mình biết

\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{2015.2016}\)

\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{2015}-\frac{1}{2016}\)

ta rút gọn được

\(\Rightarrow\frac{1}{10}-\frac{1}{2016}\)

\(\Rightarrow\frac{1003}{10080}\)

= 7.( \(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-....-\frac{1}{70}\))

= 7.( \(\frac{1}{10}-\frac{1}{70}\))

= 7.(\(\frac{7}{70}-\frac{1}{70}\))

= 7.\(\frac{6}{70}\)

= \(\frac{3}{5}\)

caau hoi tuong tu cungx cos

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(C=7.\frac{3}{35}\)

\(C=\frac{3}{5}\)

Mình nói lí thuyết cho nghe:

 Với phân số \(\frac{a-b}{a.b}\)\(\left(VD:\frac{1}{1.2};\frac{1}{2.3};\frac{1}{2015.2016};\frac{3}{15.18};\frac{3}{18.21};\frac{1}{10.11};\frac{1}{11.12};...\right)\)thì:

 \(\frac{b-a}{a.b}=\frac{b}{a.b}-\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\left(VD:\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{3}{15.18}=\frac{1}{15}-\frac{1}{18}\right)\)

ÁP dụng để tính:

 c) \(\Rightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\right)=\frac{3}{15.18}+\frac{3}{18.21}+....+\frac{3}{87.90}\)

\(\Rightarrow\frac{1}{4}C=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}=\frac{1}{15}-\frac{1}{90}\)

=> \(C=\left(\frac{1}{15}-\frac{1}{90}\right).4\)

a,\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(A=1-\frac{1}{2016}\)suy ra \(A=\frac{2015}{2016}\)

b, \(B=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)

\(B=5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(B=5\left(\frac{1}{10}-\frac{1}{70}\right)\)suy ra \(B=5.\frac{3}{35}\)

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

c,\(C=4.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)suy ra \(C=4.\frac{1}{18}\)

\(C=\frac{2}{9}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)

\(A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=1.\left(1-\frac{1}{2016}\right)\)

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

\(A=\frac{2015}{2016}\)

b) \(B=\frac{5}{10.11}+\frac{5}{11.12}+...+\frac{5}{69.70}\)

\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(B=5.\frac{3}{35}\)

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

c) \(C=\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\)

\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(C=4.\frac{1}{18}\)

\(C=\frac{2}{9}\)

A=.....

=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY

TỰ LÀM NHE

\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)

\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)

\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)

\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)

\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)

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

\(C=9-\left(1-\frac{1}{10}\right)\)

\(C=9-\frac{9}{10}=\frac{81}{10}\)

các bạn đều sai hết mà các bạn lại cho những câu hỏi đó đúng 

Lưu ý : . là dấu nhân

ai làm đúng , nhanh mình k

bình chọn đúng

A=10.11+11.12+...+19.20

3A=10.11.3+11.12.3+...+19.20.3=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+19.20(21-18)

3A=10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + 12.13.14 - 11.12.13 +...+ 19.20.21 - 18.19.20

3A=19.20.21-9.10.11 => A=(19.20.21-9.10.11)/3

A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)

=\(7.\frac{3}{35}\)

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

B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

=\(\frac{1}{2}.\frac{2}{75}\)

=\(\frac{1}{75}\)

C : có ở bên dưới rồi, còn A và B thôi

1) A=7(\(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-......+\frac{1}{69}-\frac{1}{70}\) )

 A=7 ( \(\frac{1}{10}-\frac{1}{70}\))

A=7x 6/70

A=3/5