e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

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

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

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

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

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

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)

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

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

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

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

\(\Rightarrow x=2\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

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

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

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

\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

\(A=\frac{22}{45}:2=\frac{11}{45}\)

thay A vào ta được

\(\frac{11}{45}.x=\frac{23}{45}\)

        \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

x = 23/11

bạn vào câu hỏi tương tự tham khảo nha

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

\(\Leftrightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)

\(\Leftrightarrow x\approx0,0648\)

Ta có:

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)

\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)

\(\frac{11}{45}.x=\frac{23}{45}\)

 \(x=\frac{23}{45}\div\frac{11}{45}\)

\(x=\frac{23}{11}\)

=> \(x=\frac{23}{11}\)

x = 2 nha bạn !!!

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

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

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

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

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