Упр.36.8 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (5x+3)/(x^2-16)+(6x-1)/(16-x^2 ) при x=-4,1;
2) (a^2+a)/(a^2-9)-(7a-9)/(a^2-9) при a=7.

1) При $$x=-4{,}1$$:

$$ \frac{5x+3}{x^2-16}+\frac{6x-1}{16-x^2} = \frac{5x+3}{x^2-16}-\frac{6x-1}{x^2-16} = \frac{5x+3-(6x-1)}{x^2-16} $$
$$ = \frac{-x+4}{x^2-16} = \frac{-(x-4)}{(x-4)(x+4)} = -\frac{1}{x+4} $$
$$ -\frac{1}{-4{,}1+4} = -\frac{1}{-0{,}1} = 10. $$

2) При $$a=7$$:

$$ \frac{a^2+a}{a^2-9}-\frac{7a-9}{a^2-9} = \frac{a^2+a-(7a-9)}{a^2-9} = \frac{a^2-6a+9}{a^2-9} $$
$$ = \frac{(a-3)^2}{(a-3)(a+3)} = \frac{a-3}{a+3} $$
$$ \frac{7-3}{7+3} = \frac{4}{10} = 0{,}4. $$

Ответ

1) $$10$$; 2) $$0{,}4$$.