\(\left(3x-5\right)^2=16\)
\(\left(3x-5\right)^2=4^2\) hoặc \(\left(3x-5\right)^2=\left(-4\right)^2\)
\(3x-5=4\) hoặc \(3x-5=-4\)
\(3x=4+5\) hoặc \(3x=-4+5\)
\(3x=9\) hoặc \(3x=1\)
\(x=9:3\) hoặc \(x=1:3\)
\(x=3\) hoặc \(x=\dfrac{1}{3}\)
Vậy \(x=3\) hoặc \(x=\dfrac{1}{3}\)
[3x-5]2=16
[3x-5]2=42
3x-5=4
3x=4+5
3x=9
x=9:3
x=3
[3x-5]^2=16
[3x-5]^2=(4)^2
=>3x-5=4
3x=4+5
3x=9
x=9:3
x=3
vậy....