Tìm x, biết:
a) \(2 - 25{x^2} = 0;\)
b)\({x^2} - x + \frac{1}{4} = 0\)
Câu a:
\(\begin{array}{l} 2 - 25{x^2} = 0\\ {(\sqrt 2 )^2} - {(5x)^2} = 0\\ (\sqrt 2 - 5x)(\sqrt 2 + 5x) = 0 \end{array}\)
Hoặc\(\sqrt 2 - 5x = 0 \Rightarrow 5x = \sqrt 2 \Rightarrow x = \frac{{\sqrt 2 }}{5}\)
Hoặc\(\sqrt 2 + 5x = 0 \Rightarrow 5x = - \sqrt 2 \Rightarrow x = - \frac{{\sqrt 2 }}{5}\)
Câu b:
\(\begin{array}{l} {x^2}\; - x + \frac{1}{4}\; = 0{\rm{ }}\\ \Rightarrow \;{x^2}-2.x.\;\frac{1}{2}\; + {\rm{ }}{(\frac{1}{2})^2} = 0\;{\rm{ }}\;\\ \Rightarrow {(x - \;\frac{1}{2})^2}\; = 0{\rm{ }}\\ = > x - \frac{1}{2}\; = 0{\rm{ }}\\ = > x{\rm{ }} = \frac{1}{2} \end{array}\)
-- Mod Toán 8
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