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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
\(=\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)\(< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)\(=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
Vậy ............
Ta có: 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/3 + 3/31 + 3/47 < 1/3 + 3/30 + 3/45
= 1/3 + 1/10 + 1/15 = 1/3 + (1/30) * (3+2) = 1/3 + (1/0) * 5 = 1/3 + 1/6
= (1/6) * (2+1) = (1/6) * 3 = 1/2.
=> 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/2.
Ủng hộ mk nha mina^^
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : 1/31 < 1/30 ; 1/35 < 1/30 ; 1/37 < 1/30
1/47 < 1/45 ; 1/53 < 1/45 ; 1/61 < 1/45
=> 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/3 + 1/30 + 1/30 + 1/30 + 1/45 + 1/45 + 1/45 = 1/2
=> ĐPCM
Tk mk nha
Gọi dãy số cần chứng minh là A
Ta có : \(A< \) \(\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)\)
\(A< \frac{1}{3}+\frac{3}{30}+\frac{4}{60}\)
\(A< \frac{10}{30}+\frac{3}{30}+\frac{2}{30}\)
\(A< \frac{13}{30}+\frac{2}{30}\)
\(A< \frac{15}{30}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\RightarrowĐPCM\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}\cdot3+\frac{1}{45}\cdot3=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}<\frac{1}{2}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1/3 + 1/31 + 1/35 + 1/37 + 1/53 + 1/61
A < 1/3+ ( 1/30+1/30+1/30)+( 1/45+1/45+1/45)
A < 1/3+1/10+1/15
A < 1/2
Chứng tỏ 1/3+1/31+1/35+1/37+1/53+1/61<1/2
k nhé, ủng hộ k, mk trả lời đầu tiên đó
![](https://rs.olm.vn/images/avt/0.png?1311)
\(T=\left(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)
\(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)