\(ĐKXĐ:x\ge5\\ \sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\dfrac{3\sqrt{x-5}}{3}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}-4=0\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}.3\sqrt{x-5}-4=0\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}-4=0\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
Vậy `S={9}`
ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(2\sqrt{\left(x-5\right)}+\sqrt{x-5}-\sqrt{\left(x-5\right)}=4\)
\(\sqrt{\left(x-5\right)}=2\)
\(x-5=4\)
\(x=9\) (TMĐK)
