Question

ВОПРОС НА ФОТО на фото​

Answer 1

Решение и ответ:

$\displaystyle \frac{{\sqrt{13}-\sqrt{12}}}{{\sqrt{13}+\sqrt{12}}}+\frac{{\sqrt{13}+\sqrt{12}}}{{\sqrt{13}-\sqrt{12}}}=\frac{{\left({\sqrt{13}-\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}+\frac{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}+\sqrt{12}}\right)}}{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}=$

$\displaystyle \frac{{{{\left({\sqrt{13}}\right)}^2}-\sqrt{13}\sqrt{12}-\sqrt{13}\sqrt{12}+{{\left({\sqrt{12}}\right)}^2}}}{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}+\frac{{{{\left({\sqrt{13}}\right)}^2}+\sqrt{13}\sqrt{12}+\sqrt{13}\sqrt{12}+{{\left({\sqrt{12}}\right)}^2}}}{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}=$

$\displaystyle \frac{{13-\sqrt{13}\sqrt{12}-\sqrt{13}\sqrt{12}+12+13+\sqrt{13}\sqrt{12}+\sqrt{13}\sqrt{12}+12}}{{\left({\sqrt{13}+\sqrt{12}}\right)\left({\sqrt{13}-\sqrt{12}}\right)}}=\frac{{13+13+12+12}}{{{{\left({\sqrt{13}}\right)}^2}-{{\left({\sqrt{12}}\right)}^2}}}=$

$\displaystyle \frac{{50}}{{13-12}}=\frac{{50}}{1}=\boxed{50}$