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26 tháng 8 2015

M = \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{2015^2}\right)\)

M = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)\left(-\frac{3.5}{4.4}\right)....\left(-\frac{2014.2016}{2015.2015}\right)\)

M = \(\frac{\left(1.2.3....2014\right)\left(3.4.5...2016\right)}{\left(2.3.4.....2015\right)\left(2.3.4....2015\right)}\)

M = \(\frac{2016}{2015.2}\)

M = \(\frac{1008}{2015}\)

N = \(\frac{1}{2}\)=\(\frac{1008}{2016}\)

Vì \(\frac{1008}{2015}>\frac{1008}{2016}\)

=> M > N

26 tháng 8 2015

Hồ Thu Giang nhầm ở bước thứ hai nhưng ý tưởng tốt                 

7 tháng 9 2018

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

20 tháng 8 2017

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

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

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

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

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

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

20 tháng 8 2017

Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)

Áp dụng :

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

\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)

\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)

20 tháng 8 2017

Sr còn thiếu

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

Vậy \(A< B\)

4 tháng 7 2017

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)

\(=\frac{1}{n+1}\)

\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+5+...+21}{2}=115\)

20 tháng 9 2015

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

\(y=\left(\frac{-1.3}{2.2}\right)\left(\frac{-2.4}{3.3}\right)....\left(\frac{-2013.2015}{2014.2014}\right)\)

\(y=-\left(\frac{1.2....2013.3.4...2015}{2.3....2014.2.3....2014}\right)\)

\(y=-\left(\frac{2015}{2014.2}\right)\)

\(y=\frac{-2015}{4028}\)

\(x=\frac{-1}{2}=\frac{-2014}{4028}\)

Vì \(\frac{-2015}{4028}