Упр.6.13 ГДЗ Мордкович 8 класс (Алгебра)

1. $$\frac{9n+27}{3n^2-n^3}+\left(\frac{3n+9}{n-3}\right)^2\cdot\left(\frac{1}{3n-9}+\frac{2}{9-n^2}-\frac{1}{n^2+3n}\right)$$

   Преобразуем выражение в скобках:
   $$\frac{1}{3n-9}+\frac{2}{9-n^2}-\frac{1}{n^2+3n}=\frac{1}{3(n-3)}-\frac{2}{(n-3)(n+3)}-\frac{1}{n(n+3)}$$

   $$=\frac{n(n+3)-2\cdot 3n-3(n-3)}{3n(n-3)(n+3)}=\frac{n^2-6n+9}{3n(n-3)(n+3)}=\frac{(n-3)^2}{3n(n-3)(n+3)}=\frac{n-3}{3n(n+3)}$$

   Тогда
   $$\left(\frac{3n+9}{n-3}\right)^2\cdot\frac{n-3}{3n(n+3)}=\frac{9(n+3)^2}{(n-3)^2}\cdot\frac{n-3}{3n(n+3)}=\frac{3(n+3)}{n(n-3)}$$

   А первая дробь:
   $$\frac{9n+27}{3n^2-n^3}=\frac{9(n+3)}{n^2(3-n)}=-\frac{9(n+3)}{n^2(n-3)}$$

   Складываем:
   $$-\frac{9(n+3)}{n^2(n-3)}+\frac{3(n+3)}{n(n-3)}=\frac{-9(n+3)+3n(n+3)}{n^2(n-3)}$$
   $$=\frac{3(n+3)(n-3)}{n^2(n-3)}=\frac{3(n+3)}{n^2}$$

2. $$\left(\frac{2}{2p-q}+\frac{6q}{q^2-4p^2}-\frac{4}{2p+q}\right):\left(1+\frac{4p^2+q^2}{4p^2-q^2}\right)$$

   Так как $$q^2-4p^2=-(4p^2-q^2)$$, то
   $$\frac{2}{2p-q}+\frac{6q}{q^2-4p^2}-\frac{4}{2p+q}=\frac{2}{2p-q}-\frac{6q}{4p^2-q^2}-\frac{4}{2p+q}$$

   Приведём к общему знаменателю:
   $$\frac{2(2p+q)-6q-4(2p-q)}{(2p-q)(2p+q)}=\frac{4p+2q-6q-8p+4q}{(2p-q)(2p+q)}=\frac{-4p}{4p^2-q^2}$$

   Второй множитель:
   $$1+\frac{4p^2+q^2}{4p^2-q^2}=\frac{4p^2-q^2+4p^2+q^2}{4p^2-q^2}=\frac{8p^2}{4p^2-q^2}$$

   Тогда
   $$\frac{-4p}{4p^2-q^2}:\frac{8p^2}{4p^2-q^2}=\frac{-4p}{4p^2-q^2}\cdot\frac{4p^2-q^2}{8p^2}=-\frac{1}{2p}$$

Ответ

а) $$\frac{3(n+3)}{n^2}$$;
б) $$-\frac{1}{2p}$$.