Question

tìm GTLN của :\(\dfrac{2}{3x^2-x+5}\)

Answer 1

Ta có: 3x²-x+5=3(x-\(\dfrac{1}{6}\))²+\(\dfrac{59}{12}\)

Do 3(x-\(\dfrac{1}{6}\))²\(\ge\)0 với mọi x

=> 3(x-\(\dfrac{1}{6}\))²+\(\dfrac{59}{12}\)\(\ge\) \(\dfrac{59}{12}\) với mọi x

=>\(\dfrac{2}{3\left(x-\dfrac{1}{6}\right)^2+\dfrac{59}{12}}\) \(\le\) \(\dfrac{2}{\dfrac{59}{12}}\)

=> \(\dfrac{2}{3\left(x-\dfrac{1}{6}\right)^2+\dfrac{59}{12}}\) \(\le\) \(\dfrac{24}{59}\)

Dấu (=) xảy ra <=> 3(x-\(\dfrac{1}{6}\))²=0

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