\(\left(2x-3\right)^2=9\)
\(\left(2x-3\right)^2=\pm3^2\)
\(TH1:2x-3=3\)
\(2x=3+3\)
\(2x=6\)
\(x=3\)
\(TH2:2x-3=-3\)
\(2x=-3+3\)
\(2x=0\)
\(x=0\)
(2x-3)2=9
(2x-3)2=32
=> 2x-3 = 3 hoặc 2x-3 = -3
2x = 3+3 2x = -3 +3
2x = 6 2x = 0
x = 6:2 x =0:2
x = 3 x = 0
Vậy x=3 Vậy x =0
Vậy \(x=\left\{3;0\right\}\)
(2x-3)2=9
(2x-3)2=32
=> 2x-3 = 3 hoặc 2x-3 = -3
2x = 3+3 2x = -3 +3
2x = 6 2x = 0
x = 6:2 x = 0:2
x = 3 x = 0
Vậy x = 3 Vậy x = 0
Vậy\(x\varepsilon\left\{3;0\right\}\)