Đặt: \(A=\dfrac{3}{x^2+5x+3}\)
Để A lớn nhất thì \(x^2+5x+3\) nhỏ nhất.
Có: \(x^2+5x+3=\left(x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)-\dfrac{13}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{13}{4}\)
Vì: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\)
Dấu ''='' xảy ra khi x = -5/2
=> \(A_{Max}=\dfrac{3}{-\dfrac{13}{4}}=-\dfrac{12}{3}\) khi x = -5/2
Gọi A=\(\dfrac{3}{x^2+5x+3}\) = \(\dfrac{3}{\left(x^2+5x+\dfrac{25}{4}\right)-\left(\dfrac{25}{4}-3\right)}\) = \(\dfrac{3}{\left(x+\dfrac{5}{4}\right)^2-\dfrac{13}{4}}\)
Vì (x + \(\dfrac{5}{4}\))2 ≥ 0 ∀x nên (x + \(\dfrac{5}{4}\))2 - \(\dfrac{13}{4}\) ≥ -\(\dfrac{13}{4}\)
Do đó
\(\dfrac{1}{\left(x+\dfrac{5}{4}\right)^2-\dfrac{13}{4}}\) ≤ \(\dfrac{-4}{13}\)
<=> \(\dfrac{3}{\left(x+\dfrac{5}{4}\right)^2-\dfrac{13}{4}}\) ≤ \(\dfrac{-12}{13}\)
<=> A ≤ \(\dfrac{-12}{13}\)
Vậy A đạt GTLN bằng \(\dfrac{-12}{13}\)
Khi và chỉ khi x + \(\dfrac{5}{4}\) = 0
<=> x = -\(\dfrac{5}{4}\)
