Đáp án:
Giải thích các bước giải:
$a)2-25x^2=0$
$⇔(\sqrt{2})^2-(5x)^2=0$
$⇔(\sqrt{2}-5x).(\sqrt{2}+5x)=0$
$⇔$ \(\left[ \begin{array}{l}\sqrt{2}-5x=0\\\sqrt{2}+5x=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}5x=\sqrt{2}\\5x=-\sqrt{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{\sqrt{2}}{5}\\x=-\dfrac{\sqrt{2}}{5}\end{array} \right.\)
$b)x^2-x+\dfrac{1}{4}=0$
$⇔x^2-2.\dfrac{1}{2}.x+(\dfrac{1}{2})^2=0$
$⇔(x-\dfrac{1}{2})^2=0$
$⇔x-\dfrac{1}{2}=0$
$⇔x=\dfrac{1}{2}$
