Đáp án:
`1.(2x - 1)^2 - (x+3)^2 = 0`
`<=> (2x-1 - x - 3)(2x - 1 + x + 3)=0`
`<=> (x - 4)(3x +2)=0`
`<=>` $\left[\begin{matrix} x-4=0\\ 3x+2=0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x= 4\\ x=\dfrac{-2}{3}\end{matrix}\right.$
Vậy `S={ 4 ; -2/3}`
`2.x^2y^2 + 1 - x^2 - y^2`
`= (x^2 y^2 - x^2) - (y^2 - 1)`
`= x^2 (y^2 - 1) - (y^2 - 1)`
`= (x^2 - 1)(y^2 - 1)`
`= (x-1)(x+1)(y-1)(y+1)`