Ta có: \(\left|x+5\right|\ge0,\forall x\in R\)
\(\Rightarrow-\left|x+5\right|\le0\Rightarrow12-\left|x+5\right|\le12+0=12\)
\(\Rightarrow\dfrac{4}{12-\left|x+5\right|}\ge\dfrac{4}{12}=\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x+5\right|=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\)