Question

\(\sqrt{9x-27}-4\sqrt{\dfrac{x-3}{4}}=\sqrt{3-x}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge3\\x\le3\end{matrix}\right.\Leftrightarrow x=3\)

Thay x=3 vào ta thấy TM.

Vậy x=3

Answer 2

ĐKXĐ: x=3

Ta có: \(\sqrt{9x-27}-4\cdot\sqrt{\dfrac{x-3}{4}}=\sqrt{3-x}\)

\(\Leftrightarrow3\sqrt{x-3}-4\cdot\dfrac{\sqrt{x-3}}{2}=\sqrt{3-x}\)

\(\Leftrightarrow\sqrt{x-3}=\sqrt{3-x}\)

\(\Leftrightarrow x-3=3-x\)

\(\Leftrightarrow x-3-3+x=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow2x=6\)

hay x=3(Thỏa ĐKXĐ)

Vậy: S={3}