Breuken haakjesregel (reeks 1)

Hoofdmenu Eentje per keer  🖨

Bereken

1. \(+\left(\frac{-3}{-10}\right)+\left(\frac{-3}{-10}\right)\)
2. \(-\left(\frac{-7}{8}\right)+\left(\frac{-8}{11}\right)\)
3. \(-\left(\frac{8}{-13}\right)-\left(\frac{-7}{-5}\right)\)
4. \(+\left(\frac{-2}{5}\right)+\left(\frac{-12}{-5}\right)\)
5. \(-\left(\frac{-4}{-11}\right)-\left(\frac{8}{3}\right)\)
6. \(-\left(\frac{2}{9}\right)-\left(\frac{-8}{9}\right)\)
7. \(+\left(\frac{-8}{5}\right)-\left(\frac{3}{7}\right)\)
8. \(-\left(\frac{10}{13}\right)+\left(\frac{15}{-2}\right)\)
9. \(+\left(\frac{-8}{-11}\right)-\left(\frac{5}{-6}\right)\)
10. \(-\left(\frac{8}{3}\right)+\left(\frac{-6}{13}\right)\)
11. \(+\left(\frac{-4}{3}\right)-\left(\frac{4}{3}\right)\)
12. \(-\left(\frac{-6}{-11}\right)+\left(\frac{-15}{-7}\right)\)

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Bereken

Verbetersleutel

1. \(+\left(\frac{-3}{-10}\right)+\left(\frac{-3}{-10}\right)\\= \frac{3}{10}+\frac{3}{10}\\=\frac{3+3}{10}\\=\frac{6}{10}\\=\frac{3}{5}\)
2. \(-\left(\frac{-7}{8}\right)+\left(\frac{-8}{11}\right)\\= \frac{7}{8}-\frac{8}{11}\\=\frac{77}{88}-\frac{64}{88}\\=\frac{77-64}{88}\\=\frac{13}{88}\)
3. \(-\left(\frac{8}{-13}\right)-\left(\frac{-7}{-5}\right)\\= \frac{8}{13}-\frac{7}{5}\\=\frac{40}{65}-\frac{91}{65}\\=\frac{40-91}{65}\\=\frac{-51}{65}\)
4. \(+\left(\frac{-2}{5}\right)+\left(\frac{-12}{-5}\right)\\= \frac{-2}{5}+\frac{12}{5}\\=\frac{-2+12}{5}\\=\frac{10}{5}\\=\frac{2}{1}\)
5. \(-\left(\frac{-4}{-11}\right)-\left(\frac{8}{3}\right)\\= \frac{-4}{11}-\frac{8}{3}\\=\frac{-12}{33}-\frac{88}{33}\\=\frac{-12-88}{33}\\=\frac{-100}{33}\)
6. \(-\left(\frac{2}{9}\right)-\left(\frac{-8}{9}\right)\\= \frac{-2}{9}+\frac{8}{9}\\=\frac{-2+8}{9}\\=\frac{6}{9}\\=\frac{2}{3}\)
7. \(+\left(\frac{-8}{5}\right)-\left(\frac{3}{7}\right)\\= \frac{-8}{5}-\frac{3}{7}\\=\frac{-56}{35}-\frac{15}{35}\\=\frac{-56-15}{35}\\=\frac{-71}{35}\)
8. \(-\left(\frac{10}{13}\right)+\left(\frac{15}{-2}\right)\\= \frac{-10}{13}-\frac{15}{2}\\=\frac{-20}{26}-\frac{195}{26}\\=\frac{-20-195}{26}\\=\frac{-215}{26}\)
9. \(+\left(\frac{-8}{-11}\right)-\left(\frac{5}{-6}\right)\\= \frac{8}{11}+\frac{5}{6}\\=\frac{48}{66}+\frac{55}{66}\\=\frac{48+55}{66}\\=\frac{103}{66}\)
10. \(-\left(\frac{8}{3}\right)+\left(\frac{-6}{13}\right)\\= \frac{-8}{3}-\frac{6}{13}\\=\frac{-104}{39}-\frac{18}{39}\\=\frac{-104-18}{39}\\=\frac{-122}{39}\)
11. \(+\left(\frac{-4}{3}\right)-\left(\frac{4}{3}\right)\\= \frac{-4}{3}-\frac{4}{3}\\=\frac{-4-4}{3}\\=\frac{-8}{3}\)
12. \(-\left(\frac{-6}{-11}\right)+\left(\frac{-15}{-7}\right)\\= \frac{-6}{11}+\frac{15}{7}\\=\frac{-42}{77}+\frac{165}{77}\\=\frac{-42+165}{77}\\=\frac{123}{77}\)