Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) va B = \(\frac{-1}{2}\), So sanh A va B
Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right);B=-\frac{1}{2}\). hãy so sanh a va b
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
Cho A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)..\left(\frac{1}{2014^2}-1\right)\&B=\frac{1}{2}\) so sánh A và B
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 so A=\(\frac{2013+\frac{1}{2}}{\left(2012+\frac{1}{2}\right)^2+2013+\frac{1}{2}}\)
B=\(\frac{2013+\frac{1}{3}}{\left(2012+\frac{1}{3}\right)^2+2013+\frac{1}{3}}\)
so sanh A va B
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2014^2}-1\right)\)va \(B=-\frac{1}{2}\). So sanh A va B
\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) và B=\(\frac{-1}{2}\). Hãy so sánh A và 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 \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
Và B =\(\frac{1}{2}\)
So sánh A và 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
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) và B = \(\frac{-1}{2}\). Hãy so sánh A và B