Đây nha:
\(A=27-\dfrac{12x}{x^2+9}=29-2-\dfrac{12x}{x^2+9}=29-\dfrac{2x^2+12x+18}{x^2+9}\)
\(=29-\dfrac{2\left(x+3\right)^2}{x^2+9}\le29\)
Dấu "=" xảy ra khi:\(x+3=0\Leftrightarrow x=-3\)
Vậy\(maxA=29\Leftrightarrow x=-3\)
A = [(4x^2 + 36) - (4x^2 + 12x + 9)] / (x^2 + 9) = 4 - [(2x + 3)^2]/(x^2 + 9) =< 4
GTLN là 4 khi x = -3/2
A = [(x^2 - 12x + 36) - (x^2 + 9)] / (x^2 + 9) = [(x - 6)^2]/(x^2 + 9) - 1 >= -1
GTNN là -1 khi x = 6