(1/10-1).(1/11-1).....(1/100-1)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
So sánh: mk làm luôn nè:
Ta có: \(\frac{10}{11}>\frac{10}{11+12};\frac{11}{12}>\frac{11}{11+12}\)
\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10}{11+12}+\frac{11}{11+12}\)
\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10+11}{11+12}\)
MK KO BIẾT ĐÚNG KO NỮA NÊN BN CÓ THỂ THAM KHẢO CỦA CÁC BẠN KHÁC NHÉ.!!
CHÚC BẠN HỌC TỐT. ^_^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}.\dfrac{99}{100}=\dfrac{9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{10}-1\right)\) \(\left(\dfrac{1}{11}-1\right)\)\(\left(\dfrac{1}{12}-1\right)\)...\(\left(\dfrac{1}{100}-1\right)\)
=\(\left(\dfrac{1}{10}-\dfrac{10}{10}\right)\)\(\left(\dfrac{1}{11}-\dfrac{11}{11}\right)\)\(\left(\dfrac{1}{12}-\dfrac{12}{12}\right)\)...\(\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
=\(\dfrac{-9}{10}\) . \(\dfrac{-10}{11}\) . \(\dfrac{-11}{12}\) ... \(\dfrac{-99}{100}\)
= \(\dfrac{9.\left(-10\right).11...99}{\left(-10\right).11...\left(-100\right)}\)=\(\dfrac{9}{-100}\)=\(\dfrac{-9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=-\dfrac{9}{10}\left(-\dfrac{10}{11}\right)\left(-\dfrac{11}{12}\right)...\left(-\dfrac{98}{99}\right)\left(-\dfrac{99}{100}\right)\)
Ta thấy C có \(\left(100-10\right):2+1=46\) thừa số nên số dấu âm là chẵn
Vậy \(C=\dfrac{9}{10}\cdot\dfrac{10}{11}\cdot\dfrac{11}{12}\cdot...\cdot\dfrac{99}{100}=\dfrac{9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chỉ cần 30 số hạng đầu đã lớn hơn 1.
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=>
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
\(=\frac{9}{10}.\frac{10}{11}...\frac{99}{100}=\frac{9}{100}\)