\(B=1+2+4+5+7+8+10+...+119+121+122\)

Ta có : 

\(A=1+2+3+...+121+122\)

\(A=\left[\left(122-1\right):1+1\right]\left(1+122\right):2\)

\(A=122.\left(123\right):2=7503\)

Ta lại có :

\(C=3+6+9+...+120\)

\(C=\left[\left(120-3\right):3+1\right]\left(3+120\right):2\)

\(C=40.123:2=2460\)

Ta thấy : \(B=A-C\)

\(B=7503-2460=5043\)

c: C=1*2+2*3+3*4+...+58*59+59*60

=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)

=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60

=>3*C=59*60*61

=>C=59*20*61=71980

=(3/10/99+4/11/119-5/8/299)*(3/6-2/6-1/6)

=(3/10/99+4/11/119-5/8/299)*(1/6-1/6)

=(3/10/99+4/11/119-5/8/299)*0

=0

Số các số hạng là:

(119 - 1) : 2 + 1 = 60 số

Tổng S = 60 x (119 + 1) : 2 

S= 7200: 2

S = 3600

Số số hạng là :

(119 - 1) : 2 + 1 = 60 (số)

Do đó S = (199 + 1) x 60 : 2 = 3600

\(\dfrac{22}{5}\times\dfrac{6}{121}\times\dfrac{11}{4}\times\dfrac{3}{5}\times\dfrac{1}{3}\times\dfrac{5}{4}\)

\(=\left(\dfrac{22}{5}\times\dfrac{5}{4}\right)\times\left(\dfrac{6}{121}\times\dfrac{11}{4}\right)\times\left(\dfrac{3}{5}\times\dfrac{1}{3}\right)\)

\(=\dfrac{11}{2}\times\dfrac{3}{22}\times\dfrac{1}{5}\)

\(=\dfrac{3}{20}\)

\(=\dfrac{22\times6\times11\times3\times1\times5}{5\times121\times4\times5\times3\times4}=\dfrac{11\times2\times6\times11\times1}{11\times11\times4\times5\times4}=\dfrac{2\times6\times1}{4\times5\times4}=\dfrac{18}{100}=\dfrac{9}{50}\)