Question

помогите........ ......​

Answer 1

Ответ:

-1

Объяснение:

$\displaystyle \frac{3}{2a-3} -\frac{8a^3-18a}{4a^2+9} \bigg(\frac{2a}{4a^2-12a+9} -\frac{3}{4a^2-9} \bigg)=$

$\displaystyle =\frac{3}{2a-3} -\frac{2a(4a^2-9)}{4a^2+9} \bigg(\frac{2a}{(2a-3)^2}^{\setminus 2a+3} -\frac{3}{(2a-3)(2a+3)}^{\setminus 2a-3} \bigg)=$

$\displaystyle = \frac{3}{2a-3} -\frac{2a(2a-3)(2a+3)}{4a^2+9} *\frac{2a(2a+3)-3(2a-3)}{(2a-3)^2(2a+3)} =$

$\displaystyle =\frac{3}{2a-3} -\frac{2a(2a+3)(2a-3)*(4a^2+6a-6a+9)}{(4a^2+9)(2a-3)^2(2a+3)} =$

$\displaystyle =\frac{3}{2a-3} -\frac{2a(2a+3)(2a-3)*(4a^2+9)}{(4a^2+9)(2a-3)^2(2a+3)} =$

$\displaystyle =\frac{3}{2a-3} -\frac{2a}{2a-3} =\frac{3-2a}{2a-3} =-1$