\(S.2=\frac{2}{2.4}+\frac{2}{4.6}+...........+\frac{2}{98.100}\)

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

\(S.2=\frac{1}{2}-\frac{1}{100}\)

\(S.2=\frac{49}{100}\)

\(S=\frac{49}{100}:2\)

\(S=\frac{49}{200}\)

lấy 2 S                

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

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

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)

\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)

\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)

\(=1\cdot3\cdot2\)

\(=6\)

\(A^2=6^2=36\)

\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{12\times14}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{2}\times\frac{3}{7}=\frac{3}{14}\)

:V Làm sai hết rồi sai ngay từ bước đầu tiên.

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\frac{3}{20}\)

\(=\frac{-11}{12}\)

\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)

= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

= \(-\frac{7}{30}\)

\(=-(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10})\)

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

\(=-\frac{7}{30}\)

~ Hok tốt ~

1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10

=1/2-1/10

=5/10-1/10

=4/10=2/5

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}-\frac{1}{10}\)

\(\frac{2}{5}\)

= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

= 1/2-0-0-0-0-0-0-0-0-1/10

=1/2 - 1/10 

=2/5

tớ trước nhé

Bằng 1/4 bạn nhé.

bằng 1/1/4 bạn nha

\(y=\frac{1x2x3+2x4x6+4x8x12+8x16x24}{2x3x4+4x6x8+8x12x16+16x24x32}\)

\(y=\frac{1}{4}\)

tích nha . Đúng 100% luôn.