Question

Tìm Max

A= -5x²-4x+1

B= 2x+1/x²+2

Answer 1

A=1-[(√5x)^2+2.2(√5.x)/√5+4/5]+4/5

A=9/5-(√5x+2/√5)^2<=9/5

GtlnA=9/5 khi x=-2/5

Answer 2

a) A = -5x² - 4x + 1

A = -5( x² + 2.\(\dfrac{2}{5}\)x + \(\dfrac{4}{25}-\dfrac{4}{25}\)) + 1

A = -5\(\left(x+\dfrac{2}{5}\right)^2\)+ \(\dfrac{4}{5}+1\)

A = -5\(\left(x+\dfrac{2}{5}\right)^2\) + \(\dfrac{9}{5}\)

Do : -5\(\left(x+\dfrac{2}{5}\right)^2\) ≥ 0 ∀x

⇒ -5\(\left(x+\dfrac{2}{5}\right)^2\) + \(\dfrac{9}{5}\) ≥ \(\dfrac{9}{5}\)

⇒ A_MAX = \(\dfrac{9}{5}\) ⇔ x = \(\dfrac{-2}{5}\)

Answer 3

b) B = \(\dfrac{2x+1}{x^2+2}\)

B = \(\dfrac{x^2+2-x^2+2x-1}{x^2+2}\)

B = \(\dfrac{x^2+2}{x^2+2}-\dfrac{\left(x-1\right)^2}{x^2+2}\)

B = 1 - \(\dfrac{\left(x-1\right)^2}{x^2+2}\)

Do : - \(\dfrac{\left(x-1\right)^2}{x^2+2}\) ≤ 0 ∀x

⇒ 1 - \(\dfrac{\left(x-1\right)^2}{x^2+2}\) ≤ 1

⇒ B_Max = 1 ⇔ x = 1