\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.x=1\)

=> \(x=2\)

Vậy x = 2

Chúc bạn học tốt !!!

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

\(A=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{10-8}{8.9.10}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{11}{45}\).

Phương trình tương đương với: 

\(\frac{11}{45}x=\frac{22}{45}\Leftrightarrow x=2\).

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}=1\)

\(\Rightarrow x=2\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\cdot x=\frac{22}{45}\)

\(\Rightarrow1-\frac{1}{10}.x=\frac{22}{45}\)

\(\Rightarrow\frac{9}{10}.x=\frac{22}{45}\)

\(\Rightarrow x=\frac{22}{45}:\frac{9}{10}\)

\(\Rightarrow x=\frac{22}{45}.\frac{10}{9}=\frac{22.10}{45.9}=\frac{44}{81}\)

=>x=\(\frac{44}{81}\)

\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}x=1\)

\(\Rightarrow x=2\)

\(\left(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-.........+\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\right)\).x = \(\frac{22}{45}\)

 \(\left(1-\frac{1}{10}\right).x=\frac{22}{45}\)

 \(\frac{9}{10}.x=\frac{22}{45}\)

\(x=\frac{22}{45}:\frac{9}{10}\)

\(x=\frac{44}{81}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

(Bạn lưu ý chỗ này thì 2 nhân tử ở đầu và cuối của mẫu số cách nhau bao nhiêu đơn vị thì tử là bấy nhiêu nhé )

\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{10}\right)x=\frac{44}{45}\)

\(\Leftrightarrow\left(1+\frac{1}{10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\frac{11x}{10}=\frac{44}{45}\)

\(\Leftrightarrow x=\frac{8}{9}\)