Даю 40 баллов за пример 6 класса.

Решите задачу:

$\frac{ a^{2} -49}{a+2} * \frac{1}{ a^{2}+7a } - \frac{ a^{2}+14a+49 }{ a^{3} -4 a^{2}+4a } : \frac{a+7}{a-2} + (\frac{4}{a-2}) ^{2} = ? \\ \\ 1) \frac{ a^{2} -49}{a+2} * \frac{1}{ a^{2}+7a } = \frac{(a-7)(a+7)}{a+2} * \frac{1}{a(a+7)} = \frac{a-7}{a(a+2)} \\ \\ 2) \frac{ a^{2}+14a+49 }{ a^{3} -4 a^{2}+4a } : \frac{a+7}{a-2}= \frac{(a+7) ^{2} }{a(a-2)(a-2) } * \frac{a-2}{a+7} = \frac{a+7}{a(a-2)} \\ \\$
$3)\frac{a-7}{a(a+2)}-\frac{a+7}{a(a-2)}= \frac{(a-7)(a-2)}{a(a+2)(a-2)} - \frac{(a+7)(a+2)}{a(a+2)(a-2)} = \frac{ a^{2} -9a+14-( a^{2}+9a+14) }{a(a+2)(a-2)} =\\ \\=\frac{a^{2}-9a+14- a^{2}-9a-14}{a(a+2)(a-2)}=\frac{-18a}{a(a+2)(a-2)} \\ \\ 4)(\frac{4}{a-2})^{2}=\frac{4}{a-2}*\frac{4}{a-2}=\frac{16}{(a-2)^{2}}$
$\\ 5)\frac{-18a}{a(a+2)(a-2)}+\frac{16}{(a-2)^{2}}=\frac{-18a(a-2)}{a(a+2)(a-2)^{2}}+\frac{16a(a+2)}{a(a+2)(a-2)^{2}}= \\ \\ =\frac{-18a(a-2)+16a(a+2)}{a(a+2)(a-2)^{2}}= \frac{-18a^{2}+36a+16a^{2}+32a}{a(a+2)(a-2)^{2}} = \frac{-2a^{2}+68a }{a(a+2)(a-2)^{2}} = \\ \\ = \frac{-2a(a-34)}{a(a+2)(a-2)^{2}} = \frac{-2(a-34)}{(a+2)(a-2)^{2}} = \frac{-2a+68}{ a^{3} -2a^{2} -4a+8}$