a, (3 + x)2 = 4
(3 + x)2 = 22 = (-2)2
TH1: 3 + x = 2
x = 2 - 3
x = -1
TH2: 3 + x = -2
x = -2 - 3
x = -2 + (-3)
x = -5
Vậy x ϵ {-1; -5}
b, (2x + 1) 3 = -8
(2x + 1)3 = (-2)3
2x + 1 = -2
2x = -2 - 1
2x = -2 + (-1)
2x = -3
x = \(\dfrac{-3}{2}\)
a) \(\left(3+x\right)^2\) = 4
\(\left(3+x\right)^2\)= 22
⇒ 3 + \(x\) = 2
\(x\) = 2 - 3
\(x\) = -1
b) \(\left(2x+1\right)^3\) = -8
\(\left(2x+1\right)^3\) = 23
⇒ 2\(x\) + 1 = 2
2\(x\) = 2 - 1
2\(x\) = 1
\(x\) = 1 : 2
\(x\) = 0,5