ЗПЗ §§ 5–8 Алгебра = ЗПЗ §§ 10–13 Математика
Доведіть тотожність
$\left(\frac{7}{x+7}+\frac{x^2+49}{x^2-49}-\frac{7}{7-x}\right)\cdot$
$\cdot\frac{x-7}{x^2+49+14x}=$
$= \frac{1}{x+7}.$
Розв'язок:
а) $\frac{7}{x+7}+\frac{x^2+49}{x^2-49}-\frac{7}{7-x}=$
$= \frac{7}{x+7}+\frac{x^2+49}{x^2-49}-\frac{7}{x-7}=$
$= \frac{7\left(x-7\right)+\left(x^2+49\right)+7\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}=$
$= \frac{7x-49+x^2+49+7x+49}{\left(x-7\right)\left(x+7\right)}=$
$=\frac{x^2+14x+49}{\left(x-7\right)\left(x+7\right)}=$
$= \frac{\left(x+7\right)^2}{\left(x-7\right)\left(x+7\right)}=\frac{x+7}{x-7};$
б) $\frac{x+7}{x-7}\cdot\frac{x-7}{x^2+49+14x}=$
$= \frac{\left(x+7\right)\left(x-7\right)}{{\left(x-7\right)\left(x+7\right)}^2}=\frac{1}{x+7}.$