Lời giải:
1) $\sqrt{x^2-2x+1}=3$
$⇔ \sqrt{(x-1)^2}=3$
$⇔ |x-1|=3$
$⇔ \left[ \begin{array}x-1=3\\x-1=-3\end{array}\right.$
$⇔ \left[ \begin{array}x=4\\x-1=-2\end{array}\right.$
2) $\sqrt{4x^2-4x+1}-5=0$
$⇔ \sqrt{(2x-1)^2}=5$
$⇔|2x-1|=5$
$⇔ \left[ \begin{array}2x-1=5\\2x-1=-5\end{array}\right.$
$⇔ \left[ \begin{array}x=3\\x=-2\end{array}\right.$
3) $\sqrt{x^2-2x+1}=x-1$
$⇔ \sqrt{(x-1)^2}=x-1$
$⇔ |x-1|=x-1$
$⇔ x≥1$
4) $\sqrt{16-8x+x^2}+x=4$
$⇔ \sqrt{(x-4)^2}=4-x$
$⇔ |x-4|=4-x$
$⇔x≤4$
5) $\sqrt{x^2+x+\dfrac{1}{4}}=-x-\dfrac12$
$⇔\sqrt{(x+\dfrac12)^2}=-x-\dfrac12$
$⇔ \left|x+\dfrac12\right|=-x-\dfrac12$
$⇔x≤\dfrac{-1}2$
