a) Đặt \(A=16x^2-6x+3\)
\(A=\left(16x^2-6x+\frac{9}{16}\right)+\frac{39}{16}\)
\(A=\left(4x-\frac{3}{4}\right)^2+\frac{39}{16}\)
Do \(\left(4x-\frac{3}{4}\right)^2\ge0\forall x\)
\(\Rightarrow A\ge\frac{39}{16}\)
Dấu "=" xảy ra khi :
\(4x-\frac{3}{4}=0\Leftrightarrow4x=\frac{3}{4}\Leftrightarrow x=\frac{3}{16}\)
Vậy ...
b) Đặt \(B=\frac{5}{3}x^2-x+1\)
\(\frac{5}{3}B=\frac{25}{9}x^2-\frac{5}{3}x+\frac{5}{3}\)
\(\frac{5}{3}B=\left(\frac{25}{9}x^2-\frac{5}{3}x+\frac{1}{4}\right)+\frac{17}{12}\)
\(\frac{5}{3}B=\left(\frac{5}{3}x-\frac{1}{2}\right)^2+\frac{17}{12}\)
Do \(\left(\frac{5}{3}x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\frac{5}{3}B\ge\frac{17}{12}\Leftrightarrow B\ge\frac{17}{20}\)
Dấu "=" xảy ra khi :
\(\frac{5}{3}x-\frac{1}{2}=0\Leftrightarrow\frac{5}{3}x=\frac{1}{2}\Leftrightarrow x=\frac{3}{10}\)
Vậy ...
