Question

Очень срочно нужно помогите

Answer 1

Ответ:

Пошаговое объяснение:

√x ·x^(1/3)=x^(1/2) ·x^(1/3)=x^(1/2 +1/3)=x^(3/6 +2/6)=x^(5/6)=⁶√x⁵

y^(1/6) ·∛y=y^(1/6) ·y^(1/3)=y^(1/6 +1/3)=y^(1/6 +2/6)=y^(1/2)=√y

a^(2/5) ·¹⁰√a=a^(2/5) ·a^(1/10)=a^(2/5 +1/10)=a^(4/10 +1/10)=a^(1/2)=√a

∛b² ÷b^(5/6)=b^(2/3) ÷b^(5/6)=b^(2/3 -5/6)=b^(4/6 -5/6)=b^(-1/6)=⁶√b⁻¹=⁶√(1/b)

c^(4/7) ÷(c·√c)=c^(4/7) ÷c^(1 +1/2)=c^(4/7 -3/2)=c^(8/14 -21/14)=c^(-13/14)=¹⁴√(1/c)¹³

a^(5/6) ·⁶√a ÷a^(-1/2)=a^(5/6 +1/6 +1/2)=a^(2/2 +1/2)=a^(3/2)=√a³