Question

giải các phương trình:

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

Answer 1

easy

\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

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

S=\(\left\{3;\dfrac{-1}{2};\dfrac{4}{5}\right\}\)

Answer 2

a.

\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

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

Answer 3

\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

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

Vậy \(x\in\left\{3;\dfrac{-1}{2};\dfrac{4}{5}\right\}\)