Question

giải pt \(2x^3+6x^2=x^2+3x\)

Answer 1

pt $2x^3+6x^2=x^2+3x$

giải

<=>\(2x^3+6x^2-x^2-3x=0\)

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

<=>\(x\left(2x^2+5x-3\right)\)=0

<=>\(x\left(2x^2-x+6x-3\right)=0\)

<=>\(x\left(2x-1\right)\left(x+3\right)=0\)

<=>x=0

x=\(\dfrac{1}{2}\)

x=-3

tham khảo nhé

Answer 2

\(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^3+6x^2-x^2-3x=0\)

\(\Leftrightarrow2x^3+5x^2-3x=0\)

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

\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)

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

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

=.= hok tốt!!