Đáp án+Giải thích các bước giải:
`a)`
`2-25x^2=0`
`⇔(\sqrt2)^2-(5x)^2=0`
`⇔(\sqrt2-5x)(\sqrt2+5x)=0`
\(⇔\left[ \begin{array}{l}\sqrt 2-5x=0\\\sqrt 2+5x=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}5x=\sqrt 2\\5x=-\sqrt 2\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{\sqrt 2}{5}\\x=-\dfrac{\sqrt 2}{5}\end{array} \right.\)
Vậy `x=\sqrt2/5` hoặc `x=-\sqrt2/5`
`b)`
`x^2-x+1/4=0`
`⇔x^2-2.x.(1)/2+(1/2)^2=0`
`⇔(x-1/2)^2=0`
`⇔x-1/2=0`
`⇔x=1/2`
Vậy `x=1/2`
