Question

a) Tìm Giá trị nhỏ nhất A= (2x+\(\dfrac{1}{3}\))\(^4-1\)

b) Tìm Giá trị lớn nhất B= -\(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)

Answer 1

a/ Vì: \(\left(2x+\dfrac{1}{3}\right)^4\ge0\) với mọi x

=> \(\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)

dấu ''='' xảy ra khi :

\(2x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{6}\)

Vậy Min_A = -1 <=> \(x=-\dfrac{1}{6}\)

b/ Vì: \(\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\ge0\Rightarrow-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0\)

=> \(-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)

dấu ''='' xảy ra khi :

\(-\dfrac{4}{9}x-\dfrac{2}{15}=0\Leftrightarrow x=-\dfrac{3}{10}\)

vậy Max_B = 3 khi \(x=-\dfrac{3}{10}\)