- Tính A trước:

\(A=\frac{\left(2015+1\right).2017-2}{2015+2015.2017}\)

\(A=\frac{2017.2015+2017-2}{2017.2015+2015}\)

\(A=\frac{2017.2015+2015}{2017.2015+2015}\)

\(A=1\)

- Tính B:

\(B=\frac{-2016.20172017}{2017.20162016}\)

\(B=\frac{-2016}{2017}.\frac{20172017}{20162016}\)

\(B=\frac{-10001}{10001}\)

\(B=-1\)

Vậy ta có: \(A+B=1-1=0\)

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 \(A< B\).

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

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

Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015

Ta có :

B = \(\dfrac{2015}{1}+\dfrac{2014}{2}+\dfrac{2013}{3}+...+\dfrac{2}{2014}+\dfrac{1}{2015}\) => B = \(\left(1+\dfrac{2014}{2}\right)+\left(1+\dfrac{2013}{3}\right)+...+\left(1+\dfrac{2}{2014}\right)+\left(1+\dfrac{1}{2015}\right)+1\) => B = \(\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2014}+\dfrac{2016}{2015}+\dfrac{2016}{2016}\) => B = \(2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\) Ta có :

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)}\)

=> \(\dfrac{A}{B}=\dfrac{1}{2016}\)

Vậy \(\dfrac{A}{B}=\dfrac{1}{2016}\)

Hãy cố gắng giải bài này nhé!

Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}=\dfrac{a+2b+c}{2b+c+a}=1\)

\(\dfrac{a}{2b}=1\Rightarrow a=2b\\ \dfrac{2b}{c}=1\Rightarrow c=2b\\ \dfrac{c}{a}=1\Rightarrow a=c\\ \Rightarrow a=2b=c\)

\(M=\dfrac{a^3.c^2.b^{2015}}{b^{2020}}=\dfrac{a^3.a^2}{b^5}=\dfrac{a^5}{b^5}=\dfrac{\left(2b\right)^5}{b^5}=\dfrac{32b^5}{b^5}=32\)

Có \(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}=\dfrac{a+2b+c}{2b+c+a}=1\)

=> a = 2b = c

M = \(\dfrac{a^3.c^2.b^{2015}}{b^{2020}}=\dfrac{a^3.c^2}{b^5}=\dfrac{\left(2b\right)^3.\left(2b\right)^2}{b^5}=\dfrac{32.b^5}{b^5}=32\)

A=\(\dfrac{2016.2017+1}{2016.2017}=\dfrac{2016.2017}{2016.2017}+\dfrac{1}{2016.2017}=1+\dfrac{1}{2016.2017}\)

A=\(\dfrac{2017.2018+1}{2017.2018}=\dfrac{2017.2018}{2017.2018}+\dfrac{1}{2017.2018}=1+\dfrac{1}{2017.2018}\)

Mà 1=1; \(\dfrac{1}{2016.2017}\)>\(\dfrac{1}{2017.2018}\) nên A>B