\(\left(3x-5\right)^2=16\)
\(\Rightarrow\left(3x-5\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{3;\dfrac{1}{3}\right\}\)
`(3x - 5)^2 = 16`
`=> (3x - 5 )^2 = 4^2` hoặc `(-4)^2`
`@` Trường hợp `1` `:`
`3x - 5 = 4`
`=> 3x = 4 + 5`
`=> 3x = 9`
`=> x = 9 : 3`
`=> x = 3`
`@` Trường hợp `2` `:`
`3x - 5 = -4`
`=> 3x = -4 + 5`
`=> 3x = 1`
`=> x = 1 : 3`
`=> x = 1/3`
Vậy `x in { 3 ; 1/3}`