`M=(x-1)(x-3)+11`

`=x(x-3)-(x-3)+11`

`=x^2-3x-x+3+11`

`=x^2-4x+14`

`=(x^2-4x+4)+10`

`=(x-2)^2+10`

Ta có: `(x-2)^2>=0\AAx`

`->M=(x-2)^2+10>=10\AAx`

Dấu "=" xảy ra: `x-2=0->x=2`

Vậy: `M_(min)=10` khi `x=2`

`12)N=(x-3)^2+(x-2)^2`

`=(x^2-6x+9)+(x^2-4x+4)`

`=2x^2-10x+13`

`=2(x^2-5x)+13`

`=2[(x^2-2*x*5/2+25/4)-25/4]+13`

`=2(x-5/2)^2-25/2+13`

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

Ta có: `2(x-5/2)^2>=0\AAx`

`->N=2(x-5/2)^2+1/2>=1/2\AAx`

Dấu "=" xảy ra: `x-5/2=0->x=5/2`

Vậy: `N_(min)=1/2` khi `x=5/2`