Question

\(\dfrac{3}{2}\sqrt{3x}-3x-5=-\dfrac{1}{2}\sqrt{3x}\) với ĐK \(x\ge0\)

Answer 1

\(\dfrac{3}{2}\sqrt{3x}-3x-5=-\dfrac{1}{2}\sqrt{3x}\left(đk:x\ge0\right)\)

\(\Leftrightarrow3x-2\sqrt{3x}+5=0\)

\(\Leftrightarrow\left(\sqrt{3x}-1\right)^2+4=0\)(vô lý do \(\left(\sqrt{3x}-1\right)^2+4\ge4>0\))

Vậy \(S=\varnothing\)