`a)5(x+3)-2x(3+x)=0`
`⇔5(x+3)-2x(x+3)=0`
`⇔(x+3)(5-2x)=0`
`⇔`\(\left[ \begin{array}{l}x+3=0\\5-2x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-3\\x=\dfrac{5}{2}\end{array} \right.\)
Vậy `x=-3` hoặc `x=5/2`
`b)6x(x²-2)-(2-x²)=0`
`⇔6x(x²-2)+(x²-2)=0`
`⇔(x²-2)(6x+1)=0`
`⇔`\(\left[ \begin{array}{l}x²-2=0\\6x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x²=2\\6x=-1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\sqrt[]{2}\\x=-\sqrt[]{2}\\x=-\dfrac{1}{6}\end{array} \right.\)
Vậy `x=`$\sqrt[]{2}$ hoặc `x=-`$\sqrt[]{2}$ hoặc `x=-1/6`
