\(A=\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{6.5}+\frac{1}{6.7}+\frac{1}{8.7}+...+\frac{1}{18.19}+\frac{1}{20.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{5-4}{4.5}+\frac{6-5}{6.5}+\frac{7-6}{6.7}+...+\frac{20-19}{20.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{5}\)

\(\Rightarrow A=\frac{2}{5}\)

Mình có cách giải khác:

A= \(\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)

A= \(\frac{2.1}{2.2.5}+\frac{2.1}{2.3.5}+\frac{2.1}{2.3.7}+\frac{2.1}{2.4.7}+...+\frac{2.1}{2.9.19}+\frac{2.1}{2.10.19}\)

A= \(\frac{2.1}{4.5}+\frac{2.1}{5.6}+\frac{2.1}{6.7}+\frac{2.1}{7.8}+...+\frac{2.1}{18.19}+\frac{2.1}{19.20}\)

A= \(2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

A=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

A= \(2.\left(\frac{1}{4}+0+0+0+...+0+0-\frac{1}{20}\right)\)

A=\(2.\left(\frac{1}{4}-\frac{1}{20}\right)\)

A=\(2.\left(\frac{5}{20}-\frac{1}{20}\right)\)

A= \(2.\frac{1}{5}\)

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

Xong rùi đó!!!!! :))

\(A=\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{9.19}+\dfrac{1}{10.19}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{18.19}+\dfrac{2}{19.20}\)

\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{20}\right)\)

\(A=2.\dfrac{1}{5}\)

\(A=\dfrac{2}{5}\)

\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)

= \(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)

= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

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

= \(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)

= \(2.\frac{49}{100}\)

= \(\frac{49}{50}\)

a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

= ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) +...+ ( 1/2001 - 1/2003) + (1/2003 - 1/2005)

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005

a=1/1x2+1/2x3+....+1/99x100

a=1-1/2+1/2-1/3+....+1/99-1/100

a=1-1/100

a=99/100

b=4/1x3+4/3x5+.....+4/51x53

b=2x(2/1x3+2/3x5+....+2/51x53)

b=2x(1-1/3+1/3-1/5+...+1/51-1/53)

b=2x(1-1/53)

b=2x52/53

b=104/53

đúng tick cho mình nha

Bài này cũng dễ mà

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)