Question

Tìm GTLN hoặc GTNN :

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

B = \(\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\)

Help me!!!

Answer 1

a, Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

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

Dấu " = " khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

b, Để B lớn nhất thì \(\left(x-\dfrac{2}{3}\right)^2+9\) nhỏ nhất

Ta có: \(\left(x-\dfrac{2}{3}\right)^2+9\ge9\)

\(\Leftrightarrow B=\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\le\dfrac{4}{9}\)

Dấu " = " khi \(\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(MAX_B=\dfrac{4}{9}\) khi \(x=\dfrac{2}{3}\)