Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
+ Ta có 3/10>3/15; 3/11>3/15; 3/12>3/15; 3/13>3/15; 3/14>3/15
=> S> 3/15+3/15+3/15+3/15+3/15=15/15=1
+ Ta có 3/10<3/8; 3/11<3/8; 3/12<3/8; 3/13<3/8; 3/14<3/8
=> S<3/8+3/8+3/8+3/8+3/8=15/8<2
=> 1<S<2
- Ta có:
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)
mà \(\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
\(\Rightarrow\frac{3}{10}+\frac{3}{11}+\frac{3}{13}+\frac{3}{14}>1\) (1)
- Ta có: \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
mà \(\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\) (1)
Từ (1) và (2) => 1<S<2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{15}{14}>1\left(1\right)\)
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}\)
=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}.5=\frac{15}{15}=1\)(1)
Mặt khác:\(\frac{3}{10}=\frac{3}{10};\frac{3}{11}<\frac{3}{10};\frac{3}{12}<\frac{3}{10};\frac{3}{13}<\frac{3}{10};\frac{3}{14}<\frac{3}{10}\)
=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{3}{10}.5=\frac{15}{10}<\frac{20}{10}=2\)(2)
Từ (1) và (2)
=>\(1<\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<2\)(ĐPCM)
3/10+3/11+3/12+3/13+3/14>3/15+3/15+3/15+3/15+3/15=15/15=1
mặt khác: 3/10+3/11+3/12+3/13+3/14<3/10+3/10+3/10+3/10+3/10=15/10<20/10=2
Vậy: 1<S<2
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tham khảo ở link này nhé :
Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
![](https://rs.olm.vn/images/avt/0.png?1311)
Mỗi số ahjng trong S đều lớn hơn \(\frac{3}{15}\) mà S có 5 số hạng nên :
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}.5=\frac{15}{15}=1\)
Vậy S > 1 hay 1 < S (1)
Mỗi số hạng trong S đều nhỏ hơn \(\frac{4}{10}\) mà S có 5 số hạng nên :
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}
![](https://rs.olm.vn/images/avt/0.png?1311)
* Ta có : \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
=> \(S=3\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)\)
Ta có : \(\frac{1}{10}>\frac{1}{15};\frac{1}{11}>\frac{1}{15};\frac{1}{12}>\frac{1}{15};\frac{1}{13}>\frac{1}{15};\frac{1}{14}>\frac{1}{15}\)
=> \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}>\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}=\frac{5}{15}=\frac{1}{3}\)
=> \(S=3\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)>3.\frac{1}{3}=1\)
=> S >1 (1)
** Ta có : \(\frac{1}{11}<\frac{1}{10};\frac{1}{12}<\frac{1}{10};\frac{1}{13}<\frac{1}{10};\frac{1}{14}<\frac{1}{10}\)
=> \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}<\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{5}{10}=\frac{1}{2}\)
=> \(S=3\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)<3.\frac{1}{2}=\frac{3}{2}<\frac{4}{2}=2\)
=> S < 2 (2)
Từ (1) và (2) => 1 < S < 2 (đpcm)
Vì \(\frac{3}{10}=\frac{3}{10};\frac{3}{11}<\frac{3}{10};\frac{3}{12}<\frac{3}{10};\frac{3}{13}<\frac{3}{10};\frac{3}{14}<\frac{3}{10}\)
\(\Rightarrow S<\frac{3}{10}.5\Rightarrow S<\frac{15}{10}\Rightarrow S<\frac{20}{10}\Rightarrow S<2\left(1\right)\)
Vì \(\frac{3}{10}>\frac{3}{14};\frac{3}{11}>\frac{3}{14};\frac{3}{12}>\frac{3}{14};\frac{3}{13}>\frac{3}{14};\frac{3}{14}=\frac{3}{14}\)
\(\Rightarrow S>\frac{3}{14}.5\Rightarrow S>\frac{15}{14}\Rightarrow S>1\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow1
1) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh rằng : S > 1
![](https://rs.olm.vn/images/avt/0.png?1311)
S=3.(\(\frac{1}{10}\)+\(\frac{1}{11}\)+\(\frac{1}{12}\)+\(\frac{1}{13}\)+\(\frac{1}{14}\))>3.(5.\(\frac{1}{14}\))>3.\(\frac{1}{3}\)=1
Vậy:S>1