Question

Giai pt: 2x³ - 5x² + 3x = 0

Answer 1

2x³-2x²-3x²+3x=0

<=>2x(x-1)-3x(x-1)=0

<=>(x-1)(2x-3x)=0

<=>-x(x-1)=0

Th₁:-x=0

<=>x=0

Th₂:x-1=0

<=>x=1

Vậy phương trình có tập n_o là S=(0, 1)

Answer 2

\(2x^3-5x^2+3x=0\)

\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)

\(\Leftrightarrow2x\left(x-1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-3x\right)=0\)

\(\Leftrightarrow-x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\\x=1\end{matrix}\right.\)

Answer 3

2x3-2x2-3x2+3x=0

<=>2x(x-1)-3x(x-1)=0

<=>(x-1)(2x-3x)=0

<=>-x(x-1)=0

\(\left\{{}\begin{matrix}-x=0\\x-1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy phương trình có tập no là S=(0, 1)

Answer 4