Question

Упростите:Спасибо!

Answer 1

$( \frac{2x-3}{x^2}- \frac{x+3}{x}+ \frac{x+12}{9} ) \frac{3x}{9-x^2} = \frac{3x(2x-3)}{x^2(9-x^2)}- \frac{3x(x+3)}{x(9-x^2)}+ \frac{3x(x+12)}{9(9-x^2)} = \\\ =\frac{9(2x-3)}{3x(9-x^2)}- \frac{9x(x+3)}{3x(9-x^2)}+ \frac{x^2(x+12)}{3x(9-x^2)} = \frac{9(2x-3)-9x(x+3)+x^2(x+12)}{3x(9-x^2)}= \\\ =\frac{18x-27-9x^2-27x+x^3+12x^2}{3x(9-x^2)}= \frac{x^3+3x^2-9x-27}{3x(9-x^2)}=\frac{(x-3)(x+3)^2}{3x(3-x)(x+3)}=-\frac{x+3}{3x}$

Answer 2

9*(2x-3)-9x(x+3)+x^2(x+12) / 9x^2  *3x/(9-x^2) =
= (18x-27-9x^2-27x+x^3+12x^2) / 3x  *1/(9-x^2) =
= (x^3+3x^2-9x-27) / 3x  *1/(9-x^2) =
= x^2(x+3)-9(x+3)  /3x  *  -1/(x^2-9)=
=(x+3)(x^2-9) /3x *  -1/(x^2-9)= - (x+3)/3x