Ta có : 

\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)

\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)

\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)

\(=\frac{2015}{1}.\frac{2}{2016}\)

\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)

\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)

Vậy \(x=2015\)

Ủng hộ mk nha !!! ^_^

ê cần giúp ko0

x = 2015 

\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

Chắc Sai kết quả chứ công thức đúng nha!!!...

Fighting!!!...

Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)

\(36.\left(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}-\frac{1}{29}\right).x=\frac{4}{25}\)

Triệt tiêu còn

\(36.\left(\frac{1}{1}-\frac{1}{29}\right).x=\frac{4}{25}\)

từ đây dễ rồi, tình lần lượt rồi tìm x nhé

G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)                         

=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)

=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)

=> G = \(\frac{2.50}{1.51}\)

=> G = \(\frac{100}{51}\)

公关稿黄继线长旧款您

\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)

\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)

\(=\frac{50.2}{51}=\frac{100}{51}\)

\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)

\(=0\)

Bài 1 : 

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)

\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)

\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)

\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)

\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)

\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)

Bài 2 : 

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)