`a) 5x(x-3)(x+3)-(2x-3)^2-5(x+2)^3+34(x+2)=1`
`<=> 5x(x^2-9)-(4x^2-12x+9)-5(x^2+4x+4)+34x+68=1`
`<=> 5x^3-45x-4x^2+12x-9-5x^2-20x-20+34x+67=0`
`<=> 5x^3-9x^2-19x+38=0`
`<=> 5x^3+10x^2-19x^2-38x+19x+38=0`
`<=> 5x^2(x+2)-19x(x+2)+19(x+2)=0`
`<=> (x+2)(5x^2-19x+19)=0`
Do `5x^2-19x+19=5(x^2-19/5x+19/5)=5(x-19/10)^2+19/20>0` với `AAx`
`-> x+2=0<=>x=-2`
Vậy `S={-2}`
`b) (x-2)^3+6(x+1)^2-(x-3)(x^2+3x+9)=97`
`<=> x^3-6x^2+12x-8+6(x^2+2x+1)-(x^3-27)-97=0`
`<=> x^3-6x^2+12x-105+6x^2+12x+6-x^3+27=0`
`<=> 24x-72=0`
`<=> x-3=0`
`<=> x=3`
Vậy `S={3}`