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\(5A=\frac{5^{2011}+5}{5^{2011}+1}=1+\frac{4}{5^{2011}+1}\)
\(5B=\frac{5^{2010}+5}{5^{2010}+1}=1+\frac{4}{5^{2010}+1}\)
\(5B>5A\Rightarrow B>A\)
Ta có:
A = \(\frac{5^{2010}+1}{5^{2011}+1}\)
5A = \(\frac{5^{2011}+5}{5^{2011}+1}\) = \(\frac{5^{2011}+1+4}{5^{2011}+1}\) = 1 + \(\frac{4}{5^{2011}+1}\)
B = \(\frac{5^{2009}+1}{5^{2010}+1}\)
5B = \(\frac{5^{2010}+5}{5^{2010}+1}\) = \(\frac{5^{2010}+1+4}{5^{2010}+1}\) = 1 + \(\frac{4}{5^{2010}+1}\)
Vì 1 + \(\frac{4}{5^{2011}+1}\) < \(\frac{4}{5^{2010}+1}\) => 5A < 5B
Vì 5A < 5B => A < B
Ta có: \(5A=\frac{5^{2011}+5}{5^{2011}+1}=\frac{5^{2011}+1+4}{5^{2011}+1}=1+\frac{4}{5^{2011}+16}\)
\(5B=\frac{5^{2010}+5}{5^{2010}+1}=\frac{5^{2010}+1+4}{5^{2010}+1}=1+\frac{4}{5^{2010}+1}\)
Vì \(\frac{4}{5^{2011}+1}< \frac{4}{5^{2010}+1}\Rightarrow5A< 5B\Rightarrow A< B\)
Ta có:
A = \(\frac{5^{2010}+1}{5^{2011}+1}\)
\(\Rightarrow5A=\frac{5.\left(5^{2010}+1\right)}{5^{2011}+1}\)\(=\frac{5^{2011}+5}{5^{2011}+1}=1+\frac{4}{5^{2011}+1}\)
B=\(\frac{5^{2009}+1}{5^{2010}+1}\)
\(\Rightarrow5B=\frac{5.\left(5^{2009}+1\right)}{5^{2010}+1}=\frac{5^{2010}+5}{5^{2010}+1}=1+\frac{4}{5^{2010}+1}\)
Ta thấy \(5^{2011}+1>5^{2010}+1\)
\(\Rightarrow\frac{4}{5^{2011}+1}< \frac{4}{5^{2010}+1}\)
\(\Rightarrow1+\frac{4}{5^{2011}+1}< 1+\frac{4}{5^{2010}+1}\)
Hay 5.A<5.B
Vậy A<B (đpcm)
Ta có :
\(B=\frac{5^{2009}+1}{5^{2010}+1}=\frac{\left(5^{2009}+1\right).10}{\left(5^{2010}+1\right).10}=\frac{5^{2010}+10}{5^{2011}+10}\)
Ta thấy :
\(5^{2010}=5^{2010};1< 10\Rightarrow5^{2010}+1< 5^{2010}+10\)
\(5^{2011}=5^{2011};1< 10\Rightarrow5^{2011}+1< 5^{2011}+10\)
Suy ra : \(A< B\)
Vậy \(A< B\)
\(A< 1\)
\(A< \frac{5^{2010}+1}{5^{2011}+1}\)
\(A< \frac{5^{2010}+1+4}{5^{2011}+1+4}\)
\(A< \frac{5^{2010}+5}{5^{2011}+5}\)
\(A< \frac{5\left(5^{2009}+1\right)}{5\left(5^{2010}+1\right)}\)
\(A< \frac{5^{2009}+1}{5^{2010}+1}\)
\(A< B\)
A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)
B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Do \(\frac{1}{9^{2010}}<\frac{1}{5^{2010}}\) ; \(\frac{1}{9^{2009}}<\frac{1}{5^{2009}}\) ;.....; \(\frac{1}{9}<\frac{1}{5}\)
=> \(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+...+\frac{1}{9}<\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\)
=> 1:\(\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+...+\frac{1}{9}\right)>1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Vậy A > B
Đặt M = \(1+9+9^2+......+9^{2010}\)
\(9M=9+9^2+9^3+......+9^{2011}\)
\(9M-M=8M=9^{2011}-1\)
Đặt K = \(1+9+9^2+......+9^{2009}\)
\(9K=9+9^2+9^3+.....+9^{2010}\)
\(9K-K=8K=9^{2010}-1\)
\(\Rightarrow A=\frac{9^{2011}-1}{9^{2010}-1}\)
Đặt H=\(1+5+5^2+....+5^{2010}\)
\(5H=5+5^2+......+5^{2011}\)
\(5H-H=4H=5^{2011}-1\)
ĐẶT G = \(1+5+5^2+.......+5^{2009}\)
\(5G-G=4G=5^{2010}-1\)
\(\Rightarrow B=\frac{5^{2011}-1}{5^{2010}-1}\)
Rồi bạn so sánh sẽ ra ngay
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Rightarrow B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Suy ra : \(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2009}+1}{2009^{2010}+1}\) hay \(B< A\)
Vậy \(A>B\)
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
Do 2009\(^{2010}\)-2 < 2009\(^{2011}\)-2 \(\Rightarrow\)B<1
Theo đề bài ta có:
B= \(\frac{2009^{2010}-2}{2009^{2011}-2}\)< \(\frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)= \(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)= \(\frac{2009.\left(1+2009^{2009}\right)}{2009.\left(1+2009^{2010}\right)}\)= \(\frac{2009^{2009}+1}{2009^{2010}+1}\)= A \(\Rightarrow\)B<A
Nhân 5 với B và A cho kết quả A<B