Câu hỏi của Nguyễn Minh Dương

Question

So sánh bt: $A=\dfrac{2008^{2008}+1}{2008^{2009}+1};B=\dfrac{2008^{2007}+1}{2008^{2008}+1}$

Toru · Accepted Answer

$A=\dfrac{2008^{2008}+1}{2008^{2009}+1}$$2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}$$=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}$$=1+\dfrac{2007}{2008^{2009}+1}$$B=\dfrac{2008^{2007}+1}{2008^{2008}+1}$$2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}$$=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}$$=1+\dfrac{2007}{2008^{2008}+1}$Ta có: $2008^{2009}+1>2008^{2008}+1$$\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}$$\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}$$\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}$hay $A < B$#$Toru$Câu trả lời: $A=\dfrac{2008^{2008}+1}{2008^{2009}+1}$$2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}$$=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}$$=1+\dfrac{2007}{2008^{2009}+1}$$B=\dfrac{2008^{2007}+1}{2008^{2008}+1}$$2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}$$=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}$$=1+\dfrac{2007}{2008^{2008}+1}$Ta có: $2008^{2009}+1>2008^{2008}+1$$\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}$$\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}$$\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}$hay $A < B$#$Toru$

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