Bài 2:
a. $x^2-2x+5$
$= x^2-2x+1+4$
$= (x-1)^2+4$
Vì $(x-1)^2\ge 0\;\forall x\in \mathbb{R}$
$⇒ (x-1)^2+4 \ge 4\;\forall x\in \mathbb{R}$
Vậy $\min = 4$ khi $x-1=0 ⇔ x=1$
b. $2x^2-6x$
$= 2\left(x^2-3x+\dfrac94-\dfrac94\right)$
$= 2\left[\left(x-\dfrac32\right)^2-\dfrac94\right]$
$= 2\left(x-\dfrac32\right)^2-\dfrac92$
Vì $2\left(x-\dfrac32\right)^2\ge 0 \;\forall x\in \mathbb{R}$
$⇒ 2\left(x-\dfrac32\right)^2-\dfrac92\ge -\dfrac92\;\forall x\in \mathbb{R}$
Vậy $\min = -\dfrac92$ khi $x-\dfrac32 = 0 ⇔ x=\dfrac32$
c. $x^2+y^2-x+6y+10$
$= \left(x-\dfrac12\right)^2+(y+3)^2+\dfrac34$
Vì $\left(x-\dfrac12\right)^2+(y+3)^2 \ge 0\;\forall x\in \mathbb{R}$
$⇔ \left(x-\dfrac12\right)^2+(y+3)^2+\dfrac34 \ge \dfrac54\;\forall x\in \mathbb{R}$
Vậy $\min = \dfrac34$ khi $x=\dfrac12$ và $y=-3$
