\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\)
=>(2x-3)2=27.3
2x-3=9 hoặc 2x-3= -9
x=6 hoặc x=-3
\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ \Leftrightarrow\left(2x-3\right)^2=27.3=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=9\\2x-3=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tmđk\right)\\x=-3\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-3\right\}\)
Ta có: \(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\)
\(\Leftrightarrow\left(2x-3\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=9\\2x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)
Vậy: S={6;-3}
\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\)
\(\Rightarrow\left(2x-3\right).\left(2x-3\right)=3.27\)
\(\Rightarrow\left(2x-3\right)^2=81\)
\(\Rightarrow\left(2x-3\right)^2=9^2\) hoặc \(\left(2x-3\right)^2=\left(-9\right)^2\)
\(2x-3=9\) hoặc \(2x-3=-9\)
\(x=6\) hoặc \(x=-3\)