A = \(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

= \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

= \(5.\left(1-\frac{1}{100}\right)\)

= \(5-\frac{5}{100}<5\)

=> A < 5 (Đpcm).

cho minh ket ban voi

A=5(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))

A=5\(\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\right]\)

A=\(5\left(1-\frac{1}{100}\right)\)

Ta có: 1-\(\frac{1}{100}\)<1

=>5(1-\(\frac{1}{100}\))<5.1

=>A<5      (đpcm)

A = 5/1.2 + 5/2.3 +...+ 5/99.100

 2A = 10/1.2 + 10/2.3 +...+ 10/99.100

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

2A=5/1-5/100

2A=9/2 => A=9/2:2=9/4

cho 1 đ-ú-n-g nha bạn!!

99/20 đảm bảo

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Kẻ Bí Mật sai rồi !                                                                                      

\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)

     \(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

       \(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

      \(=5.\left(1-\frac{1}{100}\right)\)

   \(=5.\frac{99}{100}\)

      \(=\frac{99}{20}\)

5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100

= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )

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

= 5 . ( 1 - 1/100 )

= 5 . 99/100

= 99/20

\(\frac{99}{100}\)nhé

bạn chép sai đề rồi 1/2.3 phai bang 1/3.4

1.

a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5.\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)

\(=2.\frac{100}{101}\)

\(=\frac{200}{101}\)

Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}.2\)

\(\Rightarrow A=\frac{200}{101}.\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\) 

A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )

A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )

A= 5. (\(1-\frac{1}{100}\))

A= 5.\(\frac{99}{100}\)

A= \(\frac{99}{20}\)

B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)

    = \(\frac{1}{2}\)-  \(\frac{1}{3}\)+\(\frac{1}{3}\)-   \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)-     \(\frac{1}{10}\)

    =  \(\frac{1}{2}\) -     \(\frac{1}{10}\)

     =       \(\frac{2}{5}\)

C = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 = 

   = 1/3 - 1/15 = 5/15 - 1/15 = 4/15

\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=5\left(1-\frac{1}{100}\right)\)

\(A=5.\frac{99}{100}\)

\(A=\frac{99}{20}\)

\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(B=\frac{1}{2}-\frac{1}{10}\)

\(B=\frac{2}{5}\)

\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(C=\frac{1}{3}-\frac{1}{15}\)

\(C=\frac{4}{15}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=5\left(1-\frac{1}{100}\right)\)

\(A=5.\frac{99}{100}\)

\(A=\frac{99}{20}\)

\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(B=\frac{1}{2}-\frac{1}{10}\)

\(B=\frac{2}{5}\)

\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(C=\frac{1}{3}-\frac{1}{15}\)

\(C=\frac{4}{15}\)