a)x² $-$ 2x $+$ 2021
=x² -2x + 1 + 2020
=(x-1)² + 2020
(x-1)² ≥ 0
⇔ (x-1)² + 2020 ≥ 2020
Dấu bằng xảy ra khi x=1
b)x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}
$=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}$
$\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}≥\dfrac{3}{4}$
Dấu bằng xảy ra khi $x=\dfrac{1}{2}$
c)2x²+y²-2xy+4x-2y+100
=x²-2xy+y² +x²+2x+1+2x-2y+99
=(x-y)²+2(x-y)+1+(x+1)²+98
=(x-y+1)²+(x+1)²+98
(x-y+1)²+(x+1)² ≥ 0
⇒(x-y+1)²+(x+1)²+98 ≥ 98
Dấu bằng xảy ra khi
$\left \{ {{x-y+1=0} \atop {x+1=0}} \right.$
⇔$\left \{ {{x=-1} \atop {y=0}} \right.$
