\(M=\dfrac{2x^2+4x-1}{x^2+1}=\dfrac{4x^2+4x+1-2x^2-2}{x^2+1}\\ M=\dfrac{\left(2x+1\right)^2}{x^2+1}-\dfrac{2\left(x^2+1\right)}{x^2+1}\\ M=\dfrac{\left(2x+1\right)^2}{x^2+1}-2\)
\(vì\:\dfrac{\left(2x+1\right)^2}{x^2+1}\ge0\left(\forall x\in R\right)\\ \Rightarrow\dfrac{\left(2x+1\right)^2}{x^2+1}-2\ge-2\\ hay\:M\ge-2\)
đẳng thức xảy ra khi 2x+1=0 => x=-1/2
vậy MINM=-2 khi x=-1/2