a)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

Ai trả lời nhanh mình tích cho nhé!

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(A=\frac{1}{2}.\frac{4949}{9900}\)

\(A=\frac{4949}{19800}\)

Ta có 99/1+98/2+97/3+...+1/99=(98/2+1)+(97/3+1)+...+(1/99+1)+1

=100/2+100/3+...+100/99+100/100

=100(1/2+1/3=1/4+1/5+...+1/99+1/100)

Vậy (1/2+1/3+...+1/100)/((99/1+98/2+...+1/99)=1/100

xét mẫu số = \(\frac{99}{1}\)+\(\frac{98}{2}\)+....+\(\frac{1}{99}\)

mẫu số = (\(1+\frac{98}{2}\))+(\(1+\frac{97}{3}\))+.......+(\(1+\frac{1}{99}\))

mẫu số = \(\frac{100}{2}\)+\(\frac{100}{3}\)+....+\(\frac{100}{99}\)

mẫu số =100 x (\(\frac{1}{2}\)+\(\frac{1}{3}\)+....+\(\frac{1}{99}\))             (1)

thay (1) vào biểu thức trên

1/2+1/3+1/4+.....+1/100  /   100 x (1/2+1/3+...+1/99)

= \(\frac{1}{100}\)

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt A = \(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)

A = \(1-\frac{1}{99}\)

A = \(\frac{98}{99}\)

Thay A vào biểu thức trên, ta được :

\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

S=(1+2+3+4+...+96+97+98+99):5

S=(99x(99+1):2):5

S=(99x100:2):5

S=(9900:2):5

S=4950:5

S=990

990

(1+2+3+4+...+96+97+98+99):5

Đặt 1+2+3+4+...+96+97+98+99=A1+2+3+4+...+96+97+98+99=S

Số số hạng của S là:

(99−1):1+1=99(99-1):1+1=99

Tổng của S là:

(99+1).99:2=4950(99+1).99:2=4950

→(1+2+3+4+...+96+97+98+99):5→(1+2+3+4+...+96+97+98+99):5

=4950:5=990