giải PT
<=>
\(\dfrac{5}{2}.4.\sqrt{x-2}-16-\dfrac{1}{5}.25.\sqrt{x-2}=0\)
<=>
\(10\sqrt{x-2}-5\sqrt{x-2}-16=0\)
<=>
\(5\sqrt{x-2}-16=0\)
=> \(5\sqrt{x-2}=16\)
=> \(\sqrt{x-2}=\dfrac{16}{5}\)
=> \(x-2=\left(\dfrac{16}{5}\right)^2=\dfrac{256}{25}\)
=> \(x=\dfrac{256}{25}+2=\dfrac{306}{25}\)
\(\dfrac{5}{2}\sqrt{4x-8}-16=\dfrac{1}{5}\sqrt{25x-50}\)
\(ĐK:x\ge2\)
\(\Leftrightarrow\dfrac{5}{2}\sqrt{4\left(x-2\right)}-16=\dfrac{1}{5}\sqrt{25\left(x-2\right)}\)
Đặt \(x-2=a;a\ge0\)
\(\Leftrightarrow\dfrac{5}{2}\sqrt{4a}-16=\dfrac{1}{5}\sqrt{25a}\)
\(\Leftrightarrow\dfrac{5}{2}.2\sqrt{a}-\dfrac{1}{5}.5\sqrt{a}=16\)
\(\Leftrightarrow5\sqrt{a}-\sqrt{a}-16=0\)
\(\Leftrightarrow4\sqrt{a}-16=0\)
\(\Leftrightarrow4\left(\sqrt{a}-4\right)=0\)
\(\Leftrightarrow\sqrt{a}-4=0\)
\(\Leftrightarrow\sqrt{a}=4\)
\(\Leftrightarrow a=16\left(tm\right)\)
\(\rightarrow x-2=16\)
\(\Leftrightarrow x=18\left(tm\right)\)
Vậy \(S=\left\{18\right\}\)