\(\Leftrightarrow18x^2\left(x+4\right)-12x\left(x+4\right)=0\)
\(\Leftrightarrow6x\left(x+4\right)\left[3x-2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
⇔18x2(x+4)−12x(x+4)=0⇔18x2(x+4)−12x(x+4)=0
⇔6x(x+4)[3x−2]=0⇔6x(x+4)[3x−2]=0
⇔⎡⎢⎣x=0x+4=03x−2=0⇔[x=0x+4=03x−2=0
Đúng 0
Bình luận (0)
