Nhận thấy \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
=> \(\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> \(2x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{6}\)
Vậy Min A = -1 <=> X = -1/6
a, \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\)
Dấu "=" xảy ra <=> 2x+1/3=0 <=> x= -1/6
b) Sửa đề \(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)
Ta có \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\forall x\)
=> \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\le3\forall x\)
Dấu "=" xảy ra <=> \(\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\)
Vậy Max B = 3 <=> x = 3/10