Đáp án:
`N=(x-1)(x-3)+11`
`=x^2-3x-x+3+11`
`=x^2-4+14`
`=(x^2-4x+4)+10`
`=(x-2)^2+10>=10`
Dấu "=" xảy ra `<=>x-2=0`
`=> x=2`
Vậy `N_(min)=10 <=> x=2`
`----`
`M=(x-2)(x-4)+4`
`=x^2-4x-2x+8+4`
`=x^2-6x+12`
`=(x^2-6x+9)+3`
`=(x-3)^2+3>=3`
Dấu "=" xảy ra `<=> x-3=0`
`=> x=3`
Vậy `M_(min)=3 <=> x=3`