Úprava zlomku v reálných číslech
Uprav v \mathbb{R} a urči podmínky:
\large \Large \frac{{5a-5b}}{{4a+4b}}\large \cdot \Large \frac{{8a+8b}}{{15a-15b}}\large
\frac{{5a\ –\ 5b}}{{4a\ +\ 4b}}\cdot\frac{{8a\ +\ 8b}}{{15a\ –\ 15b}} = \frac{5\ \cdot\ \left (a\ –\ b \right )}{4\ \cdot\ \left (a\ +\ b \right )}\cdot \frac{8\ \cdot\ \left (a\ +\ b \right )}{15\ \cdot\ \left (a\ –\ b \right )}=
=\frac{3}{2}
\large a \neq \pm b
\frac{{5a\ –\ 5b}}{{4a\ +\ 4b}}\cdot\frac{{8a\ +\ 8b}}{{15a\ –\ 15b}} = \frac{5\ \cdot\ \left (a\ –\ b \right )}{4\ \cdot\ \left (a\ +\ b \right )}\cdot \frac{8\ \cdot\ \left (a\ +\ b \right )}{15\ \cdot\ \left (a\ –\ b \right )}=
=\frac{1}{2}
\large a \neq \pm b
\frac{{5a\ –\ 5b}}{{4a\ +\ 4b}}\cdot\frac{{8a\ +\ 8b}}{{15a\ –\ 15b}} = \frac{5\ \cdot\ \left (a\ –\ b \right )}{4\ \cdot\ \left (a\ +\ b \right )}\cdot \frac{8\ \cdot\ \left (a\ +\ b \right )}{15\ \cdot\ \left (a\ –\ b \right )}=
=\frac{2}{3}
\large a \neq \pm b
\frac{{5a\ –\ 5b}}{{4a\ +\ 4b}}\cdot\frac{{8a\ +\ 8b}}{{15a\ –\ 15b}} = \frac{5\ \cdot\ \left (a\ –\ b \right )}{4\ \cdot\ \left (a\ +\ b \right )}\cdot \frac{8\ \cdot\ \left (a\ +\ b \right )}{15\ \cdot\ \left (a\ –\ b \right )}=
=\frac{4}{3}
\large a \neq \pm b