\frac{4}{60}\)

\(A<\frac{10}{30}+\frac{3}{30}+\frac{2}{30}\)

\(A<\frac{15}{30}=\frac{1}{2}\)

Vậy \(A<\frac{1}{2}\)

k nha

\(T=\left(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)

\(T

Nhận xét:

\(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{10}\)

\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}+\frac{1}{45}+\frac{1}{45}=\frac{1}{15}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)

Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\) (Đpcm)

\(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}

Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\)\(\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)

\(=\frac{1}{3}+\frac{3}{30}+\frac{3}{45}=\frac{1}{2}\)

\(\Rightarrow\)\(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)

\(\RightarrowĐPCM\)

\(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{47}+\frac{1}{47}+\frac{1}{47}\)

\(=\frac{1}{3}+3.\frac{1}{31}+3.\frac{1}{47}=\frac{1}{3}+\frac{3}{31}+\frac{3}{47}=\frac{2159}{4371}< \frac{1}{2}\left(đpcm\right).\)