Đáp án: `S={0;1;2; 5/4}`
Giải thích các bước giải:
`(x^2-3x+2)^3=x^6-(3x-2)^3`
`<=> (x^2-3x+2)^3=(x^2)^3-(3x-2)^3`
`<=> (x^2-3x+2)^3=(x^2-3x+2)[x^2+x^2 (3x-2) +(3x-2)^2]`
`<=> (x^2-3x+2)[ (x^2-3x+2)^2 - (x^2 +x^2 (3x-2) +(3x-2)^2)]`
`<=> (x^2-3x+2)[ (x-1)^2 (x-2)^2 - (x^2 +3x^3-2x^2+9x^2-12x+4)=0`
`<=> (x-1)(x-2)(x^4-9x^3+5x^2)=0`
`<=> (x-1)(x-2). x^2 (4x^2-9x+5)=0`
`<=> x^2 .(x-1)(x-2). (4x-5)(x-1)=0`
`<=>` $\begin{cases}x=0\\x=2\\x=1\\x=\dfrac{5}{4}\\\end{cases}$
Vậy `S={0;1;2; 5/4}`
