Question

Tìm giá trị lớn nhất của biểu thức
9 - 9x² + 2x - \(\dfrac{2}{9}\)

Answer 1

\(9-9x^2+2x-\dfrac{2}{9}\\ =-\left(9x^2-2x+\dfrac{1}{9}-\dfrac{80}{9}\right)\\ =-\left(3x+\dfrac{1}{3}\right)^2+\dfrac{80}{9}\le\dfrac{80}{9}\)

Dấu "=" xảy ra khi \(-\left(3x+\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow3x+\dfrac{1}{3}=0\\ \Leftrightarrow3x=-\dfrac{1}{3}\\ \Leftrightarrow x=-\dfrac{1}{9}\)

Vậy \(Max=\dfrac{80}{9}\Leftrightarrow x=-\dfrac{1}{9}\)

Answer 2

9 - 9x² + 2x - \(\dfrac{2}{9}\)
=\(\dfrac{80}{9}\)-[(3x)²-2x+(\(\dfrac{1}{3}\))²]
=\(\dfrac{80}{9}\)-(3x-\(\dfrac{1}{3}\))²
Vì (3x-\(\dfrac{1}{3}\))²≥0 ⇒-(3x-\(\dfrac{1}{3}\))²≤0⇒\(\dfrac{80}{9}\)-(3x-\(\dfrac{1}{3}\))²≤\(\dfrac{80}{9}\)
Trường hợp dấu bằng xảy ra khi: (3x-\(\dfrac{1}{3}\))²=0⇒3x-\(\dfrac{1}{3}\)=0⇒3x=\(\dfrac{1}{3}\)⇒x=\(\dfrac{1}{9}\)
Vậy GTLN của biểu thức là \(\dfrac{80}{9}\) khi x=\(\dfrac{1}{9}\)