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12 tháng 5 2017

\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...................+\dfrac{2}{97.100}\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{2}\left(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+..................+\dfrac{2}{97.100}\right)\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{97.100}\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+..............+\dfrac{1}{97}-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{3}{2}A=1-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{99}{100}\)

\(\Rightarrow A=\dfrac{99}{100}:\dfrac{3}{2}\)

\(\Rightarrow A=\dfrac{33}{50}\)

=2/3(3/1*4+3/4*7+...+3/97*100)

=2/3(1-1/4+1/4-1/7+...+1/97-1/100)

=2/3*99/100

=198/300

=66/100

=33/50

4 tháng 3 2023

cop mạng à

8 tháng 5 2022

A= 2/1.4+2/4.7+2/7.10+...+2/97.100

= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)

= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

= 2.(1/1-1/100)

= 2.(99/100)

=99/50

8 tháng 5 2022

quá dễ bạn ạ

4 tháng 5 2017

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(A=\frac{2}{3}.\left(1-\frac{1}{4}\right)+\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

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

\(A=\frac{33}{50}\)

11 tháng 9 2016

Ta có: \(A=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

Nhận xét: \(\frac{a}{x.\left(x+a\right)}=\frac{1}{x}-\frac{1}{x+a}\)

Do đó: \(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

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

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

\(=\frac{33}{50}\)

Vậy,\(A=\frac{33}{50}\)

11 tháng 9 2016

\(\text{Ta có: }A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)

\(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)

\(\Rightarrow\frac{3}{2}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

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

\(\Rightarrow\frac{3}{2}A=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}:\frac{3}{2}\)

\(A=\frac{99}{100}.\frac{2}{3}=\frac{33}{50}\)

20 tháng 9 2016

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=\frac{297}{100}\)

11 tháng 9 2016

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

12 tháng 4 2018

Ta có : 

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(=\)\(\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\)\(\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\)\(\frac{2}{3}\left(1-\frac{1}{103}\right)\)

\(=\)\(\frac{2}{3}.\frac{102}{103}\)

\(=\)\(\frac{68}{103}\)

Chúc bạn học tốt ~ 

3 tháng 8 2016

\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{61.64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{61}-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\frac{63}{64}\)

\(=\frac{21}{32}\)

3 tháng 8 2016

\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{61.61}\)

\(=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{61.64}\right)\)

\(=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{61}-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)

\(=\frac{2}{3}.\frac{63}{64}\)

\(=\frac{21}{32}\)