E1
Pt 5^(1/2), 1 se va trece ca exponent al lui 5 iar 2 semnifica ordinul radicalului
deci 5^(1/2) = √5
3^(3/2) = √3³
4^(-4/3) = ∛4⁻⁴
(1/3)^(2/3) = ∛(1/3)²
(4/9)^(-1/2) = √(4/9)⁻¹
(8/5)^(-1/3) = ∛(8/5)⁻¹
1 intreg2/3 = 5/3
(0,2)^(5/3) = ∛(0,2)⁵
-2 intregi si 1/2 = -5/2
(1,2)^(-5/2) = √(1,2)⁻⁵
E2
V7 = 7^(1/2)
∛27 = 27^(1/3)
∛0,008 = 0,008^(1/3)
V3⁷ = 3^(7/2)
V-(0,001)⁴ = -(0,001)^(4/2)
V(0,0225)⁻¹ = (0,225)^(-1/2)
Va⁶x² = (a⁶x²)^(1/2)
