`P=x/[(x+a)^2]`
`P=x/[(x+a)^2]-1/(4a)+1/(4a)`
`P=(4ax-x^2-2ax-a^2)/[4a(x+a)^2]+1/(4a)`
`P=[-(x-a)^2]/[4a(x+a)^2]+1/(4a)`
Ta có :` (x±a)^2 ≥ 0 ∀ x `
mà `a > 0 `
`⇒ [(x-a)^2]/[4a(x+a)^2] ≥ 0 ∀ x`
`⇔[-(x-a)^2]/[4a(x+a)^2] ≤ 0 ∀ x `
`⇔[-(x-a)^2]/[4a(x+a)^2]+1/(4a) ≤ 1/(4a) ∀ x`
Vậy `Max_{P}=1/(4a)` đạt khi `x=a`