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Đặt \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}............\frac{1-100^2}{100}\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}............\frac{-9999}{100^2}\)
\(\Rightarrow A=\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}...............\frac{-99.101}{100^2}\)
\(\Rightarrow A=\frac{-\left(1.2.3.............99\right).\left(3.4.5............101\right)}{\left(2.3.4......100\right).\left(2.3.4.............100\right)}\)
\(\Rightarrow A=\frac{-1.101}{100.2}=\frac{-101}{200}\)
Vậy \(A=\frac{-101}{200}\)
Chúc bn học tốt
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
Ta có : \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
\(< \frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{4}.\left(2-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{4n}< 1\)
Vậy A < 1
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}.\)
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{4n^2}.\)
\(A=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}\right)\)
\(A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
So sánh \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....\)
\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n-1\right)}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n-1}+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(2-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}\)
có \(\frac{1}{2}>\frac{1}{2}-\frac{1}{4n}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}< \frac{1}{2}\) mà \(\frac{1}{2}< 1\)
\(\Rightarrow A< 1\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
2) \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9999}{100^2}\)
\(A=-\left(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\right)\) (vì có 99 thừa số âm nên kết quả là âm)
\(A=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3.}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\right)\)
\(A=-\left(\frac{1}{100}.\frac{101}{2}\right)\)
\(A=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
Trả lời câu nào cũng được nha mấy bạn! Help me, please!!!!!!! 
\(\left(\frac{1}{-2}\right)^{40}=\left(\frac{1}{2}\right)^{40}=\frac{1}{2^{40}}=\frac{1}{\left(2^{10}\right)^4}=\frac{1}{1024^4}<\frac{1}{\left(10^3\right)^4}=\frac{1}{1000^4}=\left(\frac{1}{-10}\right)^{12}\)