a, với mọi x.Có:
|x+\(\frac{5}{6}\)| > 0
=> 2-|x+\(\frac{5}{6}\) | > 2
=> A > 2
dấu = xảy ra <=> |x+5/6|=0
<=> x+5/6=0
<=> x=-5/6
vậy GTLN A=2 khi x=-5/6
a) Vì \(\left|x+\frac{5}{6}\right|\ge0\forall x\)\(\Rightarrow2-\left|x+\frac{5}{6}\right|\le2\forall x\)
hay \(A\le2\)
Dấu " = " xảy ra \(\Leftrightarrow x+\frac{5}{6}=0\)\(\Leftrightarrow x=\frac{-5}{6}\)
Vậy \(maxA=2\Leftrightarrow x=\frac{-5}{6}\)
b) Vì \(\left|\frac{2}{3}-x\right|\ge0\forall x\)\(\Rightarrow5-\left|\frac{2}{3}-x\right|\le5\forall x\)
hay \(B\le5\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{2}{3}-x=0\)\(\Leftrightarrow x=\frac{2}{3}\)
Vậy \(maxB=5\Leftrightarrow x=\frac{2}{3}\)