Ta có

\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)

=> A<-1/2

Cho mk sửa xíu : ( 1/2)^-1 ; (1/2)^-2

(1/2)^-1=2

(1/2)^-2=4

có 2<4

=>(1/2)^-1<(1/2)^-2

Ta có :

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

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

\(\left(\frac{1}{2}\right)^{-2}=\left(2^{-1}\right)^{-2}=2^2=4\)

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

\(\left(\frac{1}{3}\right)^{-2}=\left(3^{-1}\right)^{-2}=3^2=9\)

Do đó ta có :

\(\left(\frac{1}{2}\right)^{-1}< \left(\frac{1}{3}\right)^{-1}< \left(\frac{1}{2}\right)^{-2}=\left(\frac{1}{4}\right)^{-1}< \left(\frac{1}{3}\right)^{-2}\)

Ta có:

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)

\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)

\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)

\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)

\(A=\frac{1.2018}{2017.2}\)

\(A=\frac{1009}{2017}\)

Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)

           \(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)

Vậy A>B

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)

\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)

\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)

\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)

\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)

Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)

A có: \(\frac{2014-2}{3-2}+1=2013\) ( thừa số )

Ta thấy mỗi thừa số của A đều có dạng \(\frac{1}{n^2}-1\)với \(n\inℕ^∗\)và \(n>1\)

Có \(\frac{1}{n^2}< 1\Rightarrow\frac{1}{n^2}-1< 1-1=0\)

=> Mỗi thừa số của A đều nhỏ hơn 0

=> A là tích của 2013 thừa số nhỏ hơn 0

Mà 2013 là số lẻ

=> A < 0

Mà B = \(\frac{1}{2}\)> 0

=> A < B

M=-(\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{1-100^2}{100^2}\))

=-(\(\frac{1.3}{2.2}.\frac{2.4}{3.3}\frac{3.5}{4.4}...\frac{99.100}{100.100}\))

=-(\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...100}{2.3.4..100}\))

=-(\(\frac{1}{100}.\frac{1}{2}\))

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

thanks you very very much