a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(=\dfrac{1}{3}-\dfrac{1}{39}\)

\(=\dfrac{12}{39}\)

Vậy \(A=\dfrac{12}{39}\)

b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=1-\dfrac{1}{76}\)

\(=\dfrac{75}{76}\)

Vậy \(B=\dfrac{75}{76}\)

a) Ta có :

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)

\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)

b) Ta có :

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)

\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)

~ Học tốt ~

Ta co:\(\)

\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.....+\dfrac{2}{73.76}\)

\(=>A=2.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+....+\dfrac{1}{73.76}\right)\)

\(=>A=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{73}-\dfrac{1}{76}\right)\)

\(=>A=2.\left(1-\dfrac{1}{76}\right)\)

\(=>A=2.\dfrac{75}{76}=\dfrac{2.75}{2.38}\)

\(=>A=\dfrac{75}{38}\)

Tick cho mk nha

cảm ơn bạn nhé!

\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+.....+\dfrac{5}{100.103}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\left(1-\dfrac{1}{103}\right)\)
\(B=\dfrac{5}{3}.\dfrac{102}{103}\)
\(B=1\dfrac{67}{103}\)

Ta có: \(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}\right)+\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+...+\dfrac{5}{3}\left(\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}.\dfrac{102}{103}=\dfrac{170}{103}\)

Vậy \(B=\dfrac{170}{103}\).

A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

\(B=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)=\dfrac{5}{3}\left(\dfrac{102}{103}\right)=\dfrac{170}{103}\)

\(B=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{100.103}+\dfrac{4}{103.106}\)

\(B=4\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}+\dfrac{1}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}+\dfrac{3}{103.106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}\left(\dfrac{1}{3}-\dfrac{1}{106}\right)\)

\(B=\dfrac{4}{3}.\dfrac{103}{318}\)

\(B=\dfrac{412}{954}\)

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

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

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

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

Câu 1 :

1/n - 1/n + a = a + n/a ( a + n ) = a + n - a/a ( n + a ) = n/a ( a + n )

Câu 2 :

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

= 1/1 - 1/100 = 99/100