Question

Tính:

a) A= 1/1.2+1/2.3+1/3.4+...+1/199.200

b) B= 3/2.4+ 3/4.6+3/6.8+...+3/2018.2020

Giúp mk với, mk đag cần gấp. Gấp lắm lun í

Mk cảm ơn trước nhaaaaa

Answer 1

a,A =\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{199.200}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{199}-\frac{1}{200}\)

= 1-\(\frac{1}{200}\)

=\(\frac{199}{200}\)

Answer 2

b, B=\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{2018.2020}\)

=3.(\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+..+\frac{1}{2018.2020}\))

=3(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2018}-\frac{1}{2020}\))

= 3.(\(\frac{1}{2}-\frac{1}{2020}\))

=\(\frac{6057}{2020}\)

Answer 3

bạn nhớ tick và theo dõi mk nhé

Answer 4

\(B=3\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2018.2020}\right)\)

\(B=\frac{3}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)

\(B=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

\(B=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)=...\)