Question

Tìm Max của A = \(\dfrac{3x^2-6x+17}{x^2-2x+5}\)

Answer 1

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

Mà \(x^2-2x+5\ge4\)

=> \(\dfrac{2}{x^2-2x+5}\le\dfrac{1}{2}\)

=> A ≤ 7/2

Dấu "=" xảy ra ⇔ x = 1

Answer 2

Ta có : \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}=\dfrac{3x^2-6x+15+2}{x^2-2x+5}=\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

- Thấy : \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

Lại có : \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow\dfrac{2}{x^2-2x+5}\le\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Rightarrow3+\dfrac{2}{x^2-2x+5}\le\dfrac{7}{2}\)

\(HayA\le\dfrac{7}{2}\)

Vậy Max_A = \(\dfrac{7}{2}\) Dấu " = " xảy ra <=> x - 1 = 0

<=> x = 1 .