Cho: \(A=\left(2x+\dfrac{4}{5}\right)\times\left(x^2+\dfrac{1}{4}\right)=0\)
Để biểu thức A=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{4}{5}=0\\x^2+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=-\dfrac{4}{5}\\x^2=-\dfrac{1}{4}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{5}:2=-\dfrac{2}{5}\\x=\sqrt{-\dfrac{1}{4}}\end{matrix}\right.\)
Vậy: \(x=-\dfrac{2}{5}\) hoặc \(x=\sqrt{-\dfrac{1}{4}}\)
Cho: A=\(\left(2x+\dfrac{4}{5}\right)\times\left(x^2+\dfrac{1}{4}\right)=0\)
Để biểu thức : \(A=0\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{4}{5}=0\\x^2+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=-\dfrac{4}{5}\\x^2=-\dfrac{1}{4}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-\dfrac{4}{5}:2=-\dfrac{8}{5}\\x=\sqrt{-\dfrac{1}{4}}=\dfrac{1}{2}\end{matrix}\right.\)
