`[-1]/2+[-1]/6+[-1]/12+[-1]/20+[-1]/30+[-1]/42+[-1]/56+[-1]/72+[-1]/90`

`=(-1)(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`

`=(-1)(1-1/2+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)(1-1/10)`

`=(-1). 9/10=-9/10`

A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{6}\)+ \(\dfrac{-1}{12}\)+ \(\dfrac{-1}{20}\)+ \(\dfrac{-1}{30}\)+ \(\dfrac{-1}{42}\)+ \(\dfrac{-1}{56}\)+ \(\dfrac{-1}{72}\)+ \(\dfrac{-1}{90}\)

A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{2\times3}\)+ \(\dfrac{-1}{3\times4}\)+ \(\dfrac{-1}{4\times5}\)+ \(\dfrac{-1}{5\times6}\)+ \(\dfrac{-1}{6\times7}\)+ \(\dfrac{-1}{7\times8}\)+ \(\dfrac{-1}{8\times9}\)+ + \(\dfrac{-1}{9\times10}\)

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

A = -(1-\(\dfrac{1}{10}\))

A = \(\dfrac{-9}{10}\)

ok luôn.hay thì like nha

ta có

1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72

=1/1*2+1/2*3+1/3*4+...+1/8*9

=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

=1-1/9

=8/9

\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)

\(x=\dfrac{2}{7}\)

ta nhận thấy

1/2=1-1/2

1/6=1/2-1/3

1/12=1/3-1/4

1/20=1/4-1/5

1/30=1/5-1/6

1/42=1/6-1/7

ta có:

1/2+1/6+1/12+1/20+1/30+1/42

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

bn tự hiểu nha

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

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)

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

\(A=\dfrac{367}{420}\approx0,87\)

=9/10-(1/2+1/6+...+1/90)

=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)

=9/10-9/10=0

\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

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

\(B=1-\dfrac{1}{7}\)

\(B=\dfrac{6}{7}\)

Quy đồng ik em

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

câu b, đâu ạ?

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}=\dfrac{8}{9}-\left(\dfrac{1}{8.9}+\dfrac{1}{7.8}+\dfrac{1}{6.7}+\dfrac{1}{5.6}+\dfrac{1}{4.5}+\dfrac{1}{3.4}+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)

8/9−1/72−1/56−1/42−1/30−1/20−1/12−1/6−1/2sao bằng \(\dfrac{8}{9}\)