a) \(A=-\left|x+\frac{3}{4}\right|-3\)
Vì \(\left|x+\frac{3}{4}\right|\ge0\Rightarrow-\left|x+\frac{3}{4}\right|\le0\Rightarrow A=-\left|x+\frac{3}{4}\right|-3\le-3\)
=>\(A_{max}=-3\)=> \(\left|x+\frac{3}{4}\right|=0\Rightarrow x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)
Vậy Amax = -3 khi x=-3/4
b) \(B=2-\left(x+\frac{5}{6}\right)^2\)
Vì \(\left(x+\frac{5}{6}\right)^2\ge0\Rightarrow B=2-\left(x+\frac{5}{6}\right)^2\le2\)
=>\(B_{max}=2\Rightarrow\left(x+\frac{5}{6}\right)^2=0\Rightarrow x+\frac{5}{6}=0\Rightarrow x=-\frac{5}{6}\)
Vậy Bmax=2 khi x=-5/6