a. $x^2-4x+5$
$= (x^2-4x+4)+1$
$= (x-2)^2+1$
Vì $(x-2)^2\ge 0 \; \forall x$
$\Rightarrow (x-2)^2+1 \ge 1\;\forall x$
Vậy $\min = 1$ khi $x-2=0 \Leftrightarrow x=2$
b. $2x^2+3x+5$
$= 2\left(x^2+\dfrac32x+\dfrac52\right)$
$= 2\left(x^2+\dfrac32x+\dfrac9{16}+\dfrac{31}{16}\right)$
$= 2\left[\left(x+\dfrac34\right)^2+\dfrac{31}{16}\right]$
$= 2\left(x+\dfrac34\right)^2+\dfrac{31}8$
Vì $2\left(x+\dfrac34\right)^2 \ge 0 \; \forall x$
$\Rightarrow 2\left(x+\dfrac34\right)^2+\dfrac{31}8 \ge \dfrac{31}8\;\forall x$
Vậy $\min =\dfrac{31}8$ khi $x+\dfrac34=0 \Leftrightarrow x=-\dfrac34$
c. $3x^2+6x+2$
$= 3(x^2+2x+\dfrac23)$
$= 3(x^2+2x+1-\dfrac13)$
$= 3(x+1)^2-1$
Vì $3(x+1)^2\ge 0 \; \forall x$
$\Rightarrow 3(x+1)^2-1 \ge -1\;\forall x$
Vậy $\min =- 1$ khi $x+1=0 \Leftrightarrow x=-1$
d. $3x^2-5x+1$
$= 3\left(x^2-\dfrac{5}{3}x+25/36-\dfrac{13}{36}\right)$
$= 3\left[\left(x-\dfrac56\right)^2-\dfrac{13}{36}\right]$
$= 3\left(x-\dfrac56\right)^2-\dfrac{13}{12}$
Vì $3\left(x-\dfrac56\right)^2 \ge 0 \; \forall x$
$\Rightarrow 3\left(x-\dfrac56\right)^2-\dfrac{13}{12} \ge -\dfrac{13}{12}\;\forall x$
Vậy $\min =-\dfrac{13}{12}$ khi $x-\dfrac56=0 \Leftrightarrow x=\dfrac56$
