Question

Tìm max B=2:(x²-6x+17)

min C=3:(4x-x²+10)

Answer 1

1 ) Ta có : \(x^2-6x+17=x^2-6x+9+8=\left(x-3\right)^2+8\ge8\forall x\)

\(\Rightarrow\dfrac{2}{x^2-6x+17}\le\dfrac{2}{8}\)

\(\Rightarrow B\le\dfrac{1}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Max B là : \(\dfrac{1}{4}\Leftrightarrow x=3\)

2 ) Ta có : \(4x-x^2+10=-\left(x^2-4x+4\right)+14=14-\left(x-2\right)^2\le14\forall x\)

\(\Rightarrow\dfrac{3}{4x-x^2+10}\ge\dfrac{3}{14}\)

\(\Rightarrow C\ge\dfrac{3}{14}\)

Dấu " = " xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy Min C là : \(\dfrac{3}{14}\Leftrightarrow x=2\)