Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{2018^2}-1\right)\)
B= \(-\frac{1}{2}\)
So sánh A và B
Cho A = \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\cdot\left(\frac{1}{2018^2}-1\right)\) và B = \(-\frac{1}{2}\)
Hãy so sánh A và B
cho A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)......\left(\frac{1}{2018^2}-1\right)\) và B= \(-\frac{1}{2}\).So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
Cho A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)..\left(\frac{1}{2014^2}-1\right)\&B=\frac{1}{2}\) 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
A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)............\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
B=\(-\frac{1}{2}\)
So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
\(A=\frac{\left(1-2^2\right)\left(1-3^2\right)\left(1-4^2\right)...\left(1-2018^2\right)}{2^23^24^2...2018^2}\)
\(A=\frac{-1\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot...\cdot\left(-2016\right)\cdot2018}{2018!^2}\)
\(A=\frac{2016!\cdot3\cdot4\cdot5\cdot...\cdot2018}{2018!^2}=\frac{2016!\cdot2018!}{2018!^2\cdot2!}=\frac{2016!}{2!2018!}=\frac{1}{2!\cdot2017\cdot2018}>0>-\frac{1}{2}=B\)
A = (1/2+1)(1/2-1)(1/3+1)(1/3-1)....(1/2018+1)(1/2018-1) đặt các tích phần tử có dấu + là X, tích các phần tử có dấu - là Y => A= X.Y
X = 3/2.4/3.5/4.....2019/2018 = 2019/2
Y= (-1/2)(-2/3)(-3/4)...(-2017/2018) = -1/2018 (tích của 2017 số <0)
A= X.Y = -2019/2018.1/2 < B= -1/2
cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2016^2}-1\right)\left(\frac{1}{2017^2}-1\right)\)và b=-1/2
Hãy so sánh A với B
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
Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{2018^2}-1\right)\)
B= \(-\frac{1}{2}\)
So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4072323}{2018\cdot2018}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2017\cdot2019\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(2018\cdot2018\right)}\)
\(-A=\frac{\left(1\cdot2\cdot...\cdot2017\right)\left(3\cdot4\cdot...\cdot2019\right)}{\left(2\cdot3\cdot...\cdot2018\right)\left(2\cdot3\cdot...\cdot2018\right)}\)
\(-A=\frac{1\cdot2019}{2018\cdot2}\)
\(-A=\frac{2019}{4036}\)
\(A=-\frac{2019}{4036}< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right)\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right)...\left(\frac{1}{2018}-\frac{2018^2}{2018^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-4072323}{2018^2}\)
\(\Rightarrow\frac{-\left(3.8.....4072323\right)}{\left(2.3.4...2018\right).\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(1.3.2.4....2017.2019\right)}{\left(2.3.4...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(\left(1.2.3...2017\right).\left(3.4.5..2019\right)\right)}{\left(2.3...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-2019}{2018.2}< -\frac{2018}{2018.2}=\frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\)
P/s: mk ko copy baì của bn uyên đâu nha
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 = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\) và B=\(\frac{-1}{2}\)
Hãy so sánh A và B
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 :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)
\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)
Sr còn thiếu
\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)
Vậy \(A< B\)