Giải:

\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}-\left(\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\right)\)

\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-1\right)\)

\(=\dfrac{1}{99}-\dfrac{-98}{99}\)

\(=\dfrac{1}{99}+\dfrac{98}{99}\)

\(=\dfrac{99}{99}=1\)

Chúc bạn học tốt!

\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}+\dfrac{1}{2.1}\)

=\(\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{96}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

=\(0+1\)

=\(1\)

Bạn học tốt^^

\(P=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(P=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(P=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(P=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(P=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

Xong !

\(A=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ =\dfrac{1}{100}+1=\dfrac{101}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(A=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

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

\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(A=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-49}{50}\)

A=1/100−1/100+1/99−1/99+1/98−1/98+1/97−...−1/3+1/2−1/2+1

=1

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-...-\dfrac{1}{2.1}\\ =\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\\ =\dfrac{1}{100}-\dfrac{99}{100}\\ =\dfrac{-98}{100}\\ =-\dfrac{49}{100}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

\(=-\dfrac{49}{50}\)

a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)

câu b và c xem lại đề nha

Chúc bạn học tốt!!!

b, \(\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+......+\dfrac{1}{3.2}+\dfrac{1}{2.1}\)

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

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\))

\(=\dfrac{1}{99}+\dfrac{1}{99}-1=\dfrac{2}{99}-1=\dfrac{-97}{99}\)

Chúc bạn học tốt!!!

c/ \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1}{155\cdot257}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}+\dfrac{2}{255\cdot257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}+...+\dfrac{1}{255}-\dfrac{1}{257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{257}\right)=\dfrac{1}{2}\cdot\dfrac{256}{257}=\dfrac{128}{257}\)

p/s: thấy bn kia hơn 20' r` k lm tiếp nên t lm lun nhé!

Lời giải:

Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow A+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}=\frac{1}{99.100}\)

\(\Leftrightarrow A+1-\frac{1}{98}=\frac{1}{99.100}\Rightarrow A=\frac{1}{9900}-\frac{97}{98}\)

\(B=\dfrac{-1}{99}+\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{2\cdot1}\)

\(=\dfrac{-1}{99}+\dfrac{1}{98\cdot99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{97\cdot98}\right)\)

\(=-\dfrac{2}{99}+\dfrac{1}{98}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)

\(=-\dfrac{2}{99}+\dfrac{1}{98}-\dfrac{97}{98}=\dfrac{-2}{99}-\dfrac{23}{49}=\dfrac{-2375}{4851}\)

C= \(\dfrac{1}{100}-\)(\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{98.99}\)+\(\dfrac{1}{99.100}\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

= \(\dfrac{1}{100}-\dfrac{99}{100}\)

=\(\dfrac{-98}{100}=-\dfrac{49}{50}\)

Ta có:

\(=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+......+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

sau khi giản ước ta được như sau:

=\(\dfrac{1}{100}-1\)=\(\dfrac{-99}{100}\)

Chúc bạn học tốt nhé

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

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

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

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

\(=1-1+\dfrac{1}{72}\)

\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)

\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)

\(=-\left(-\dfrac{173}{1287}\right)\)

\(=\dfrac{173}{1287}\)

c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

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

\(=\dfrac{-49}{50}\)