`a)A=1/(1.2)+1/(2.3)+....+1/(2017.2018)`
`=1-1/2+1/2-1/3+.....+1/2017-1/2018`
`=1-1/2018=2017/2018`
`b)B=3/(1.3)+3/(3.5)+.....+3/(199.201)`
`=3/2(3/(1.3)+2/(3.5)+.....+2/(199.201))`
`=3/2(1-1/3+1/3-1/5+.....+1/199-1/201)`
`=3/2(1-1/201)`
`=3/2*200/201`
`=100/67`
Giải:
a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2017.2018}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2018}\)
\(A=\dfrac{2017}{2018}\)
b) \(B=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...+\dfrac{3}{199.201}\)
\(B=\dfrac{3}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{199.201}\right)\)
\(B=\dfrac{3}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
\(B=\dfrac{3}{2}.\left(\dfrac{1}{1}-\dfrac{1}{201}\right)\)
\(B=\dfrac{3}{2}.\dfrac{200}{201}\)
\(B=\dfrac{100}{67}\)
Chúc bạn học tốt!
