a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)

b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)

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

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

= \(1-\frac{1}{50}=\frac{49}{50}\)

B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)

= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

= \(\frac{4}{13}\)

C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)

= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\) 

= \(\frac{9}{38}\)

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

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

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

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

A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51

 A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51

A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51

A=1-1/51

A=50/51

Cảm ơn bn

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 ~

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sau 3 phút có kết quả tùy bạn

A=\(\frac{37}{78}\)

B=\(\frac{9}{38}\)

a, 1/1.2+1/1.3+...+1/99.100

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

=1-1/100

=99/100

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)

\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)

Ta có: \(B=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

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

\(=\frac{1}{3}-\frac{1}{8}\)

\(=\frac{5}{24}\)

Vậy \(B=\frac{5}{24}\)

Ta có: \(C=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

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

\(=\frac{1}{4}-\frac{1}{9}\)

\(=\frac{5}{36}\)

Vậy \(C=\frac{5}{36}\)

A= \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{35}+\frac{1}{99}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.6}+...+\frac{2}{9.11}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)

\(2A=1-\frac{1}{11}=\frac{10}{11}\)

\(A=\frac{10}{11}:2=\frac{5}{11}\)

\(D=\frac{3^2}{1.4}+\frac{3^2}{4.7}+...+\frac{3^2}{13.16}\)

\(D=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{13.16}\right)\)

\(D=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)

\(D=3.\left(1-\frac{1}{16}\right)=3.\frac{15}{16}=2\frac{13}{16}\)