$\displaystyle \begin{array}{{>{\displaystyle}l}} x^{2} -4x+1\\ x^{2} -4x+4-3\\ =( x-2)^{2} -3\ \\ Ta\ có\ :\ ( x-2)^{2} \geqslant 0\ với\ mọi\ x\ \\ Do\ đó\ :\ ( x-2)^{2} -3\geqslant -3\ \\ Dấu\ =\ xảy\ ra\ khi\ x-2=0\ \rightarrow x=2\ \\ Vậy\ min=-3\ tại\ x=2\ \\ x^{2} -7x+2\ \\ =x^{2} -2.\frac{7}{2} +\frac{49}{4} -\frac{41}{4}\\ =\left( x-\frac{7}{2}\right)^{2} -\frac{41}{4} \ \\ Ta\ có\ :\ \left( x-\frac{7}{2}\right)^{2} \geqslant 0\ \\ do\ đó\ :\ \left( x-\frac{7}{2}\right)^{2} -\frac{41}{4} \geqslant \frac{-41}{4} \ \\ Dấu\ =\ xảy\ ra\ khi\ x-\frac{7}{2} =0\ \rightarrow x=\frac{7}{2} \ \\ Vậy\ min=\frac{-41}{4} \ tại\ x=\frac{7}{2} \ \\ x^{2} -3x+1\\ =x^{2} -2.\frac{3}{2} x+\frac{9}{4} -\frac{5}{4} \ \\ =\left( x-\frac{3}{2}\right)^{2} -\frac{5}{4} \ \\ Ta\ có\ :\left( x-\frac{3}{2}\right)^{2} \geqslant 0\ \\ \rightarrow \left( x-\frac{3}{2}\right)^{2} -\frac{5}{4} \geqslant \frac{-5}{4} \ \\ Dấu\ =\ xảy\ ra\ khi\ x-\frac{3}{2} =0\ \rightarrow x=\frac{3}{2} \ \\ Vậy.....\ \end{array}$

`x^2-4x+1=x^2-4x+4-3=(x-2)^2-3`

`\text{Có}` `(x-2)^2\ge0` $\text{với mọi $x$}$

`⇒(x-2)^2-3\ge-3` $\text{với mọi $x$}$

$⇒GTNN=-3.$ $\text{Dấu = xảy ra}$

`⇔x-2=0⇔x=2`

$\text{Vậy}$ $GTNN=-3$ $\text{khi}$ $x=2$

`x^2-7x+2=x^2-7x+49/4-41/4=(x-7/2)^2-41/4`

`\text{Có}` `(x-7/2)^2\ge0` $\text{với mọi $x$}$

`⇒(x-7/2)^2-41/4\ge-41/4` $\text{với mọi $x$}$

$⇒GTNN=$`-41/4.` $\text{Dấu = xảy ra}$

`⇔x-7/2=0⇔x=7/2`

$\text{Vậy}$ $GTNN=$ `-41/4` $\text{khi}$ `x=7/2`

`x^2-3x+1=x^2-3x+9/4-5/4=(x-3/2)^2-5/4`

`\text{Có}` `(x-3/2)^2\ge0` $\text{với mọi $x$}$

`⇒(x-3/2)^2-5/4\ge-5/4` $\text{với mọi $x$}$

$⇒GTNN=$`-5/4.` $\text{Dấu = xảy ra}$

`⇔x-3/2=0⇔x=3/2`

$\text{Vậy}$ $GTNN=$ `-5/4` $\text{khi}$ `x=3/2`