Question

Cho 0 < x < \(\dfrac{1}{2}\). Tìm GTNN của A = \(\dfrac{2-x}{1-2x}+\dfrac{1+2x}{3x}\)

Answer 1

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

\(A=\dfrac{7}{6}+\dfrac{3}{2-4x}+\dfrac{1}{3x}=\dfrac{7}{6}+\dfrac{3}{2-4x}+\dfrac{\dfrac{4}{3}}{4x}\ge\dfrac{7}{6}+\dfrac{\left(\sqrt{3}+\sqrt{\dfrac{4}{3}}\right)^2}{2-4x+4x}=\dfrac{16}{3}\)

\(A_{min}=\dfrac{16}{3}\) khi \(\dfrac{\sqrt{3}}{2-4x}=\dfrac{\sqrt{\dfrac{4}{3}}}{4x}\Rightarrow x=\dfrac{1}{5}\)