Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)

=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)

=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)

=\(\frac{1}{2}-\frac{1}{509}\)

=\(\frac{507}{1018}\)

MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))

làm 2 câu còn lại đi câu đó làm rồi

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=2.\left(1-\dfrac{1}{100}\right)\)

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=3.\left(1-\dfrac{1}{101}\right)\)

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp 

mình đang cần gấp

\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)

Tự tính

\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\frac{102}{103}\)

= \(\frac{374}{103}\)

giup minh nhe

Chị sẽ giúp em nốt mấy bài này, em còn nhận ra chị ko vậy?

\(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{99x101}\)

\(A=2x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{99x101}\right)\)

\(A=2x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=2x\left(1-\frac{1}{101}\right)=2x\frac{100}{101}=\frac{200}{101}\)

------------------------------

\(B=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{2016}\right)\)

\(B=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{2017}{2016}\) (rút gọn từ trên tử xuống dưới mẫu nhé)

\(B=\frac{2017}{2}\)

-------------------------------

\(C=\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{64x67}\)

\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\)

\(C=1-\frac{1}{67}=\frac{67}{67}-\frac{1}{67}=\frac{66}{67}\)

--------------------------------

\(D=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{20}\right)\)

\(D=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{19}{20}\)(chỗ này cũng rút gọn từ trên tử xuống dưới mẫu)

\(D=\frac{1}{20}\)

Đề  phải như này nhé

\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{99.100}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)

  \(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

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

    \(=2.\frac{99}{100}\)

Em cần phần nào nhỉ .

A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)

A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)

A = \(\dfrac{105}{106}\)

B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)

B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)

C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))

C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)

C = \(\dfrac{5}{51}\) 

D = \(\dfrac{1}{2}\) +   \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)

D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)

D = \(\dfrac{8}{9}\)

E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))

E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)

E = \(\dfrac{147}{200}\)

100km

cần lm hộ ko 

mà rem và rem sai xùi