Bài 3:
*) Ta có:
`x+3/x=7`
`⇒x^2+3=7x`
`⇒x^2+3+3x=10x`
`⇒x^2+3-3x=4x`
Xét `x^6+27`
`⇔(x^2+3)(x^4-3x^2+9)`
`⇔(x^2+3)[(x^4+3x^2)+(3x^2+9)-9x^2]`
`⇔(x^2+3)[x^2(x^2+3)+3(x^2+3)-9x^2]`
`⇔(x^2+3)[(x^2+3)^2-9x^2]`
`⇔(x^2+3)(x^2+3+3x)(x^2+3-3x)`
`⇔7x . 10x . 4x `
`⇔280x^3`
Vậy ta có `x^6+27=280x^3`
`⇒x^3/(x^6+27)=x^3/(280x^3)=1/280 `