Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cho A = (1/2^2 - 1)(1/3^2 - 1) (1/4^2 - 1) ... (1/2013^2 -1)(1/2014^2 - 1) Và B = -1/2
So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(\Leftrightarrow A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{4052169}-1\right)\left(\frac{1}{\text{}\text{}4056196}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-4056195}{\text{}4056196}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3}{2^2}.\frac{\left(-2\right)4}{3^3}.\frac{\left(-3\right)5}{4^2}.....\frac{\left(-2013\right)2015}{\text{}2014^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(-2\right)....\left(-2013\right)}{2.3...1014}.\frac{3.4......2015}{2.3......2014}\)
\(\Leftrightarrow A=\frac{-1}{1014}.\frac{2015}{2}=\frac{-2015}{4028}\)
VÌ \(\frac{-2015}{4028}< \frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\Leftrightarrow A< B\)
Ta có \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)
\(=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4...2013.2015}{2.2.3.3...2014.2014}\right)\)
\(=-\left(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< \frac{-2014}{4028}< \frac{1}{2}=B\)
=> 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 \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
\(\Rightarrow A-B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4026}\)
\(B>1+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{4026}=\frac{1}{2}+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{4026}\right)=\frac{1}{2}+\left(A-B\right)\)
\(\Rightarrow B>\frac{1}{2}+\left(A-B\right)\left(1\right)\)
\(A-B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4026}< \frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}=\frac{2013}{2}\)
\(\Rightarrow A-B< \frac{2013}{2}\Rightarrow\frac{A-B}{2013}< \frac{1}{2}\left(2\right)\)
Cộng (1) với (2)
\(\Rightarrow\frac{A-B}{2013}+\frac{1}{2}+\left(A-B\right)< \frac{1}{2}+B\Rightarrow\frac{A-B}{2013}+\left(A-B\right)< B\Rightarrow\frac{2014\left(A-B\right)}{2013}< B\Rightarrow\frac{A-B}{B}< \frac{2013}{2014}\)
\(\Rightarrow\frac{A-B}{B}+1< \frac{2013}{2014}+1\Rightarrow\frac{A}{B}< 1\frac{2013}{2014}\left(đpcm\right)\)
Ax1007x1008=A1= 1007x(1+1/3+...+1/2013)
Bx1007x1008=B1=1008x(1/2+1/4+...+1/2014)
A1-B1=1007x(1-1/2+1/3-1/4+..+1/2013-2/1014) - ( 1/2+1/4+..1/2014)
=1007x(1/2+1/3x4+..1/1007x1008)- (1/2+1/4+..1/2014)
Xet' (1/2+1/4+..1/2014) < (1/2 + 1/2 + .... 1/2) (co' 1007 so' ) = 1007/2
xet' 1007x(1/2 +1/3x4 +... 1/1007x1008 ) > 1007/2
=> A> 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 \(A< B\)
Bạn tham khảo tại đây nhé: https://hoc247.net/hoi-dap/toan-7/so-sanh-a-1-2-2-1-1-3-2-1-1-4-2-1-1-2014-2-1-va-b-1-2-faq417898.html
Trang web này hơi dài, mong bạn thông cảm.
Chúc bạn học tốt!