xét tính đơn điệu của hs y= √x ²-6x+5

$$\eqalign{ & y = \sqrt {{x^2} - 6x + 5} \cr & DKXD:\,\,{x^2} - 6x + 5 \ge 0 \Leftrightarrow \left[ \matrix{ x \ge 5 \hfill \cr x \le 1 \hfill \cr} \right. \cr & y' = {{2x - 6} \over {2\sqrt {{x^2} - 6x + 5} }} = {{x - 3} \over {\sqrt {{x^2} - 6x + 5} }} \cr & Cho\,\,y' = 0 \Leftrightarrow x = 3 \cr & BXD \cr & - \infty \,\,\,\,\, + \,\,\,\,\,1\,\,\,\,\, + \,\,\,\,\,\,3\,\,\,\,\,\, - \,\,\,\,\,\,5\,\,\,\,\,\, - \,\,\,\,\, + \infty \cr & \Rightarrow Ham\,\,so\,\,DB/\left( { - \infty ;1} \right),\,\,NB/\left( {5; + \infty } \right) \cr} $$