Question

Tìm GTNN của phân thức \(A=\dfrac{2030x^2+8x+1}{x^2}\left(x\ne0\right)\)

Answer 1

\(A=2030+\dfrac{8}{x}+\dfrac{1}{x^2}=\left(\dfrac{1}{x}\right)^2+8.\dfrac{1}{x}+16+2014\)

\(\Rightarrow A=\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)

\(\Rightarrow A_{min}=2014\) khi \(\dfrac{1}{x}+4=0\Rightarrow x=-\dfrac{1}{4}\)

Answer 2

\(A=\dfrac{2030x^2+8x+1}{x^2}\\ =\dfrac{2030x^2}{x^2}+\dfrac{8x}{x^2}+\dfrac{1}{x^2}\\ =2030+\dfrac{8}{x}+\dfrac{1}{x^2}\\ =\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot4+16+2014\\ =\left(\dfrac{1}{x}+4\right)^2+2014\)

Do \(\left(\dfrac{1}{x}+4\right)^2\ge0,2014>0\)

\(\Rightarrow\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)

\(\Rightarrow Min\left(A\right)=2014\Leftrightarrow\dfrac{1}{x}+4=0\Rightarrow x=\dfrac{-1}{4}\)