a) \(\left(3x+4\right)^2-64=0\)
\(\left(3x+4\right)^2=0+64\)
\(\left(3x+4\right)^2=64\)
=> \(3x+4=8\) hay \(3x+4=-8\)
\(3x=8-4\) hay \(3x=-8-4\)
\(3x=4\) hay \(3x=-12\)
\(x=4:3\) hay \(x=-12:3\)
\(x=\dfrac{4}{3}\) hay \(x=-4\)
Vậy \(x=\dfrac{4}{3}\)hoặc \(x=-4\)
a, (3x+4)2-64=0
(3x+4)2 = 0-64
(3x+4)2 = -64
3x+16 =-64
3x = 64-16
3x =48
x = 48: 3
x = 16
ý b mik ko biết làm