Question

giải phương trình tích (4-3x) (10x-5)=0

Answer 1

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-3x=0\\10x-5=0\end{matrix}\right.\)

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

\(S=\left\{\dfrac{4}{3},\dfrac{1}{2}\right\}\)

Answer 2

Ta có: \(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-3x=0\\10x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\10x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{4}{3};\dfrac{1}{2}\right\}\)