`A=x^2 -3x+5 =x^2 -2*x*3/2 +(3/2)^2 + 5- (3/2)^2`
`A = (x-3/2)^2 +7/2 >= 7/2 AA x`
Dấu ''=''xảy ra khi `x-3/2 =0`
`=> x=3/2`
Vậy \(Min_A=\dfrac{7}{2}\) khi `x=3/2`
`B=(2x-1)^2+(x+2)^2 = 4x^2 -4x +1 +x^2 +4x+4`
`B = 5x^2 +5 >= 5 AA x`
Dấu ''='' xảy ra khi : `5x^2 =0`
`=> x=0`
Vậy \(Min_B=5\) khi `x=0`
a: \(=x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}\)
Dấu '=' xảy ra khi x=3/2
b: \(=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5>=5\)
Dấu '=' xảy ra khi x=0