Answer 1

\(\dfrac{\sqrt{16x-48}}{2}-\sqrt{x-3}+\dfrac{1}{3}\sqrt{36x-108}=9\left(x\ge3\right)\)

\(\Rightarrow\dfrac{4\sqrt{x-3}}{2}-\sqrt{x-3}+\dfrac{1}{3}.6\sqrt{x-3}=9\)

\(\Rightarrow2\sqrt{x-3}-\sqrt{x-3}+2\sqrt{x-3}=9\Rightarrow3\sqrt{x-3}=9\)

\(\Rightarrow\sqrt{x-3}=3\Rightarrow x-3=9\Rightarrow x=12\)

Answer 2

Ta có: \(\dfrac{\sqrt{16x-48}}{2}-\sqrt{x-3}+\dfrac{1}{3}\sqrt{36x-108}=9\)

\(\Leftrightarrow2\sqrt{x-3}-\sqrt{x-3}+2\sqrt{x-3}=9\)

\(\Leftrightarrow3\sqrt{x-3}=9\)

\(\Leftrightarrow x-3=9\)

hay x=12