Упр.8.23 ГДЗ Мордкович 8 класс (Алгебра)

8.23

а)

$$ \left(\frac{x+4}{3x+3}-(x+1)^{-1}\right)\cdot\left(\frac{x+1}{3}\right)^{-1}+\frac{2}{x^2-1} $$
$$ =\left(\frac{x+4}{3(x+1)}-\frac{1}{x+1}\right)\cdot\frac{3}{x+1}+\frac{2}{x^2-1} $$
$$ =\frac{x+4-3}{3(x+1)}\cdot\frac{3}{x+1}+\frac{2}{x^2-1} =\frac{x+1}{(x+1)^2}+\frac{2}{(x-1)(x+1)} $$
$$ =\frac{1}{x+1}+\frac{2}{(x-1)(x+1)} =\frac{x-1+2}{(x-1)(x+1)} =\frac{x+1}{(x-1)(x+1)} =\frac{1}{x-1}. $$

б)

$$ \left(\frac{x+10}{5x+25}-(x+5)^{-1}\right)\cdot\left(\frac{x-5}{5}\right)^{-1}-\frac{10}{x^2-25} $$
$$ =\left(\frac{x+10}{5(x+5)}-\frac{1}{x+5}\right)\cdot\frac{5}{x-5}-\frac{10}{x^2-25} $$
$$ =\frac{x+10-5}{5(x+5)}\cdot\frac{5}{x-5}-\frac{10}{x^2-25} =\frac{x+5}{5(x+5)}\cdot\frac{5}{x-5}-\frac{10}{(x-5)(x+5)} $$
$$ =\frac{1}{x-5}-\frac{10}{(x-5)(x+5)} =\frac{x+5-10}{(x-5)(x+5)} =\frac{x-5}{(x-5)(x+5)} =\frac{1}{x+5}. $$

Ответ

а) $$\frac{1}{x-1}$$; б) $$\frac{1}{x+5}$$.