\(a,\sqrt{4u-20}+3\sqrt{\dfrac{u-5}{9}}-\dfrac{1}{3}\sqrt{9u-45}=4\left(dk:u\ge5\right)\)
\(\Leftrightarrow2\sqrt{u-5}+3.\dfrac{\sqrt{u-5}}{3}-\dfrac{1}{3}.3\sqrt{u-5}=4\\ \Leftrightarrow2\sqrt{u-5}+\sqrt{u-5}-\sqrt{u-5}=4\\ \Leftrightarrow\sqrt{u-5}\left(2+1-1\right)=4\\ \Leftrightarrow\sqrt{u-5}=2\\ \Leftrightarrow u-5=4\\ \Leftrightarrow u=9\left(tm\right)\)
Vậy \(S=\left\{9\right\}\)
\(b,\dfrac{2}{3}\sqrt{9u-9}-\dfrac{1}{4}\sqrt{16u-16}+27\sqrt{\dfrac{u-1}{81}}=4\left(dk:u\ge1\right)\)
\(\Leftrightarrow2\sqrt{u-1}-\sqrt{u-1}+3\sqrt{u-1}=4\\ \Leftrightarrow\sqrt{u-1}.\left(2-1+3\right)=4\\ \Leftrightarrow\sqrt{u-1}=1\\ \Leftrightarrow u-1=1\\ \Leftrightarrow u=2\left(tm\right)\)
Vậy \(S=\left\{2\right\}\)
