Đặt:
\(min=5+\dfrac{15}{4\left|3x+7\right|+3}\)
Ta có: \(\left|3x+7\right|\ge0\Rightarrow4\left|3x+7\right|\ge0\forall x\)
\(\Rightarrow4\left|3x+7\right|+3\ge3\)
\(\Rightarrow\dfrac{15}{4\left|3x+7\right|+3}\le5\)
\(5+\dfrac{15}{4\left|3x+7\right|+3}\le10\)
Vậy \(max_{min}=10\)