`\qquad y=\sqrt{x^2-5x+4}`
ĐKXĐ: `x^2-5x+4>=0`
`<=> x^2-4x-x+4>=0`
`<=> x(x-4)-(x-4)>=0`
`<=> (x-4)(x-1)>=0`
`<=> [({(x-4>=0),(x-1>=0):}),({(x-4<=0),(x-1<=0):}):}`
`<=> [({(x>=4),(x>=1):}),({(x<=4),(x<=1):}):}`
`<=> [(x>=4),(x<=1):}`
Vậy `D=(-oo;1]∪[4;+oo)`