Câu hỏi của PhamHoangDuong

Question

So sánh :A=$\frac{2005^{2005}+1}{2005^{2006}+1}$và B=$\frac{2005^{2004}+1}{2005^{2005}+1}$

𝐓𝐡𝐮𝐮 𝐓𝐡𝐮𝐲𝐲 · Accepted Answer

Ta có VẾ A$A=\frac{2005^{2005}+1}{2005^{2006}+1}$$2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}$$2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}$$2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}$$2005\cdot A=1+\frac{2004}{2005^{2006}+1}$Ta lại có Vế B :$B=\frac{2005^{2004}+1}{2005^{2005}+1}$$2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}$$2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}$$2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}$$2005\cdot B=1+\frac{2004}{2005^{2005}+1}$Nhìn vào trên , suy ra A < B . Câu trả lời: Ta có VẾ A$A=\frac{2005^{2005}+1}{2005^{2006}+1}$$2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}$$2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}$$2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}$$2005\cdot A=1+\frac{2004}{2005^{2006}+1}$Ta lại có Vế B :$B=\frac{2005^{2004}+1}{2005^{2005}+1}$$2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}$$2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}$$2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}$$2005\cdot B=1+\frac{2004}{2005^{2005}+1}$Nhìn vào trên , suy ra A < B .

Wall HaiAnh · Answer

$2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}$$2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}$Ta thấy $2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}$$\Rightarrow A< B$Câu trả lời: $2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}$$2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}$Ta thấy $2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}$$\Rightarrow A< B$

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