a)\(C=-\left|2-3x\right|+\frac{1}{2}\)
Vì \(\left|2-3x\right|\ge0\Rightarrow-\left|2-3x\right|\le0\Rightarrow-\left|2-3x\right|+\frac{1}{2}\le0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow C\le\frac{1}{2}\)
\(\Rightarrow MAX_C=\frac{1}{2}\Leftrightarrow-\left|2-3x\right|=0\Leftrightarrow2-3x=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
b)\(D=-3-\left|2x+4\right|\)
\(D=-\left(3+\left|2x+4\right|\right)\)
Vì \(\left|2x+4\right|\ge0\Rightarrow3+\left|2x+4\right|\ge3+0=3\Rightarrow-\left(3+\left|2x+4\right|\right)\le-3\)
\(\Rightarrow D\le-3\)
\(\Rightarrow MAX_D=-3\Leftrightarrow\left|2x+4\right|=0\Leftrightarrow2x+4=0\Leftrightarrow2x=-4\Leftrightarrow x=-2\)
a) \(C=-\left|2-3x\right|+\frac{1}{2}\)
Vì: \(\left|2-3x\right|\ge0,\forall x\)
=> \(-\left|2-3x\right|\le0\)
=>\(-\left|2-3x\right|+\frac{1}{2}\le\frac{1}{2}\)
Vậy GTLN của C là \(\frac{1}{2}\) khi \(x=\frac{2}{3}\)
b)\(D=-3-\left|2x+4\right|\)
Vì: \(\left|2x+4\right|\ge0,\forall x\)
=> \(-\left|2x+4\right|\le0\)
=> \(-3-\left|2x+4\right|\le-3\)
Vậy GTLN của D là -3 khi \(x=-2\)
