a, Với mọi x ta có :
\(\left|x-4\right|\ge0\)
\(\Leftrightarrow-\left|x-4\right|\le0\)
\(\Leftrightarrow0,5-\left|x-4\right|\le0,5\)
Dấu "=" xảy ra khi :
\(\left|x-4\right|=0\)
\(\Leftrightarrow x=4\)
Vậy \(C_{Max}=0,5\Leftrightarrow x=4\)
d, Với mọi x ta có :
\(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\ge0\)
\(\Leftrightarrow-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0\)
\(\Leftrightarrow-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)+3\le3\)
\(\Leftrightarrow D\le3\)
Dấu "=" xảy ra khi :
\(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6=0\)
\(\Leftrightarrow\dfrac{4}{9}x-\dfrac{2}{15}=0\)
\(\Leftrightarrow x=\dfrac{3}{10}\)
Vậy \(D_{Max}=3\Leftrightarrow x=\dfrac{3}{10}\)
a, Với mọi x ta có :
\(\left|4,3-x\right|\ge0\)
\(\Leftrightarrow\left|4,3-x\right|+3,7\ge3,7\)
\(\Leftrightarrow A\ge3,7\)
Dấu "=" xảy ra khi :
\(\left|4,3-x\right|=0\)
\(\Leftrightarrow x=4,3\)
Vậy \(A_{Min}=3,7\Leftrightarrow x=4,3\)
b/ Với mọi x ta có :
\(\left(2x+\dfrac{1}{3}\right)^4\ge0\)
\(\Leftrightarrow\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)
\(\Leftrightarrow B\ge-1\)
Dấu "=" xảy ra khi :
\(\left(2x+\dfrac{1}{3}\right)^4=0\)
\(\Leftrightarrow2x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x=-\dfrac{1}{6}\)
Vậy \(B_{Min}=-1\Leftrightarrow x=-\dfrac{1}{6}\)
