`#huy`
`b)x(x+3)-x-3=0`
`<=>x^2(x+3)-(x+3)=0`
`<=>(x+3)(x^2-1)=0`
`<=>(x+3)(x-1)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x+3=0\\x-1=0\\x+1=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=-3\\x=1\\x=-1\end{array} \right.\)
`c)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-8)(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x-8=0\\x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=8\\x=2\end{array} \right.\)
Vậy `x∈{2;8}`
