CHÚC BẠN HỌC TỐT!!!
Trả lời:
ĐKXĐ: $x\ge 2$
$x^2+\sqrt{2x+3}+\sqrt{x-2}=3x+4$
$⇔(x^2-3x)+(\sqrt{2x+3}-3)+(\sqrt{x-2}-1)=0$
$⇔x(x-3)+\dfrac{2x+3-9}{\sqrt{2x+3}+3}+\dfrac{x-2-1}{\sqrt{x-2}+1}=0$
$⇔(x-3).\bigg{(}x+\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x-2}+1}\bigg{)}=0$
$⇔\left[ \begin{array}{l}x-3=0\\x+\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x-2}+1}=0\,\,(*)\end{array} \right.$
Ta có: $x\ge 2⇒x\ge 0$
$x+\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{1}{\sqrt{x-2}+1}>0$
$⇔Pt(*)$ vô nghiệm
$⇔x-3=0$
$⇔x=3$
Vậy $S=\{3\}$.
