Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
thế bài này bạn hỏi hay là tớ hỏi vậy
cậu chẳng ghi đề bài thì ai làm
Ta có: \(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Xét tử : \(2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+1+...+1\right)+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)( có 2008 số hạng 1 )
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)+1\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Ghép tử và mẫu....
Vậy A = 2009
\(\frac{2009^{2008}-1}{2009^{2009}-1}< \frac{2009^{2007}+1}{2009^{2008}+1}\)
Xét hiệu:
\(\frac{2009^{2007}+1}{2009^{2008}+1}-\frac{2009^{2008}-1}{2009^{2009}-1}\)
\(=\frac{\left(2009^{2007}+1\right)\cdot\left(2009^{2009}-1\right)-\left(2009^{2008}+1\right)\cdot\left(2009^{2008}-1\right)}{\left(2009^{2008}+1\right)\cdot\left(2009^{2009}-1\right)}\)
\(=\frac{\left(2009^{2016}+2009^{2009}-2009^{2007}-1\right)-\left(2009^{2016}-1\right)}{\left(2009^{2008}+1\right)\cdot\left(2009^{2009}-1\right)}\)
\(=\frac{2009^{2009}-2009^{2007}}{\left(2009^{2008}+1\right)\cdot\left(2009^{2009}-1\right)}>0\)
\(\Rightarrow\frac{2009^{2008}-1}{2009^{2009}-1}< \frac{2009^{2007}+1}{2009^{2008}+1}\left(đpcm\right)\)
Trả lời
Vế trái 0 không nhỏ hơn vế phải 0,câu cho sai
CM đúng sai đúng không ?
Đặt \(A=\frac{2009^{2008}-1}{2009^{2009}-1}\)
\(\Rightarrow2009A=\frac{2009.\left(2009^{2008}-1\right)}{2009^{2009}-1}=\frac{2009^{2009}-2009}{2009^{2009}-1}\)
\(=\frac{2009^{2009}-1-2008}{2009^{2009}-1}=1-\frac{2008}{2009^{2009}-1}\)
Đặt \(B=\frac{2009^{2007}+1}{2009^{2008}+1}\)
\(\Rightarrow2009B=\frac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\frac{2009^{2008}+2009}{2009^{2008}+1}\)
\(=\frac{2009^{2008}+1+2008}{2009^{2008}+1}=1+\frac{2008}{2009^{2008}+1}\)
Vì : \(\frac{2008}{2009^{2009}-1}< \frac{2008}{2009^{2008}+1}\)
\(\Rightarrow A=1-\frac{2008}{2009^{2009}-1}< B=1+\frac{2008}{2009^{2008}+1}\)
Vậy \(\frac{2009^{2008}-1}{2009^{2009}-1}< \frac{2009^{2007}+1}{2009^{2008}+1}\)