\(\sqrt{4u-20}+3\sqrt{\dfrac{u-5}{9}}-\dfrac{1}{3}\sqrt{9u-45}=4\) (ĐK: \(u\ge5\))
\(\Leftrightarrow2\sqrt{u-5}+3\cdot\dfrac{\sqrt{u-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{u-5}=4\)
\(\Leftrightarrow2\sqrt{u-5}+\sqrt{u-5}-\sqrt{u-5}=4\)
\(\Leftrightarrow2\sqrt{u-5}=4\)
\(\Leftrightarrow\sqrt{u-5}=2\)
\(\Leftrightarrow u-5=4\)
\(\Leftrightarrow u=9\) (tm)