Answer: C
Step-by-step explanation:
3x-3 ≥ 0 ==> x ≥ 1
3x+6 ≥ 0 ==> x ≥ -2
So: condition x ≥ 1
[tex]\sqrt{3x-3} +1=\sqrt{3x+6} \\\\\sqrt{3x+6} -\sqrt{3x-3} =1 \ squaring\\\\3x+6+3x-3-2\sqrt{(3x+6)(3x-3)} =1\\\\6x+2=2\sqrt{(3x+6)(3x-3)}\\\\\sqrt{(3x+6)(3x-3)}=3x+1\ squaring \\\\(3x+6)(3x-3)=(3x+1)^2\\\\9x^2+18x-9x-18=9x^2+6x-1\\\\3x=19\\\\x=\dfrac{19}{3} \\\\sol=\{\dfrac{19}{3}\}\\\\Answer\ C\\[/tex]
