Question

помогите срочно алгебру ​

Answer 1

Объяснение:

6.

$a)\ (\frac{x+3}{x-3}-\frac{x}{x+3}):\frac{x+1}{x+3} =\frac{(x+3)^2-x*(x-3)}{(x-3)*(x+3)}*\frac{x+3}{x+1} =\frac{(x^2+6x+9-x^2+3x)*(x+3)}{(x-3)*(x+3)*(x+1)} =\\= \frac{9x+9}{(x-3)*(x+1)} = \frac{9*(x+1)}{(x-3)*(x+1)} =\frac{9}{x-3}.\\b)\ (\frac{2x}{5y^4} )^3=\frac{2^3*x^3}{5^3*(y^4)^3}= \frac{8x^3}{125y^{12}} .$