Đáp án + Giải thích các bước giải:
`a)` `(3x+1)^2-(2x-3)^2=0` (sửa lại đề)
`<=>(3x+1-2x+3)(3x+1+2x-3)=0`
`<=>(x+4)(5x-2)=0`
`<=>`\(\left[ \begin{array}{l}x+4=0\\5x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-4\\x=\dfrac{2}{5}\end{array} \right.\)
Vậy `S={-4;2/5}`
`b)` `x^2+x+3x+3=0`
`<=>x(x+1)+3(x+1)=0`
`<=>(x+1)(x+3)=0`
`<=>[(x+1=0),(x+3=0):}`
`<=>[(x=-1),(x=-3):}`
Vậy `S={-1;-3}`
`c)` `x^3+2x^2y+xy^2-4x`
`=x(x^2+2xy+y^2-4)`
`=x[(x+y)^2-2^2]`
`=x(x+y-2)(x+y+2)`
`d)` `x^3+3x^2y+3xy^2+y^3-x-y`
`=(x+y)^3-(x+y)`
`=(x+y)[(x+y)^2-1]`
`=(x+y)(x+y-1)(x+y+1)`.