Breuken haakjesregel (reeks 2)

Hoofdmenu Eentje per keer  🖨

Bereken

1. \(+\left(\frac{10}{-3}\right) +\left(\frac{-2}{4}\right) -\left(\frac{-3}{-5}\right)\)
2. \(+\left(\frac{-5}{6}\right) +\left(\frac{-4}{8}\right) +\left(\frac{-2}{-3}\right)\)
3. \(+\left(\frac{7}{-8}\right) +\left(\frac{-8}{-4}\right) +\left(\frac{-5}{3}\right)\)
4. \(-\left(\frac{5}{-2}\right) -\left(\frac{-8}{6}\right) +\left(\frac{-8}{7}\right)\)
5. \(+\left(\frac{4}{-3}\right) -\left(\frac{7}{8}\right) +\left(\frac{2}{-7}\right)\)
6. \(+\left(\frac{-4}{-7}\right) +\left(\frac{-6}{-10}\right) +\left(\frac{4}{-9}\right)\)
7. \(+\left(\frac{3}{-7}\right) +\left(\frac{-7}{5}\right) +\left(\frac{12}{-11}\right)\)
8. \(+\left(\frac{8}{-9}\right) +\left(\frac{-3}{10}\right) +\left(\frac{-12}{9}\right)\)
9. \(-\left(\frac{4}{-3}\right) -\left(\frac{-3}{-6}\right) +\left(\frac{3}{-7}\right)\)
10. \(-\left(\frac{-7}{-2}\right) +\left(\frac{9}{-5}\right) +\left(\frac{8}{-12}\right)\)
11. \(+\left(\frac{-3}{4}\right) +\left(\frac{-7}{-3}\right) -\left(\frac{-8}{-3}\right)\)
12. \(+\left(\frac{-4}{-5}\right) -\left(\frac{6}{6}\right) -\left(\frac{6}{12}\right)\)

---

Bereken

Verbetersleutel

1. \(+\left(\frac{10}{-3}\right) +\left(\frac{-2}{4}\right) -\left(\frac{-3}{-5}\right)\\= \frac{-10}{3}- \frac{2}{4}-\frac{3}{5}\\=\frac{-10}{3}-\frac{1}{2}-\frac{3}{5}\\=\frac{-100}{30}-\frac{15}{30}-\frac{18}{30}\\=\frac{-100-15-18}{30}\\=\frac{-133}{30}\)
2. \(+\left(\frac{-5}{6}\right) +\left(\frac{-4}{8}\right) +\left(\frac{-2}{-3}\right)\\= \frac{-5}{6}- \frac{4}{8}+\frac{2}{3}\\=\frac{-5}{6}-\frac{1}{2}+\frac{2}{3}\\=\frac{-5-1+2}{6}\\=\frac{-4}{6}\\=\frac{-2}{3}\)
3. \(+\left(\frac{7}{-8}\right) +\left(\frac{-8}{-4}\right) +\left(\frac{-5}{3}\right)\\= \frac{-7}{8}+ \frac{8}{4}-\frac{5}{3}\\=\frac{-7}{8}+\frac{2}{1}-\frac{5}{3}\\=\frac{-21}{24}+\frac{48}{24}-\frac{40}{24}\\=\frac{-21+48-40}{24}\\=\frac{-13}{24}\)
4. \(-\left(\frac{5}{-2}\right) -\left(\frac{-8}{6}\right) +\left(\frac{-8}{7}\right)\\= \frac{5}{2}+ \frac{8}{6}-\frac{8}{7}\\=\frac{5}{2}+\frac{4}{3}-\frac{8}{7}\\=\frac{105}{42}+\frac{56}{42}-\frac{48}{42}\\=\frac{105+56-48}{42}\\=\frac{113}{42}\)
5. \(+\left(\frac{4}{-3}\right) -\left(\frac{7}{8}\right) +\left(\frac{2}{-7}\right)\\= \frac{-4}{3}- \frac{7}{8}-\frac{2}{7}\\=\frac{-224}{168}-\frac{147}{168}-\frac{48}{168}\\=\frac{-224-147-48}{168}\\=\frac{-419}{168}\)
6. \(+\left(\frac{-4}{-7}\right) +\left(\frac{-6}{-10}\right) +\left(\frac{4}{-9}\right)\\= \frac{4}{7}+ \frac{6}{10}-\frac{4}{9}\\=\frac{4}{7}+\frac{3}{5}-\frac{4}{9}\\=\frac{180}{315}+\frac{189}{315}-\frac{140}{315}\\=\frac{180+189-140}{315}\\=\frac{229}{315}\)
7. \(+\left(\frac{3}{-7}\right) +\left(\frac{-7}{5}\right) +\left(\frac{12}{-11}\right)\\= \frac{-3}{7}- \frac{7}{5}-\frac{12}{11}\\=\frac{-165}{385}-\frac{539}{385}-\frac{420}{385}\\=\frac{-165-539-420}{385}\\=\frac{-1124}{385}\)
8. \(+\left(\frac{8}{-9}\right) +\left(\frac{-3}{10}\right) +\left(\frac{-12}{9}\right)\\= \frac{-8}{9}- \frac{3}{10}-\frac{12}{9}\\=\frac{-8}{9}-\frac{3}{10}-\frac{4}{3}\\=\frac{-80}{90}-\frac{27}{90}-\frac{120}{90}\\=\frac{-80-27-120}{90}\\=\frac{-227}{90}\)
9. \(-\left(\frac{4}{-3}\right) -\left(\frac{-3}{-6}\right) +\left(\frac{3}{-7}\right)\\= \frac{4}{3}- \frac{3}{6}-\frac{3}{7}\\=\frac{4}{3}-\frac{1}{2}-\frac{3}{7}\\=\frac{56}{42}-\frac{21}{42}-\frac{18}{42}\\=\frac{56-21-18}{42}\\=\frac{17}{42}\)
10. \(-\left(\frac{-7}{-2}\right) +\left(\frac{9}{-5}\right) +\left(\frac{8}{-12}\right)\\= \frac{-7}{2}- \frac{9}{5}-\frac{8}{12}\\=\frac{-7}{2}-\frac{9}{5}-\frac{2}{3}\\=\frac{-105}{30}-\frac{54}{30}-\frac{20}{30}\\=\frac{-105-54-20}{30}\\=\frac{-179}{30}\)
11. \(+\left(\frac{-3}{4}\right) +\left(\frac{-7}{-3}\right) -\left(\frac{-8}{-3}\right)\\= \frac{-3}{4}+ \frac{7}{3}-\frac{8}{3}\\=\frac{-9}{12}+\frac{28}{12}-\frac{32}{12}\\=\frac{-9+28-32}{12}\\=\frac{-13}{12}\)
12. \(+\left(\frac{-4}{-5}\right) -\left(\frac{6}{6}\right) -\left(\frac{6}{12}\right)\\= \frac{4}{5}- \frac{6}{6}-\frac{6}{12}\\=\frac{4}{5}-\frac{1}{1}-\frac{1}{2}\\=\frac{8}{10}-\frac{10}{10}-\frac{5}{10}\\=\frac{8-10-5}{10}\\=\frac{-7}{10}\)