`b) ( - 2 )/x = ( - x )/( 8/25 )`
`⇔ x . ( - x ) = ( - 2 ) . 8/25`
`⇔ - x^2 = ( -16 )/25`
`⇔ x^2 = 16/25`
`⇔` $\left[\begin{matrix}x^2 = ( \frac{4}{5} )^2\\x^2 = ( \frac{- 4}{5} )^2\end{matrix}\right.$
`⇔` $\left[\begin{matrix}x = \frac{4}{5}\\x = \frac{- 4}{5}\end{matrix}\right.$
Vậy `, x ∈ { 4/5 ; ( - 4 )/5 } .`