Đặt A = \(2x^2-2x+1=2\left(x^2-x+\frac{1}{2}\right)=2\left(x^2-x+\frac{1}{4}+\frac{1}{4}\right)=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
=> Min A = 1/2
Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2
Vậy Min A = 1/2 <=> x = 1/2
b) Đặt B = \(x^2-x+5=x^2-x+\frac{1}{4}+\frac{19}{4}=\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)
=> Min B = 19/4
Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2
Vậy Min B = 19/4 <=> x =1/2
c) Đặt C = \(3x^2-4x+5=3\left(x^2-\frac{4}{3}x+\frac{5}{3}\right)=3\left(x-\frac{2}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\)
=> Min C = 11/3
Dấu "=" xảy ra <=> x - 2/3 = 0 <=> x = 2/3
Vậy Min C = 11/3 <=> x = 2/3
d) Đặt D = \(2x^2+3x+5=2\left(x^2+\frac{3}{2}x+\frac{5}{2}\right)=2\left(x+\frac{3}{4}\right)^2+\frac{31}{8}\ge\frac{31}{8}\)
=> Min D = 31/8
Dấu "=" xảy ra <=> x + 3/4 = 0 <=> x =-3/4
Vậy Min D = 31/8 <=> x = -3/4
