$a)\ ( 2x - 3 )^3 + (6x -17)^3 =0$
$⇔(2x-3+6x-17)(4x^2-12x+9-12x^2+52x-51+36x^2-204x+289)=0$
$⇔(8x-20)(28x^2-164x+247)=0$
$⇔4(2x-5)[28.(x^2-2.\frac{41}{14}.x+\frac{1681}{196})+\frac{48}{7}]=0$
$⇔4(2x-5)[28.(x-\frac{41}{14})^2+\frac{48}{7}]=0$
Vì $28.(x-\frac{41}{14})^2≥0∀x⇒28.(x-\frac{41}{14})^2+\frac{48}{7}>0∀x$
$⇒2x-5=0$
$⇔x=\frac{5}{2}$
Vậy $S=\{\frac{5}{2}\}$.
