\(\left\{{}\begin{matrix}x^2-6x+9\ge x^2+7x+1\\x\ge\dfrac{2m-8}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{8}{13}\\x\ge\dfrac{2m-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow(-\text{∞};\dfrac{8}{13}]\cap[\dfrac{2m-8}{5};+\text{∞})=\phi\Leftrightarrow\dfrac{8}{3}< \dfrac{2m-8}{5}\Leftrightarrow m>\dfrac{72}{13}\)
Bất phương trình \(( x − 3 ) 2 ≥ x 2 + 7 x + 1 ⇔ x 2 − 6 x + 9 ≥ x 2 + 7 x + 1\)
\(⇔ − 6 x + 9 ≥ 7 x + 1 ⇔ 8 ≥ 13 x ⇔ x ≤\)\(\dfrac{8}{13}\)\(⇒ S 1 = ( − ∞ \)\(;\dfrac{8}{13}\)\(]\)
Bất phương trình \(2 m ≤ 8 + 5 x ⇔ 5 x ≥ 2 m − 8 ⇔ x ≥ \)\(\dfrac{2m-8}{5}\)
\(⇒ S 2 = [ \)\(\dfrac{2m-8}{3}\)\(; + ∞ ) .\)
Để hệ bất phương trình vô nghiệm \(⇔ S 1 ∩ S 2 = ∅ ⇔\)\(\dfrac{8}{13}\)<\(\dfrac{2m-8}{5}\)\(⇔ m >\)\(\dfrac{72}{13}\)
Vậy \(m>\)\(\dfrac{72}{13}\)