\(\Leftrightarrow log_5\left(5x^2+5\right)\ge log_5\left(mx^2+4x+m\right)\)
BPT nghiệm đúng với mọi x khi và chỉ khi:
\(\left\{{}\begin{matrix}5x^2+5\ge mx^2+4x+m\\mx^2+4x+m>0\end{matrix}\right.\) ;\(\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x^2-4x+5\ge m\left(x^2+1\right)\\m\left(x^2+1\right)>-4x\end{matrix}\right.\) ;\(\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{5x^2-4x+5}{x^2+1}\\m>-\dfrac{4x}{x^2+1}\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}m\le\min\limits_{x\in R}\left(\dfrac{x^2-4x+1}{x^2+1}\right)\\m>\max\limits_{x\in R}\left(-\dfrac{4x}{x^2+1}\right)\end{matrix}\right.\)
Ta có: \(\dfrac{5x^2-4x+5}{x^2+1}=\dfrac{3\left(x^2+1\right)+2\left(x-1\right)^2}{x^2+1}=3+\dfrac{2\left(x-1\right)^2}{x^2+1}\ge3\)
\(\Rightarrow\min\limits_{x\in R}\left(\dfrac{5x^2-4x+5}{x^2+1}\right)=3\)
\(-\dfrac{4x}{x^2+1}=\dfrac{2\left(x^2+1\right)-2\left(x+1\right)^2}{x^2+1}=2-\dfrac{2\left(x+1\right)^2}{x^2+1}\le2\)
\(\Rightarrow\max\limits_{x\in R}\left(\dfrac{-4x}{x^2+1}\right)=2\)
\(\Rightarrow2< m\le3\)
