\(P=-4x^2-5x+1\\ =-4\left(x^2+\dfrac{5}{4}x-\dfrac{1}{4}\right)\\ =-4\left(x^2+2.\dfrac{5}{8}x+\dfrac{25}{64}-\dfrac{41}{64}\right)\\ =-4\left[\left(x+\dfrac{5}{8}\right)^2-\dfrac{41}{64}\right]\\ =-4\left(x+\dfrac{5}{8}\right)^2+\dfrac{41}{16}\le\dfrac{41}{16}\)
Dấu "=" xảy ra `<=>x= -5/8`
Vậy `P_(max)= 41/16 <=> x= -5/8`
`= -4(x^2 + 5/4 - 1/4)`
`= -4(x^2 + 2. 5/8 + 25/64 - 41/64)`
`= -4(x+5/8)^2 + 41/64`
Vì `-4(x+5/8)^2 <=0 => -4(x+5/8)^2 + 41/64 <= 41/64`.
Dấu bằng xảy ra `<=> x = -5/8`
`P = -(4x^2 + 5x - 1)`
`= -((2x)^2 + 2x .5/2 + 25/4 - 29/4)`
`= -((2x+5/2)^2 - 29/4)`
`= -(2x+5/2)^2 + 29/4`.
Vì `-(2x + 5/2)^2 <= 0 => -(2x+5/2)^2 + 29/4 <= 0 + 29/4 = 29/4`.
Vậy `Mi``n P = 29/4 <=> x = -5/4`.
