Question

Tìm số tự nhiên x : 
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

Answer 1

x=2 đó bạn mình nha

Answer 2

x=2 đó duyệt đi

Answer 3

Đặt A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{44}{90}=\frac{22}{45}\)

Ta có : \(\frac{22}{45}.x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{22}{45}=\frac{1035}{990}=\frac{23}{22}\)

Vậy \(x=\frac{23}{22}\)

 

Answer 4

2 , ủng hộ mk nha