`a) ( 2x - 3)^2 = 16`
`(2x - 3)^2 = (+-4)^2`
`=>`\(\left[ \begin{array}{l}2x-3=4\\2x-3=-4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x=4+3\\2x=-4+3\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}2x=7\\2x=-1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=7:2\\x=-1:2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x= \frac{7}{2}\\x=\frac{-1}{2}\end{array} \right.\)
Vậy `x \in { 7/2 ; -1/2}`
``
`b) ( 3x - 2)^5 = -243`
`(3x - 2)^5 = (-3)^5`
`=> 3x - 2 = -3`
`3x = -3 + 2`
`3x = -1`
`x = -1 : 3`
`x = -1/3`
Vậy `x = -1/3`
`#caty09`
