Question

пожалуйста решите В2, В7, С1​ (хотя бы 2)

Answer 1

Ответ:

Пошаговое объяснение:

B2. (a^(1/4) +b^(1/4))²-(a^(1/4) -b^(1/4))²=(a^(1/4) +b^(1/4) -a^(1/4) +b^(1/4))(a^(1/4) +b^(1/4) +a^(1/4) -b^(1/4))=2b^(1/4) ·2a^(1/4)=4∜(ab)

B7. √⁹√a⁶ +(2a)/∛a²=(a^(2/3))^(1/2) +(2a)/(a^(2/3))=(a^(1/3) ·a^(2/3) +2a)/(a^(2/3))=(a^(1/3 +2/3) +2a)/(a^(2/3))=(a+2a)/(a^(2/3))=(3a)/(a^(2/3))=3a^(3/3 -2/3)=3∛a

C1. x=16/81

(√((√x+4)²-16√x))/(∜x -4/∜x)=∜x=∜(16/81)=2/3

1) √((√x+4)²-16√x)=√(x+8√x +16-16√x)=√(x-8√x +16)=√(√x-4)²=√x-4

2) ∜x -4/∜x=x^(1/4) -4=(x^(1/4 +1/4) -4)/x^(1/4)=(√x-4)/∜x

3) (√x-4)/((√x-4)/∜x)=(√x-4)·(∜x)/(√x-4)=∜x