Ta có
\(B=10-3\left|x^2-9\right|-2\left|x+3\right|\)
\(\Rightarrow B=10-3\left|x-3\right|\left|x+3\right|-2\left|x+3\right|\)
Ta có :
\(\left|x+3\right|\ge0\) với mọi x
\(\Rightarrow\begin{cases}3\left|x-3\right|.\left|x+3\right|\ge0\\2\left|x+3\right|\ge0\end{cases}\)
\(\Rightarrow3\left|x-3\right|\left|x+3\right|+2\left|x+3\right|\ge0\)
\(\Rightarrow-3\left|x-3\right|\left|x+3\right|-2\left|x+3\right|\le0\)
\(\Rightarrow10-3\left|x-3\right|\left|x+3\right|-2\left|x+3\right|\le10\)
\(\Rightarrow B\le10\)
Dấu " = " xảy ra khi \(\begin{cases}3\left|x+3\right|\left|x-3\right|=0\\2\left|x-3\right|=0\end{cases}\)
\(\Rightarrow x=-3\)
Vậy MAX B = 10 khi x = - 3
