\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)

\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)

a) = 1-1/2+1/2-1/3+...+1/99-1/100 =1 - 1/100 = 99/100

a) a = \(\dfrac{599}{3}\)

b) B = -12417

1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)

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

\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)

\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)

\(=\dfrac{1}{n+1}\)

2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)

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

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

C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)

=-12405

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\)

=\(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\).....\(\dfrac{n}{n+1}\)=\(\dfrac{1}{n+1}\)

Vậy...

\(C=-66\times(\frac{1}{2}-\frac{1}{3}+\frac{1}{11})+124\times-37+63\times-124\)

\(C=-66\times\frac{17}{66}-124\times100\)

\(C=-12417\)

#)Giải :

\(C=\left(-66\right).\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

\(C=\left(-66\right).\frac{17}{66}-4699-7812\)

\(C=\left(-17\right)-4699-7812\)

\(C=-12528\)

C = -66.(1/2 - 1/3 + 1/11) + 124.(-37) + 63.(-124)

C = -66.(1/2 - 1/3 + 1/11) - 124.37 - 63.124

C = -17 - 4588 - 7812

C = -12417