Question

[thanh tâm giang]

(1/1⋅2⋅3​+1/2⋅3⋅4​+⋯+1/8⋅9⋅10​)⋅x=2000/2002​

Answer 1

[14456125]

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\ldots+\frac{1}{8.9.10}\right).x=\frac{2000}{2002}\)

\(2\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\cdots+\frac{1}{8.9.10}\right).x=\frac{4000}{2002}\)

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\cdots+\frac{2}{8.9.10}\right).x=\frac{2000}{1001}\)

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\cdots+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{2000}{1001}\)

\(\left(\frac12-\frac{1}{90}\right).x=\frac{2000}{1001}\)

\(\frac{22}{45}.x=\frac{2000}{1001}\)

\(x=\frac{2000}{1001}:\frac{22}{45}\)

\(x=\frac{45000}{11011}\)

Answer 2

[Roblox Player]

**Trả lời:

Đề bài: \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{2000}{2002}\).

Giải:
+ Ta xét tổng \(A=\frac{1}{1 \cdot2 \cdot3}+\frac{1}{2 \cdot3 \cdot4}+\ldots+\frac{1}{8 \cdot9 \cdot10}\).
+ Ta có công thức tổng quát: \(\frac{1}{n \left(\right. n + 1 \left.\right) \left(\right. n + 2 \left.\right)} = \frac{1}{2} \left(\right. \frac{1}{n \left(\right. n + 1 \left.\right)} - \frac{1}{\left(\right. n + 1 \left.\right) \left(\right. n + 2 \left.\right)} \left.\right)\) 
+ Áp dụng công thức này, ta có: 
\(A=\frac{1}{2}\left(\right.\frac{1}{1 \cdot2}-\frac{1}{2 \cdot3}+\frac{1}{2 \cdot3}-\frac{1}{3 \cdot4}+\ldots+\frac{1}{8 \cdot9}-\frac{1}{9 \cdot10}\left.\right)\) 
\(A = \frac{1}{2} \left(\right. \frac{1}{1 \cdot 2} - \frac{1}{9 \cdot 10} \left.\right) = \frac{1}{2} \left(\right. \frac{1}{2} - \frac{1}{90} \left.\right) = \frac{1}{2} \left(\right. \frac{45}{90} - \frac{1}{90} \left.\right) = \frac{1}{2} \cdot \frac{44}{90} = \frac{22}{90} = \frac{11}{45}\) 
\(\Rightarrow A=\frac{11}{45}\).

+ Ta có: \(\frac{11}{45} \cdot x = \frac{2000}{2002} = \frac{1000}{1001}\) \(x = \frac{1000}{1001} : \frac{11}{45} = \frac{1000}{1001} \cdot \frac{45}{11} = \frac{45000}{11011}\) + Vậy \(x = \frac{45000}{11011}\).