\(A=\frac{2004x+1}{2005x-2005}=\frac{2004x+1}{2005\left(x-1\right)}=\frac{2004\left(x-1\right)+2005}{2005\left(x-1\right)}=\frac{2004}{2005}+\frac{1}{x-1}\)

\(A_{max}\Leftrightarrow\frac{1}{x-1}max\)

Nếu x > 1 thì x-1 < 0 \(\Rightarrow\frac{1}{x-1}>0\)

Nếu x<1 thì x-1 <0 \(\Rightarrow\frac{1}{x-1}< 0\)

Xét \(x>1;\)ta có

\(\frac{1}{x-1}max\)=> x-1 là số nguyên dương nhỏ nhất 

\(\Rightarrow x-1=1\Rightarrow x=2\left(t/m\right)\)

Vậy \(B_{max}=1\frac{2004}{2005}\Leftrightarrow x=2\)

1, \(x^2-2003x-2004=0\)

\(\Rightarrow x^2+x-\left(2004x+2004\right)=0\)

\(\Rightarrow x\left(x+1\right)-2004\left(x+1\right)=0\)

\(\Rightarrow\left(x-2004\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2004=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2004\\x=-1\end{matrix}\right.\)

Vậy x = 2004 hoặc x = -1

2, \(2005x^2-2004x-1=0\)

\(\Rightarrow2005x^2-2005x+x-1=0\)

\(\Rightarrow2005x\left(x-1\right)+\left(x-1\right)=0\)

\(\Rightarrow\left(2005x+1\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2005x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2005}\\x=1\end{matrix}\right.\)

Vậy \(x=\dfrac{-1}{2005}\) hoặc x = 1

Cảm ơn bạn nhiều nha!

a) 0,(26)<0,261

b) 0,15>0,14(9)

a: 0,(26)<0,261

b: 0,15>0,14(9)