\(y=\left|4x-m\right|-x^2\)
\(TH1:4x-m\ge0\)
\(y=4x-m-x^2=-\left(x^2-4x+4\right)+4-m=-\left(x-2\right)^2+4-m\le4-m,\forall x\in R\)
\(\Rightarrow y\left(max\right)=4-m\left(x=2\right)\)
Để \(y\left(max\right)=5\Leftrightarrow4-m=5\Leftrightarrow m=-1\)
\(TH1:4x-m< 0\)
\(y=-4x+m-x^2=-\left(x^2+4x+4\right)+4+m=-\left(x+2\right)^2+4+m\le4+m,\forall x\in R\)
\(\Rightarrow y\left(max\right)=4+m\left(x=-2\right)\)
Để \(y\left(max\right)=5\Leftrightarrow4+m=5\Leftrightarrow m=1\)
Vậy với \(m=\pm1\) thỏa mãn đề bài