Question

Tìm GTLN :

a) A = ( x + 2)( 3 - x) biết - 2 ≤ x ≤ 3

b) B = x³ + \(\dfrac{3}{x^2}\) với x > 0

Answer 1

b/ B = \(x^3+\dfrac{3}{x^2}=\dfrac{x^3}{2}+\dfrac{x^3}{2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}\ge5\sqrt[5]{\dfrac{x^3}{2}\cdot\dfrac{x^3}{2}\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{x^2}}=5\sqrt[5]{\dfrac{1}{4}}\)

Dấu ''='' xảy ra khi \(\dfrac{x^3}{2}=\dfrac{1}{x^2}\Leftrightarrow x^5=2\Leftrightarrow x=\sqrt[5]{2}\)

Vậy: \(MIN_B=5\sqrt[5]{\dfrac{1}{4}}\Leftrightarrow x=\sqrt[5]{2}\)

Answer 2

Ta có : - 2 ≤ x ≤ 3

⇒ x + 2 ≥ 0 và 3 - x ≥ 0

Áp dụng BĐT Cô - Si , ta có :

a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ ( a + b)² ≥ 4ab

⇔\(\dfrac{\left(a+b\right)^2}{4}\)≥ ab

⇒ A = ( x + 2)( 3 - x) ≤ \(\left[\dfrac{\left(x+2\right)+\left(3-x\right)}{2}\right]^2=\left(\dfrac{5}{2}\right)^2=\dfrac{25}{4}\)

⇒ A_MAX = \(\dfrac{25}{4}\) ⇔ x + 2 = 3 - x ⇔ x = \(\dfrac{1}{2}\)