Question

(2x+4/5)(3x-1/2)=0

Answer 1

\(\left(2x+\dfrac{4}{5}\right)\left(3x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+\dfrac{4}{5}=0\\3x-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Answer 2

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Answer 3

Ta có: \(\left(2x+\dfrac{4}{5}\right)\left(3x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{4}{5}=0\\3x-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{4}{5}\\3x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)