\(~~~~~~~HD~~~~~~~\)
\(Tacó\)
\(\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)
\(\Rightarrow C\ge-1\Rightarrow C_{min}=-1\)
Dấu "=" xảy ra khi:
\(2x+\frac{1}{3}=0\Leftrightarrow2x=-\frac{1}{3}\Leftrightarrow x=-\frac{1}{6}\)
Vậy GTNN của C là: -1 khi: x=-1/6
\(C=\left(2x+\frac{1}{3}\right)^4-1\)
\(\text{Vì }\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)
\(\Rightarrow C=\left(2x+\frac{1}{3}\right)^4-1\ge-1\)
Dấu '' = '' xảy ra khi :
\(\left(2x+\frac{1}{3}\right)^4=0\)
\(\Rightarrow2x+\frac{1}{3}=0\)
\(\Rightarrow2x=-\frac{1}{3}\)
\(\Rightarrow x=-\frac{1}{6}\)
Vậy MinC = -1 <=> x = - 1/6
