Question

помогите!!!!!!! молю!!!​

Answer 1

$1) \frac{t^{2}-3t+9}{9t^{2}-1 }*\frac{3t^{2}+t}{t^{3}+27 }=\frac{(t^{2}-3t+9)*t(3t+1) }{(3t+1)(3t-1)(t+3)(t^{2}-3t+9)}=\frac{t}{(3t-1)(t+3)}\\\\2)\frac{t}{(3t-1)(t+3)}-\frac{t+3}{3t^{2}-t}=\frac{t}{(3t-1)(t+3)}-\frac{t+3}{t(3t-1)}=\frac{t^{2}-(t+3)^{2}}{t(3t-1)(t+3)}=\frac{t^{2}-t^{2}-6t-9}{t(3t-1)(t+3)}=\\\\=\frac{-3(2t+3)}{t(3t-1)(t+3)}$

$3)\frac{-3(2t+3)}{t(3t-1)(t+3)} :\frac{7}{t^{2}+3t }=\frac{-3(2t+3)}{t(3t-1)(t+3)} *\frac{t(t+3)}{7}=\frac{-3(2t+3)}{7(3t-1)}\\\\4)\frac{-3(2t+3)}{7(3t-1)}-\frac{15t+6}{7-21t} =\frac{6t-9}{7(1-3t)}-\frac{15t+6}{7(1-3t)}=\frac{6t-9-15t-6}{7(1-3t)}=\frac{-15-9t}{7(1-3t)}=\frac{9t+15}{21t-7}$

Answer 2

Ответ:

3/7

Объяснение:

решение не прикрепляю так как тебе нужен только ответ