Ta gọi biểu thức đó là A

Ta có công thức        \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức ta có  

\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)

\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)

\(....................\)

\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)

Ai thấy đúng thì ủng hộ nha !!!

sai rồi kết quả phải bằng 9/10 chứ 

\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{18\cdot20}\)

\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=2\left(\frac{9}{20}\right)\)\(=\frac{18}{20}=\frac{9}{10}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{3}{7}\)

A==2/2x4+2/4x6+2/6x8+2/8x10

A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10

A=1/2-1/10=2/5.

Mk lam xong dau tien, k nhe

Cx2=2/2-2/4+2/4-2/6+2/6-2/8+2/8-2/10

Cx2=2/2-2/10=10/10+-2/10=4/5

Suy ra:C=4/5:2=4/5x1/2=2/5

=1/2-1/4+1/4-1/6+1/61/8+1/81/10+1/10-1/12

=1/2-1/12

5/12

tk nha

Đặt : \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)

\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{10}-\frac{1}{12}\)

\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow A=\frac{5}{12}\)

bài duoi giải sai nha

Dặt biểu thức trên là A

ta có\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)

\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{12}\)

\(2A=1-\frac{1}{12}=\frac{11}{12}\)

\(A=\frac{11}{12}\)/ 2=\(\frac{11}{24}\)

\(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).y=\frac{1}{3}\)

\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).y=\frac{1}{3}\)

\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}:\frac{1}{2}=\frac{2}{3}\)

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

\(\frac{2}{5}.y=\frac{2}{3}\)

=> \(y=\frac{2}{3}:\frac{2}{5}\)

=>\(y=\frac{3}{5}\)

sao ra 1phan 3

sao ra 1/2

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{3}{7}\)

\(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}+\frac{2}{10\times12}+\frac{2}{12\times14}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}+\frac{1}{10}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{7}{14}-\frac{1}{14}\)

\(=\frac{6}{14}=\frac{3}{7}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)

\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)

\(\Rightarrow S=\frac{11}{20}\)

ko bao giờ 323445465

=1/2-1/4+1/4-1/6+1/6-1/8+...+1/2014-1/2016

=1/2-1/2016

=1007/2016

TK va ket ban nhe

=2/2-2/2016

=1007/1008 : 2

= 1007/2016

\(\frac{1}{2x4}+\frac{1}{4x6}+...+\frac{1}{96x98}+\frac{1}{98x199}=\frac{2}{2x4}+\frac{2}{4x6}+...+\frac{2}{99x100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

A x2 = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{100}\)

A x2 = \(\frac{49}{100}\)

A = \(\frac{49}{200}\)