= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15
= 1/3-1/15
=4/15

\(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}=1-\frac{1}{15}=\frac{14}{15}\)

Bài làm:

Ta có: \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(A=1-\frac{1}{15}\)

\(A=\frac{14}{15}\)

có câu tương tự đó bn

Gọi tổng trên là A

Đề sai rồi đề đúng đây : A = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15

A = 1/3 - 1/5 + 1/5 - 1/7 +........+ 1/13 - 1/15

A = 1/3 - 1/15

A = 4/15

1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)

\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-200}{133}\)

2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)

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

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

\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)

\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)

3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)

\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)

\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)

\(=\frac{27}{70}\)

4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)

\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)

\(=\frac{2}{7}+1-1=\frac{2}{7}\)

\(a,\frac{-1}{9}.\frac{15}{22}.\frac{-9}{25}\)

\(=\frac{-1.15.\left(-9\right)}{9.22.25}\)

\(=\frac{3}{110}\)

\(b,\frac{-2}{7}.\left(\frac{5}{13}-\frac{9}{15}\right)-\frac{2}{7}.\frac{8}{13}\)

\(=\frac{-2}{7}.\left(\frac{5}{13}+\frac{8}{13}-\frac{3}{5}\right)\)

\(=\frac{-2}{7}.\left(1-\frac{3}{5}\right)\)

\(=\frac{-2}{7}.\frac{2}{5}\)

\(=\frac{-4}{35}\)

\(c,\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}\right)-\frac{3}{10}.\left(\frac{5}{9}-\frac{3}{5}\right)\)

\(=\frac{3}{10}.\left[\left(\frac{-4}{9}+\frac{2}{5}\right)-\left(\frac{5}{9}-\frac{3}{5}\right)\right]\)

\(=\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}-\frac{5}{9}+\frac{3}{5}\right)\)

\(=\frac{3}{10}.\left[\left(\frac{-4}{9}-\frac{5}{9}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)

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

\(=\frac{3}{10}.0\)

\(=0\)

\(d,\frac{4}{11}-\frac{5}{13}+\frac{7}{11}-\frac{8}{13}\)

\(=\left(\frac{4}{11}+\frac{7}{11}\right)+\left(\frac{-5}{13}-\frac{8}{13}\right)\)

\(=1-1\)

\(=0\)

Học tốt

các bạn giúp mình nhé

bằng 2/15 mình làm trên violimpic rồi

a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)

\(=\dfrac{9}{16}\)

a: =-5/9-4/9+8/15+7/15-2/11=-2/11

b: =10/17+7/17-5/13-8/13+11/25

=11/25

c: =(9/12-2/12)*3/2=7/12*3/2=21/24=7/8

d: =(31/10-25/10)*3-2

=3/5*3-2

=9/5-2

=-1/5