A=\(\frac{3}{12-2x-x^2}\)
=\(\frac{3}{-\left(x^2+2x\right)+12}=\frac{3}{-\left(x^2+2.\frac{1}{2}x+\frac{1}{4}\right)+12+\frac{1}{4}}=\frac{3}{-\left(x+\frac{1}{2}\right)^2+\frac{49}{4}}\le\frac{3}{\frac{49}{4}}=\frac{12}{49}\)
Dấu = xảy ra khi : -(x+\(\frac{1}{2}\))2=0
\(\Leftrightarrow\)x=\(-\frac{1}{2}\)
Vậy Max A = \(\frac{12}{49}\Leftrightarrow x=-\frac{1}{2}\)