Question

Cho A=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\right)\)

B=\(\dfrac{2013.2015.2017}{2018.2013.\left(2014+1\right)}\)

a/Tính Avà B

b/So sánh A và B

Answer 1

a)

\(A=2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2015\cdot2017}\right)\)

\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(=1-\dfrac{1}{2017}\)

\(=\dfrac{2017}{2017}-\dfrac{1}{2017}\)

\(=\dfrac{2016}{2017}\)

\(B=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot\left(2014+1\right)}\)

\(=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot2015}\)

\(=\dfrac{2017}{2018}\)

b)

Ta có:

\(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\)

\(B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\)

Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}\)

\(\Rightarrow1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

Answer 2

Anh làm nhé!!

Bài làm:

a) Tính A và B

\(A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\right)\\ =\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2015.2017}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\\ =1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(B=\dfrac{2013.2015.2017}{2018.2013.\left(2014+1\right)}\\ =\dfrac{2013.2015.2017}{2018.2013.2015}=\dfrac{2017}{2018}\)

b) So sánh A và B.

Ta có: \(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\\ B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\\ Mà:\dfrac{1}{2017}>\dfrac{1}{2018}\\ =>1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\\ =>A< B\)

Answer 3

a/có: A = 2.\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{2015.2017}\right)\)

=> 2A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)

=> 2A=\(\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\right)\)

=> 2A=\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2015.2017}\right)\)

=> 2A=\(\dfrac{1}{2}.\left(\dfrac{2}{1}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+...+\dfrac{2}{2015}-\dfrac{2}{2017}\right)\)

=> 2A=\(\dfrac{1}{2}.\left(\dfrac{2}{1}-\dfrac{2}{2017}\right)=\dfrac{1}{2}.\dfrac{4032}{2017}\)

=> A=\(\dfrac{4032}{2017}.\dfrac{1}{2}:2=\dfrac{4032}{2017}.\dfrac{1}{2}x2\)

=> A=\(\dfrac{4032}{2017}\).

Theo đề ta tính đc B=\(\dfrac{2017}{2018}\)

b/ Vì \(\dfrac{4032}{2017}>1\) và \(\dfrac{2017}{2018}< 1\) nên \(\dfrac{4032}{2017}>\dfrac{2017}{2018}\) nên A>B