M=4\1.5+4\5.9+4\.13+,.............+4\213.2017
Giúp mik vs ạ
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)
D=1/1.5+1/5.9+...+1/41.45
4D=4/1.5+4/5.9+...+4/41.45
4D=1-1/5+1/5-1/9+...+1/41-1/45
4D=1-1/45
D=44/45:4=11/45
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Đề bài sai : }\frac{4}{\left(n-4\right)^n}->\frac{4}{\left(n-4\right)^n}\)
\(\text{Ta có :}\)
\(S=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right)n}\)
\(=\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-...-\frac{1}{n-4}+\frac{1}{n}\)
\(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
\(=\frac{3n+5}{5n}\)
\(\text{Vậy ...}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x=\dfrac{13}{2}-2\\ x=\dfrac{9}{2}\\ b,x=\dfrac{4}{5}\times\dfrac{3}{4}\\ x=\dfrac{12}{20}=\dfrac{3}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có dạng tổng quát như thế này nhé:
\(\frac{k}{n\left(n+k\right)}=\frac{1}{n}-\frac{1}{k+n}\)
Trong trường hợp này là \(\frac{-4}{1.5}-\frac{4}{5.9}-...-\frac{4}{\left(n+4\right)n}=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)
Đáp án là: \(\frac{1}{n+4}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2021}{1\cdot5}+\dfrac{2021}{5\cdot9}+...+\dfrac{2021}{x\cdot\left(x+4\right)}=505\)
\(2021\cdot\left(\dfrac{1}{1.5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{x\cdot\left(x+4\right)}\right)=505\)
\(\dfrac{2021}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{x\cdot\left(x+4\right)}\right)=505\)
\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)
\(1-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)
\(\dfrac{1}{x+4}=\dfrac{1}{2021}\)
=> \(x+4=2021\)
=> \(x=2017\)
vậy \(x=2017\)
Ta có: \(\dfrac{2021}{1\cdot5}+\dfrac{2021}{5\cdot9}+...+\dfrac{2021}{x\left(x+4\right)}=505\)
\(\Leftrightarrow\dfrac{2021}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{x\left(x+4\right)}\right)=505\)
\(\Leftrightarrow1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)
\(\Leftrightarrow-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)
\(\Leftrightarrow x+4=\dfrac{-2021}{2020}\)
hay \(x=-\dfrac{10101}{2020}\)
$#trúc$
`4/(1.5) + 4/(5.9) + 4/(9.13) + .... + 4/(2013 . 2017)`
=`1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/2013 - 1/2017`
= `1 - 1/2017`
= `2017/2017 - 1/2017`
= `2016/2017`