\(C=\dfrac{15\left|x+1\right|+32}{6\left|x+1\right|+8}\)
\(\left\{{}\begin{matrix}15\left|x+1\right|\ge0\forall x\\6\left|x+1\right|\ge0\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}15\left|x+1\right|+32\ge32\\6\left|x+1\right|+8\ge8\end{matrix}\right.\)
\(\Rightarrow C=\dfrac{15\left|x+1\right|+32}{6\left|x+1\right|+8}\le\dfrac{32}{8}\)
\(\Rightarrow C\le4\)
Dấu "=" xảy ra khi:
\(\left|x+1\right|=0\Rightarrow x=-1\)
Đặt: \(\left|x+1\right|=t\ge0\) ta có:
\(pt\Leftrightarrow C=\dfrac{15t+32}{6t+8}=\dfrac{12t+16}{6t+8}+\dfrac{3t+4}{6t+8}+\dfrac{12}{6t+8}\)
\(=\dfrac{2\left(6t+8\right)}{6t+8}+\dfrac{3t+4}{2\left(3t+4\right)}+\dfrac{12}{6t+8}\)
\(=2+\dfrac{1}{2}+\dfrac{12}{6t+8}\le2+\dfrac{1}{2}+\dfrac{12}{8}=4\)
Dấu "=" khi \(t=0\Leftrightarrow x=-1\)
