a) \(C=-\left|2-3x\right|+\frac{1}{2}\)
Ta có: \(\left|2-3x\right|\ge0\) \(\forall x.\)
\(\Rightarrow-\left|2-3x\right|\le0\) \(\forall x.\)
\(\Rightarrow-\left|2-3x\right|+\frac{1}{2}\le\frac{1}{2}\) \(\forall x.\)
\(\Rightarrow C\le\frac{1}{2}.\)
Dấu '' = '' xảy ra khi:
\(2-3x=0\)
\(\Rightarrow3x=2-0\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\frac{2}{3}.\)
Vậy \(MAX_C=\frac{1}{2}\) khi \(x=\frac{2}{3}.\)
b) \(D=-3-\left|2x+4\right|\)
Ta có: \(\left|2x+4\right|\ge0\) \(\forall x.\)
\(\Rightarrow-\left|2x+4\right|\le0\) \(\forall x.\)
\(\Rightarrow-3-\left|2x+4\right|\le-3\) \(\forall x.\)
\(\Rightarrow D\le-3.\)
Dấu '' = '' xảy ra khi:
\(2x+4=0\)
\(\Rightarrow2x=0-4\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2.\)
Vậy \(MAX_D=-3\) khi \(x=-2.\)
Chúc bạn học tốt!
a/ Với mọi x,y ta có :
\(\left|2-3x\right|\ge0\)
\(\Leftrightarrow-\left|2-3x\right|\le0\)
\(\Leftrightarrow\frac{1}{2}-\left|2-3x\right|\le\frac{1}{2}\)
\(\Leftrightarrow C\le\frac{1}{2}\)
Dấu "=" xảy ra :
\(\Leftrightarrow2-3x=0\Leftrightarrow x=\frac{2}{3}\)
Vậy...
d/ Với mọi x ta có :
\(\left|2x+4\right|\ge0\)
\(\Leftrightarrow-\left|2x+4\right|\le0\)
\(\Leftrightarrow-3-\left|2x+4\right|\le-3\)
\(\Leftrightarrow D\le-3\)
Dấu "=" xảy ra \(\Leftrightarrow2x+4=0\Leftrightarrow x=-2\)
Vậy...
