Đáp án:
$S=\{-1\}$
Giải thích các bước giải:
$(x^3+5x+5)^3+5x^3+24x+30=0$
$⇔(x^3+5x+5)^3+1+5x^3+24x+29=0$
$⇔(x^3+5x+6).[(x^3+5x+5)^2-(x^3+5x+5)+1]+5x^3+24x+29=0$
$⇔(x^3+x^2-x^2-x+6x+6).[(x^3+5x+5)^2-(x^3+5x+5)+1]+5x^3+5x^2-5x^2-5x+29x+29=0$
$⇔(x+1)(x^2-x+6).[(x^3+5x+5)^2-(x^3+5x+5)+1]+(x+1)(5x^2-5x+29)=0$
$⇔(x+1).\{(x^2-x+6)[(x^3+5x+5)^2-(x^3+5x+5)+1]+(5x^2-5x+29)\}=0$
$⇔\left[ \begin{array}{l}x+1=0\\(x^2-x+6)[(x^3+5x+5)^2-(x^3+5x+5)+1]+(5x^2-5x+29)=0\,(*)\end{array} \right.$
$(*)⇔(x^2-x+6)[(x^3+5x+5)^2-(x^3+5x+5)+1]+(5x^2-5x+29)=0$
Ta có:
$x^2-x+6=x^2-x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0$
$(x^3+5x+5)^2-(x^3+5x+5)+1=\left(x^3+5x+5-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0$
$5x^2-5x+29=5\left(x^2-x+\dfrac{29}{5}\right)=5\left(x-\dfrac{1}{2}\right)+\dfrac{63}{2}>0$
$⇒(x^2-x+6)[(x^3+5x+5)^2-(x^3+5x+5)+1]+(5x^2-5x+29)>0$
$⇒Pt(*)$ vô nghiệm
$⇔x+1=0$
$⇔x=-1$
Vậy $S=\{-1\}$.
