Question

va rog ajutati-maaaaaaa, deoarece sunt praf
ex 1 si 2​

Answer 1

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)