`a)4x(x+1)+(3-2x)(3+2x)=15`
`→4x^2+4x+9+6x-6x-4x^2=15`
`→(4x^2-4x^2)+(4x+6x-6x)+9=15`
`→4x=15-9`
`→4x=6`
`→x=3/2`
Vậy `x=3/2`
`b)3x(x-20012)-x+20012=0`
`→3x(x-20012)-(x-20012)=0`
`→(3x-1)(x-20012)=0`
`→` \(\left[ \begin{array}{l}3x-1=0\\x-20012=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}3x=1=>x=\dfrac{1}{3}\\\\x=20012\end{array} \right.\)
Vậy `x∈{1/2;20012}`
`c)(x+3)^2=(x+3)(x-3)`
`→x^2+6x+9=x^2-9`
`→x^2-x^2+6x=-9-9`
`→6x=-18`
`→x=-3`
Vậy `x=-3`
