Tìm x, biết:
a) 2 – 25x\(^2\) = 0
b) \(x^2 - x + \dfrac{1}{4} =0\)
a) 2 – 25x\(^2\) = 0 \(\Leftrightarrow (\sqrt{2})^2 - (5x)^2 =0\)
\(\Leftrightarrow ( \sqrt{2} - 5x )(\sqrt{2} + 5x ) =0\)
\(\Leftrightarrow \sqrt{2} - 5x = 0\) hoặc \(\sqrt{2}+5x =0\)
\(\Leftrightarrow x = \dfrac{\sqrt{2}}{5} \) hoặc \(x = - \dfrac{\sqrt{2}}{5}\)
b) \(x^2 - x + \dfrac{1}{4} =0\) \(\Leftrightarrow x^2 - 2.x.\dfrac{1}{2} + (\dfrac{1}{2})^2=0\)
\((x - \dfrac{1}{2})^2=0 \Leftrightarrow x - \dfrac{1}{2} =0 \Leftrightarrow x = \dfrac{1}{2}\)
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