Question

6.Giải các phương trình

\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

Answer 1

\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

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

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

\(\Leftrightarrow x^2-x-6=0\)

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

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

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

Vậy \(S=\left\{3;-2\right\}\)

Answer 2

ĐKXĐ: x<>3

PT=>x^2+2x-3x-6=0

=>x^2-x-6=0

=>(x-3)(x+2)=0

=>x=3(loại) hoặc x=-2(nhận)