![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(B=2.\left(1-\dfrac{1}{10}\right)\)
\(B=2.\dfrac{9}{10}\)
\(B=\dfrac{9}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi A=................................
=>2A=2/1.2.3+2/2.3.4+.....+2/8.9.10
2A=1/1.2-1/2.3+1/2.3-...+1/8.9-1/9.10
2A=1/1.2-1/9.10=22/45 =>A=11/45
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của Kudo Shinichi - Toán lớp 6 - Học toán với OnlineMath
Tham khảo tại link trên chỉ cần nhấn vào .
Chúc bạn học tốt
- Gọi \(Z=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)
\(2Z=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2Z=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2Z=\frac{1}{1.2}-\frac{1}{9.10}\)
\(2Z=\frac{22}{45}\)
\(\Rightarrow\frac{22}{45}.x=\frac{22}{45}\)
\(x=\frac{22}{45}:\frac{22}{45}\)
\(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Gọi tổng trong ngoặc là $A$
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{10-8}{8.9.10}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$
$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{22}{45}$
Vậy $\frac{22}{45}x=\frac{23}{45}$
$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$
![](https://rs.olm.vn/images/avt/0.png?1311)
$x$ ở cuối là sao đây bạn? Nhân riêng với $\frac{1}{8.9.10}$ à?
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\cdot x=\frac{22}{45}\)
\(\Rightarrow1-\frac{1}{10}.x=\frac{22}{45}\)
\(\Rightarrow\frac{9}{10}.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{9}{10}\)
\(\Rightarrow x=\frac{22}{45}.\frac{10}{9}=\frac{22.10}{45.9}=\frac{44}{81}\)
=>x=\(\frac{44}{81}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(A=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{11}{45}\).
Phương trình tương đương với:
\(\frac{11}{45}x=\frac{22}{45}\Leftrightarrow x=2\).
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}...+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{11}{45}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)
\(=\frac{1}{2}.\frac{22}{45}\)
\(=\frac{11}{45}\)