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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 \(A< B\)
\(A< 1\)
\(A< \frac{5^{2010}+1}{5^{2011}+1}\)
\(A< \frac{5^{2010}+1+4}{5^{2011}+1+4}\)
\(A< \frac{5^{2010}+5}{5^{2011}+5}\)
\(A< \frac{5\left(5^{2009}+1\right)}{5\left(5^{2010}+1\right)}\)
\(A< \frac{5^{2009}+1}{5^{2010}+1}\)
\(A< B\)
Đầu tiên chúng ta sẽ so sánh như sau
5^2010 và 5^2009
vì 2010>2009 nên 5^2010>5^200 (1)
1/5^2011+1 và 1/5^2010+1
vì 2011+1=2012
2010+1=2011
mà 2012>2011 nên 1/5^2011+1>1/5^2010+1 (2)
Từ 1 và 2 ta có thể suy ra A>B
Vậy A>B
ta có 2010 >2009 suy ra 5^2010 >5^2009 suy ra 5^2010 + 1>5^2009 +1 (1)
2011>2010 suy ra 5^2011 >5^2010 suy ra 1/5^2011<1/5^2010 suy ra 1/5^2011 +1 <1/5^2010 + 1 (2)
từ (1) và (2) => 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}}
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
áp dụng tc \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{a+m}< 1\left(m\in N\right)\)
Ta có: \(A=\frac{5^{2010}+1}{5^{2011}+1}< \frac{5^{2010}+1+4}{5^{2011}+1+4}\)\(=\frac{5^{2010}+5}{5^{2011}+5}=\frac{5.\left(5^{2009}+1\right)}{5.\left(5^{2010}+1\right)}=\frac{5^{2009}+1}{5^{2010}+1}\)
\(\Rightarrow A< B\)
#)Giải :
Đầu tiên ta so sánh :
52010 và 52009
Vì 2010 > 2009 => 52010 > 52009 (1)
Tiếp theo :
1/52011 + 1 và 1/52010 + 1
Vì 2011 + 1 = 2012 và 2010 + 1 = 2011
Mà 2012 > 2011 => 1/52011 + 1 > 1/52010 + 1 (2)
Từ (1) và (2) => 52010 + 1/52011+1 > 52009+1/52010+1 => A > B
Vậy : A > B
#)Nếu đúng thì bn bảo mk nha :D
#~Will~be~Pens~#
ta có:5A = \(\frac{5^{2011}+5}{5^{2011}+1}\) = 1+\(\frac{4}{5^{2011}+1}\)
5B=\(\frac{5^{2010}+5}{5^{2010}+1}\)=1+\(\frac{4}{5^{2010}+1}\)
\(\frac{4}{5^{2011}+1}\)<\(\frac{4}{5^{2010}+1}\)=>1+\(\frac{4}{5^{2011}+1}\)<1+\(\frac{4}{5^{2010}+1}\)
=>5A<5B=>A<B
vậy:A<B
chúc pn hok tốt ^_^
Nhân 5 với B và A cho kết quả A<B