a) (x+5)3 = -64
=> (x+5)3 = (-4)3
=> x+5 = -4
=> x = -4-5
=> x = -9
b) (2x-3)2 = 9
=> (2x-3)2 = 32 hay (-3)2
=> 2x-3 = 3 hay 2x-3 = -3
=> 2x = 3+3 hay 2x = -3+3
=> 2x = 6 hay 2x = 0
=> x = 6:2 hay x = 0:2
=> x = 3 hay x = 0
=>
a)\(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(x=-9\)
b) \(\left(2x-3\right)^2=9\)
\(\left(2x-3\right)^2=3^2=\left(-3\right)^2\)
TH1 \(\left(2x-3\right)^2=3^2\) 2x-3=3 2x=6 x =3
| TH2\(\left(2x-3\right)^2=\left(-3\right)^2\) 2x-3=-3 2x = 0 x = 0 |
