Question

Tìm x ?

\(x^3+27+\left(x+3\right).\left(x-9\right)=0\)

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

Answer 1

x^3+27+(x+3).(x-9)=0

<=> x^3+3^3+(x+3).(x+9)=0

<=> (x+3).(x^2-3x+9)+(x+3).(x-9)=0

<=> (x+3).[(x^2 -3x+9)+(x-9)]=0

<=> (x+3).(x^2- 3x+9+x-9)=0

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

<=> x+3=0

hay x^2-2x=0

<=> x=-3

hay x=2

Vậy, x=-3; x=2

Answer 2

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

\(\Leftrightarrow2x^2+3x-4x-6=0\)

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

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

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

Vậy .................