Question

Tìm giá trị nhỏ nhất của D với \(D=\dfrac{x^2-3x+3}{x^2-2x+1}\)

Answer 1

\(D=\dfrac{x^2-3x+3}{x^2-2x+1}=\dfrac{3}{4}+\dfrac{x^2-3x+3-\dfrac{3}{4}x^2+\dfrac{9}{4}x-\dfrac{9}{4}}{x^2-2x+1}=\dfrac{3}{4}+\dfrac{\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{3}{4}}{x^2-2x+1}\)\(=\dfrac{3}{4}+\dfrac{\left(\dfrac{1}{2}x-\sqrt{\dfrac{3}{4}}\right)^2}{\left(x-1\right)^2}\le\dfrac{3}{4}\)

Vậy \(Min_D=\dfrac{3}{4}\)khi \(\dfrac{1}{2}x-\sqrt{\dfrac{3}{4}}=0\Rightarrow\dfrac{1}{2}x=\sqrt{\dfrac{3}{4}}\Rightarrow x=\sqrt{\dfrac{3}{4}}-\dfrac{1}{2}\)