\(3\left(2+\sqrt{x+2}\right)=2x+\sqrt{x+6}\Leftrightarrow3\sqrt{x+2}=2x-6+\sqrt{x+6}\) ==> \(2x-6\ge0\Leftrightarrow x\ge3\)
ĐKXĐ: x \(\ge\) 3
xin lỗi mình có chút nhầm lẫn về Đk
ĐK: x \(\ge\) 2
\(3\left(2+\sqrt{x-2}\right)=2x+\sqrt{x+6}\Leftrightarrow6+3\sqrt{x-2}-2x-\sqrt{x+6}=0\Leftrightarrow3\left(\sqrt{x-2}-1\right)-\left(\sqrt{x+6}-3\right)-2x+6=0\Leftrightarrow3.\dfrac{x-3}{\sqrt{x-2}+1}-\dfrac{x-3}{\sqrt{x+6}+3}-2\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(\dfrac{3}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{x+6}+3}-2\right)=0\) \(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(tm\right)\)
Vì \(\dfrac{3}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{x+6}+3}-2< 0\)
Vậy nhiệm của pt là x = 3
