`a)`
`P = -2x^2 + 3x + 1`
`= -2 (x^2 - 3/2x + 9/16) + 17/8`
` = -2 (x - 3/4)^2 + 17/8`
`\forall x` ta có :
`(x-3/4)^2 \ge 0`
`=> -2 (x-3/4)^2 \le 0`
`=> -2 (x-3/4)^2 + 17/8 \le 17/8`
`=> P \le 17/8`
Dấu `=` xảy ra `<=> x - 3/4 =0`
`<=> x = 3/4`
Vậy `text{Max}_P = 17/8 <=> x =3/4`
`b)`
`Q = -5x^2 - 4x - 19/5`
`= -5 (x^2 + 4/5x + 4/25) - 3`
`= -5 (x + 2/5)^2 - 3`
`\forall x` ta có :
`(x+2/5)^2 \ge 0`
`=> -5 (x+2/5)^2 \le 0`
`=> -5 (x+ 2/5)^2 - 3 \le -3`
`=> Q \le -3`
Dấu `=` xảy ra `<=> x + 2/5 =0`
`<=> x =-2/5`
Vậy `\text{Max}_Q = -3 <=> x = -2/5`
