Cho hàm số $y=f\left( x \right)$ có đạo hàm ${f}'\left( x...

Ta có: ${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=1\left( k\acute{e}p \right) \\
& x=0 \\
& x=2 \\
\end{aligned} \right.$$\Rightarrow y=f\left( x \right)y=f\left( {{x}^{2}}-8x+m \right){y}'=\left( 2x-8 \right){f}'\left( {{x}^{2}}-8x+m \right){y}'=0\Leftrightarrow \left[ \begin{aligned}
& 2x-8=0 \\
& {f}'\left( {{x}^{2}}-8x+m \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=4 \\
& {{x}^{2}}-8x+m=0 \\
& {{x}^{2}}-8x+m=2 \\
& {{x}^{2}}-8x+m=1\left( k\acute{e}p \right) \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=4 \\
& g\left( x \right)={{x}^{2}}-8x+m=0 \\
& h\left( x \right)={{x}^{2}}-8x+m-2=0 \\
\end{aligned} \right.y=f\left( {{x}^{2}}-8x+m \right)5\left\{ \begin{aligned}
& g\left( 4 \right)\ne 0 \\
& h\left( 4 \right)\ne 0 \\
& {{\Delta }_{g\left( x \right)}}={{4}^{2}}-{{m}^{2}}>0 \\
& {{\Delta }_{h\left( x \right)}}={{4}^{2}}-\left( m-2 \right)>0 \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& m-16\ne 0 \\
& m-18\ne 0 \\
& m<16 \\
& m<18 \\
\end{aligned} \right.$$\Leftrightarrow m<16$
Do nguyên dương nên . Vậy có giá trị của