`2/( 2x + 1 ) = ( 2x + 1 )/18`
`⇔ ( 2x + 1 ) . ( 2x + 1 ) = 2 . 18`
`⇔ ( 2x + 1 )^2 = 36`
`⇔` \(\left[ \begin{array}{l}( 2x + 1 )² = 6²\\( 2x + 1 )² = ( - 6 )²\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x + 1 = 6\\2x + 1 = - 6\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x = 5\\2x = - 7\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = \frac{5}{2} \\x = \frac{- 7}{2}\end{array} \right.\)
Vậy , `x ∈ { 5/2 ; ( - 7 )/2 } .`