Đề là: \(2x+\dfrac{1}{-27}=\dfrac{-3}{2x}+1\)
Hay: \(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\) vậy em?
ĐKXĐ: x<>-1/2
\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
=>\(\left(2x+1\right)^2=\left(-27\right)\cdot\left(-3\right)=81\)
=>\(\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
=>(2x+1).(2x+1)=(-27).(-3)
=>\(\left(2x+1\right)^2\)=81=\(9^2=\left(-9\right)^2\)
th1:
2x+1=9
2x=8
x=4
th2 :
2x+1=-9
2x =-10
x =-5
vậy x thuộc{4;-5}