So sánh m và n:
M=2007/2008+2008+2009
N=2007+2008/2008+2009
so sánh M và N biết M=2007/2008+2008/2009 và N=2007+2008/2008+2009
so sánh M và N biết M =2007/ 2008 +2008/ 2009 và N= 2007 +2008 /2008+ 2009
M=4017,99
N=4017
NÊN M>N
so sánh M và N
M= 2007/2008 + 2008/2009 N= 2007+2008/2008+2009
So sánh
M bằng 2007/ 2008 công 2008/2009
N bằng 2007/2008 công 2008/2009
So sánh 2 biểu thức:
A = 2006/2007 + 2007/2008 + 2008/2009
B = 2006 + 2007 + 2008/2007 + 2008 + 2009
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}\)
\(=3-\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>1\).
\(B=\frac{2006+2007+2008}{2007+2008+2009}< \frac{2007+2008+2009}{2007+2008+2009}=1\).
Suy ra \(A>B\).
so sánh 2008 với tổng 2009 số hạng sau\(s=\frac{2008+2007}{2009+2008}+\frac{^{2008^2+2007^2}}{2009^2+2008^2}+.....+\frac{2008^{2009}+2007^{2009}}{2009^{2009}+2008^{2009}}\)
2. So sánh A và B:
A= 2006/2007 - 2007/2008 + 2008/2009 - 2009/2010
B=-1/2006*2007 - 1/2008*2009
So sánh:
2007/2008 + 2008/2009 + 2009/2007 và 3
2007/2008+2008/2009+2009/2007=3.000000744>3
so sánh 2008^2008+1 / 2008^2009+1 và 2008^2007+1 / 2008^2008+1
ta có Đặt \(A=\frac{2008^{2008}+1}{2008^{2009}+1}\)
\(B=\frac{2008^{2007}+1}{2008^{2008}+1}\)
Xét A trước ta có
\(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}\)\(2008A=\frac{2008^{2009}+2008}{2008^{2009}+1}\)
\(2008A=\frac{2008^{2009}+1+2007}{2008^{2009}+1}\)suy ra \(2008A=1+\frac{2007}{2008^{2009}+1}\)
Xét B ta có
\(2008B=\frac{2008.\left(2008^{2007}+1\right)}{2008^{2008}+1}\)suy ra \(2008B=\frac{2008^{2008}+2008}{2008^{2008}+1}\)
\(2008B=\frac{2008^{2008}+1+2007}{2008^{2008}+1}\)suy ra \(2008B=1+\frac{2007}{2008^{2008}+1}\)
VÌ \(1+\frac{2007}{2008^{2009}+1}
Đặt \(a=2008^{2007};\)
\(A=\frac{2008^{2008}+1}{2008^{2009}+1}=\frac{2008a+1}{2008^2.a+1};\text{ }B=\frac{2008^{2007}+1}{2008^{2008}+1}=\frac{a+1}{2008a+1}\)
Quy đồng mẫu ta có:
\(A=\frac{\left(2008a+1\right)\left(2008a+1\right)}{\left(2008^2a+1\right)\left(2008a+1\right)}=\frac{2008^2a^2+2.2008a+1}{\left(2008^2a+1\right)\left(2008a+1\right)}\)
\(B=\frac{\left(a+1\right)\left(2008^2a+1\right)}{\left(2008a+1\right)\left(2008^2a+1\right)}=\frac{2008^2a^2+\left(2008^2+1\right)a+1}{\left(2008a+1\right)\left(2008^2a+1\right)}\)
So sánh ở tử ta thấy \(2.2008