1)
`a(x-3)+6=a^3-2(a^2-x)`
`⇔ax-3a+6=a^3-2a^2+2x`
`⇔ax-2x=a^3-2a^2+3a-6`
`⇔x(a-2)=(a-2)(a^2+3)`
`⇒x=a^2+3>0`(với mọi x) ( vì `a \neq 2` nên `a-2 \ne 0`)
`⇒đpcm`
2)
`a(x+a+1)=a^3+2x-2`
`⇔ax+a^2+a=a^3+2x-2`
`⇔ax-2x=a^3-a^2-a-2`
`⇔x(a-2)=(a-2)(a^2+a+1)`
`⇔x=a^2+a+1` ( vì `a \neq 2` nên `a-2 \ne 0`)
`⇔x=(a+1/2)^2+3/4≥3/4 `
dấu "=" xảy ra `⇔a=-1/2`
Vậy `minx=3/4 ⇔a=-1/2`