\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

                                          Bài làm :

Ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

TA CÓ: \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

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

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

Ta có : \(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

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

= \(1-\frac{1}{100}=\frac{99}{100}\)

Bài này lớp 6 phải không bạn

A=1/1-1/2+1/2-1/3+1/3-1/4+1/5-1/6+......................+1/99-1/100

A=1/1-1/100

A=99/100

Nếu bạn cảm thấy bài mình đúng thì cho mình một "lai"

=\(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+.........\frac{100-99}{99\cdot100}\)

=\(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}.-\frac{2}{2\cdot3}+.........\frac{100}{99\cdot100}-\frac{99}{99\cdot100}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}.-\frac{1}{4}+..........+\frac{1}{99}-\frac{1}{100}\)

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

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

1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100

=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100

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

= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0

=99/100

vậy GTBT = 99/100

bn vào câu hỏi tương tự là có

= 1/1 + 1/100

=1011/100

Tích nha

A = 5(1/1.2 + 1/2.3 +......+ 1/99.100)

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

A = 5( 1 - 1/100)

A = 5 . 99/100

A = 99/20

** k mk nha!

\(\frac{5}{1\times2}+\frac{5}{2\times3}+...+\frac{5}{99\times100}=5\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\right)=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5\times\frac{99}{100}=\frac{99}{20}=4\frac{19}{20}\)

A=5/1-5/2+5/2-5/3+...+5/99-5/100

A=5-(5/2-5/2+5/3-5/3+...5/99-5/99)-5/100

A=5-0+0+0+...+0-5/100

A=5-5/100

A=49/10=4,9

c: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{49\cdot50}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{49}+\frac{1}{50}-2\left(\frac12+\frac14+\cdots+\frac{1}{50}\right)\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{50}-1-\frac12-\cdots-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\)

giúp em câu a b nx dc hem tại khó quá em chx học kiểu chấm than ở mẫu số

Uh dễ quá

\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99.100}\)

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

\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

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

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)

\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

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

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)

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

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

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