Bài `1:`
`3xy + 15y^2 = 3y (x + 5y)`
`3x^2 + 6xy + 3y^2 = 3 (x^2 + 2xy + y^2) = 3 (x+y)^2`
Bài `2:`
`(2x-3)(4x^2 + 6x+9) - x^2 (8x+3)`
` = (2x-3) [ (2x)^2 + 2x . 3 + 3^2] - (x^2 . 8x + x^2 . 3)`
` = (2x)^3 - 3^3 - (8x^3 + 3x^2)`
` = 8x^3 - 27 - 8x^3 - 3x^2`
` = (8x^3 - 8x^3) - 3x^2 - 27`
` = -3x^2 - 27`
Bài `3:`
`(x-3)^2 + 2 (x^2-9) + (x+3)^2`
`= (x-3)^2 + 2 (x-3) (x+3) + (x+3)^2`
` = [ (x-3) + (x+3)]^2`
` = (x-3+x+3)^2`
` = (2x)^2`
` = 4x^2`
Bài `4:`
`x (3x-2) - 4 + 6x = 0`
`=> x (3x-2) + 2 (3x-2) = 0`
`=> (x+2)(3x-2) =0`
`=> x+2=0` hoặc `3x-2=0`
`+)x+2=0=>x=-2`
`+)3x-2=0=>3x=2=>x=2/3`
Vậy `x \in {-2;2/3}`