so sánh \(\frac{2009^{2008}+1}{2009^{2009}+1}\)và \(\frac{2009^{2008}+5}{2009^{2008}+9}\)
so sánh\(\frac{2009^{2008+1}}{2009^{2009+1}}\) và \(\frac{2009^{2008+5}}{2009^{2009+9}}\)
Ta có:
\(\frac{2009^{2008+1}}{2009^{2009+1}}=\frac{2009^{2009}}{2009^{2010}}=\frac{1}{2009}\)
\(\frac{2009^{2008+5}}{2009^{2009+9}}=\frac{2009^{2013}}{2009^{2018}}=\frac{1}{2009^5}\)
=>Đẳng thức trên lớn hơn đẳng thức dứi(vì 2009<2009^5)
Vậy.......
so sánh 2008 với tổng 2009 số hạng sau\(s=\frac{2008+2007}{2009+2008}+\frac{^{2008^2+2007^2}}{2009^2+2008^2}+.....+\frac{2008^{2009}+2007^{2009}}{2009^{2009}+2008^{2009}}\)
so sánh 2 phân số : \(A=\frac{2008^{2009}+2}{2008^{2009}-1};B=\frac{2008^{2009}}{2008^{2009}-3}\)
So sánh \(\frac{2008}{2009}+\frac{2009}{2010}và\frac{2008+2009}{2009+2010}\)
So sánh \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}\)và \(\sqrt{2008}+\sqrt{2009}\)
Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}=\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)
Vì \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\) nên \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)
\(\Rightarrow\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)>\sqrt{2009}+\sqrt{2008}\)
Hay \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}>\sqrt{2008}+\sqrt{2009}\)
So sánh
(2008^2009+1)/ (2008^2009-1) và (2008^2009)/ (2008^2009-3)
So sánh : \(A=\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}vàB=\frac{2008+2009+2010}{2009+2010+2011}\)
thực hiện tính và so sánh A=\(\frac{2008^{2009}+1}{2009^{2009}+1}\)và B=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)
So Sánh
\(\frac{2007}{2008}+\frac{2008}{2009}\)và \(\frac{2007}{2008}+\frac{2008}{2009}\)
\(\frac{2007}{2008}\)\(+\)\(\frac{2008}{2009}\)\(=\)\(\frac{2007}{2008}\)\(+\)\(\frac{2008}{2009}\)
k mk nha!!! *o~
\(\frac{2007}{2008}+\frac{2008}{2009}=\frac{2007}{2008}+\frac{2008}{2009}\)
nha ^_^
Nguyễn Vân AnhMọi người giúp mjk cách lm nữa nhé
Thank moj ng nhìu