$\sqrt{x^2-x+\dfrac{1}{4}}-2x=-x-\dfrac{1}{2}(x≥\dfrac{1}{2})$
$⇔\sqrt{(x-\dfrac{1}{2})^2}=x-\dfrac{1}{2}$
$⇔|x-\dfrac{1}{2}|=x-\dfrac{1}{2}$
$+,x-\dfrac{1}{2}=x-\dfrac{1}{2}$
$⇒$ Vô số nghiệm
$+,x-\dfrac{1}{2}=\dfrac{1}{2}-x$
$⇔x=\dfrac{1}{2}$
Vậy $x≥\dfrac{1}{2}$
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$\sqrt{x^2-x+\dfrac{1}{4}}-2x=-x-\dfrac{1}{2}(x≥\dfrac{1}{2})$
$VT=\sqrt{(x-\dfrac{1}{2})^2}-2x$
$VT=|x-\dfrac{1}{2}|-2x$
$VT=x-\dfrac{1}{2}-2x$
$VT=-x-\dfrac{1}{2}=VP(đpcm)$
