ĐK: $x\ge -\dfrac{1}{3}$
$\sqrt{48x+16}-5\sqrt{27x+9}+3\sqrt{75x+25}=8\\↔\sqrt{16(3x+1)}-5\sqrt{9(3x+1)}+3\sqrt{25(3x+1)}=8\\↔\sqrt{16}.\sqrt{3x+1}-5.\sqrt 9.\sqrt{3x+1}+3.\sqrt{25}.\sqrt{3x+1}=8\\↔4\sqrt{3x+1}-5.3.\sqrt{3x+1}+3.5.\sqrt{3x+1}=8\\↔4\sqrt{3x+1}=8\\↔\sqrt{3x+1}=2\\↔3x+1=4\\↔3x=3\\↔x=1(TM)$
Vậy phương trình có tập nghiệm $S=\{1\}$