Bài 2:
a: \(=x^2-3x+\dfrac{9}{4}+\dfrac{31}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}>=\dfrac{31}{4}\)
Dấu '=' xảy ra khi x=3/2
b: \(=y^2+8y+16-1=\left(y+4\right)^2-1>=-1\)
Dấu = xay ra khi y=-4
Bài 2
`a)M=x^2 -3x +10 = [x^2 - 2 * 3/2 *x + (3/2)^2 ] +10 -3/2`
`M = (x-3/2)^2 + 17/2 >= 17/2∀x`
Dấu ''='' xảy ra khi : `x-3/2 =0`
`=> x=3/2`
`b) P=y^2 +8y +15 = (y^2 + 2*y*4 + 4^2 ) +15 -4^2`
`P= (y+4)^2 -1 >= -1 ∀x`
Dấu ''='' xảy ra khi `y+4 =0`
`=> y =0-4=-4`
Bài 3
`A=12a -4a^2 +3 = - ( 4a^2 -12a -3) = -{[(2a)^2 - 2*2a*3 + 3^2] -3 -3^2}`
`A = - [(2a-3)^2 -12] = 12 - (2a-3)^2 <= 12`
dấu ''='' xảy ra khi `2a-3 =0 => 2a =3`
`=> a = 3/2`
`B =3a-a^2 +3 = -(a^2 -3a -3) = -{[a^2 - 2*a*3/2 + (3/2)^2 ] - 3-(3/2)^2}`
`B = -[(a-3/2)^2 - 21/4] = 21/4 -(a-3/2)^2 <= 21/4`
dấu ''='' xảy ra khi : ` a-3/2 =0 => a =3/2`
