Question

\(\dfrac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)

\(\left[x-2\right]+\dfrac{\sqrt{x^2-4x+4}}{x-2}\left(x< 2\right)\)

Answer 1

`@`\(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)

`@`\(\left(x-2\right)+\dfrac{\sqrt{x^2-4x+4}}{x-2}=\left(x-2\right)+\dfrac{\sqrt{\left(x-2\right)^2}}{x-2}=\left(x-2\right)+\dfrac{\left|x-2\right|}{x-2}=x-2+\dfrac{2-x}{x-2}=x-2-\dfrac{x-2}{x-2}=x-2-1=x-3\)

Answer 2

`+)x>1`

`<=>x-1>0`

`(\sqrt{x^2 -2x+1})/(x-1)`

`=(\sqrt{(x-1)^2})/(x-1)`

`=(|x-1|)/(x-1)~`

`=(x-1)/(x-1)`

`=1`

Vậy...

`+)x<2`

`<=>x-2<0`

`|x-2|+(\sqrt{x^2-4x+4})/(x-2)`

`=2-x+(\sqrt{(x-2)^2})/(x-2)`

`=2-x+(|x-2|)/(x-2)`

`=2-x+(2-x)/(-(2-x))`

`=2-x-1`

`=1-x`

Vậy...