Giả sử: \(\dfrac{(ax+b)^2}{x^2}+k=\dfrac{2x+1}{x^2}\)
Ta có: \(\dfrac{(ax+b)^2}{x^2}+k\ge k(x\ne 0)\)
\(\dfrac{(ax+b)^2}{x^2}+k\\=\dfrac{a^2x^2+2abx+b^2+kx^2}{x^2}\\=\dfrac{x^2(a^2+k)+2abx+b^2}{x^2}\\=\dfrac{2x+1}{x^2}\)
Đồng nhất hệ số:
\(→\begin{cases}a^2+k=0\\ab=1\\b^2=1\end{cases}\\↔\begin{cases}k=-a^2\\ab=1\\b=±1\end{cases}\\↔\begin{cases}k=-a^2\\a=±1\\b=±1\end{cases}\\→k=-1\\→\dfrac{2x+1}{x^2}\ge -1\)
\(→\) Dấu "=" xảy ra khi \(\dfrac{2x+1}{x^2}=-1\\↔-x^2=2x+1\\↔-x^2-2x-1=0\\↔x^2+2x+1=0\\↔(x+1)^2=0\\↔x+1=0\\↔x=-1\)
Vậy \(\min =-1\) khi \(x=-1\)