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\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)
\(A=\frac{1}{2011}\)
A= \(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
B=\(\left(\frac{2012}{1}-1\right)+\left(\frac{2012}{2}-1\right)+...+\left(\frac{2012}{2011}-1\right)\)
= \(\frac{2012}{1}-\frac{2012}{2012}+\frac{2012}{2}-\frac{2012}{2012}+...+\frac{2012}{2011}-\frac{2012}{2012}\)
=\(2012\left(1-\frac{1}{2012}+\frac{1}{2}-\frac{1}{2012}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}\)=\(\frac{2012\left(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)= 2012
Thôi, cho phép mình góp ý bài mình đã làm bằng cách đơn giản hơn nha ^^.
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có:
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)
\(=A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2009}-\frac{1}{2010}\)
\(\Rightarrow A< 1-\frac{1}{2010}\)
\(\Rightarrow A< 1\)
\(\Rightarrow A< \frac{3}{4}\)
Có: \(\frac{1}{2^2}< \frac{1}{1.2}\); \(\frac{1}{3^2}< \frac{1}{2.3}\);...;\(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}=1-\frac{1}{2010}=\frac{2009}{2010}\)Mà \(\frac{2009}{2010}>\frac{3}{4}\) -> Sai đề
\(A=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2010^2}>1-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}-\frac{1}{2010}=\frac{1004}{2010}>\frac{1}{2010}\Rightarrow A>\frac{1}{2010}\)
A=1-(1/2^2+1/3^2+...+1/2010^2)
A=1-(1/2*2+1/3*3+...+1/2010*2010)>1-(1/2*3+1/3*4+...+1/2010*2011)
A>1-(1/2-1/3+1/3-1/4+...+1/2010-1/2011)
A>1-(1/2-1/2011)=2013/4022>1/2010
=>A>1/2010
Sai thì em xin lỗi nhé
Có A = 1 - (3/4) + (3/4)2- (3/4)2+ ... - (3/4)2009 + (3/4)2010
=>A.(3/4) = (3/4) - (3/4)2 + (3/4)3 - (3/4)4 +... - (3/4)2010 + (3/4)2011
=>A + A.(3/4) = 1 + (3/4)2011
=> 7A/4 = 1 + (3/4)2011
=> 7A = 4 + 4.(3/4)2011 không là số nguyên
=> A không nguyên
vậy A ko phải là số nguyên
\(A=1-\frac{1}{2^2}-...-\frac{1}{2010^2}\)
\(=1-\left(\frac{1}{2^2}+...+\frac{1}{2010^2}\right)\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
Ta có: \(A=1-\left(\frac{1}{2^2}+...+\frac{1}{2010^2}\right)\)\(>\)\(B=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=1-\left(1-\frac{1}{2010}\right)=1-1+\frac{1}{2010}=\frac{1}{2010}\)
cảm ơn bn >.<!
bài bn vik thiếu nhưng mik hiểu nên vẫn tick