Question

Tìm m để \(f\left(x\right)=\frac{2x^2-4x+2}{x^2-3x+5}-m< 0\) đúng với moi x thuộc R

Answer 1

\(\Leftrightarrow g\left(x\right)=\frac{2x^2-4x+2}{x^2-3x+5}< m\) ; \(\forall x\in R\)

\(\Leftrightarrow m>\max\limits_{x\in R}\frac{2x^2-4x+2}{x^2-3x+5}\)

Ta có: \(\frac{2x^2-4x+2}{x^2-3x+5}=\frac{22x^2-44x+22}{11\left(x^2-3x+5\right)}=\frac{24\left(x^2-3x+5\right)-2\left(x^2-14x+49\right)}{11\left(x^2-3x+5\right)}=\frac{24}{11}-\frac{2\left(x-7\right)^2}{11\left[\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\right]}\le\frac{24}{11}\)

\(\Rightarrow m>\frac{24}{11}\)