Breuken haakjesregel (reeks 2)

Hoofdmenu Eentje per keer  🖨

Bereken

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

---

Bereken

Verbetersleutel

1. \(+\left(\frac{3}{-7}\right) +\left(\frac{-8}{2}\right) +\left(\frac{-9}{-7}\right)\\= \frac{-3}{7}- \frac{8}{2}+\frac{9}{7}\\=\frac{-3}{7}-\frac{4}{1}+\frac{9}{7}\\=\frac{-3-4+9}{7}\\=\frac{2}{7}\)
2. \(+\left(\frac{-4}{5}\right) -\left(\frac{10}{3}\right) -\left(\frac{-11}{6}\right)\\= \frac{-4}{5}- \frac{10}{3}+\frac{11}{6}\\=\frac{-24}{30}-\frac{100}{30}+\frac{55}{30}\\=\frac{-24-100+55}{30}\\=\frac{-69}{30}\\=\frac{-23}{10}\)
3. \(+\left(\frac{-6}{7}\right) +\left(\frac{-8}{10}\right) +\left(\frac{12}{9}\right)\\= \frac{-6}{7}- \frac{8}{10}+\frac{12}{9}\\=\frac{-6}{7}-\frac{4}{5}+\frac{4}{3}\\=\frac{-90}{105}-\frac{84}{105}+\frac{140}{105}\\=\frac{-90-84+140}{105}\\=\frac{-34}{105}\)
4. \(-\left(\frac{10}{-3}\right) +\left(\frac{-7}{7}\right) +\left(\frac{-5}{-4}\right)\\= \frac{10}{3}- \frac{7}{7}+\frac{5}{4}\\=\frac{10}{3}-\frac{1}{1}+\frac{5}{4}\\=\frac{40}{12}-\frac{12}{12}+\frac{15}{12}\\=\frac{40-12+15}{12}\\=\frac{43}{12}\)
5. \(-\left(\frac{7}{-6}\right) +\left(\frac{-4}{-8}\right) +\left(\frac{10}{-2}\right)\\= \frac{7}{6}+ \frac{4}{8}-\frac{10}{2}\\=\frac{7}{6}+\frac{1}{2}-\frac{5}{1}\\=\frac{7+1-5}{6}\\=\frac{3}{6}\\=\frac{1}{2}\)
6. \(-\left(\frac{4}{-5}\right) +\left(\frac{8}{8}\right) -\left(\frac{-4}{-7}\right)\\= \frac{4}{5}+ \frac{8}{8}-\frac{4}{7}\\=\frac{4}{5}+\frac{1}{1}-\frac{4}{7}\\=\frac{28}{35}+\frac{35}{35}-\frac{20}{35}\\=\frac{28+35-20}{35}\\=\frac{43}{35}\)
7. \(-\left(\frac{6}{-5}\right) +\left(\frac{3}{-9}\right) -\left(\frac{7}{3}\right)\\= \frac{6}{5}- \frac{3}{9}-\frac{7}{3}\\=\frac{6}{5}-\frac{1}{3}-\frac{7}{3}\\=\frac{18}{15}-\frac{5}{15}-\frac{35}{15}\\=\frac{18-5-35}{15}\\=\frac{-22}{15}\)
8. \(-\left(\frac{7}{9}\right) +\left(\frac{4}{5}\right) +\left(\frac{-12}{11}\right)\\= \frac{-7}{9}+ \frac{4}{5}-\frac{12}{11}\\=\frac{-385}{495}+\frac{396}{495}-\frac{540}{495}\\=\frac{-385+396-540}{495}\\=\frac{-529}{495}\)
9. \(+\left(\frac{3}{-4}\right) -\left(\frac{-5}{2}\right) +\left(\frac{-12}{10}\right)\\= \frac{-3}{4}+ \frac{5}{2}-\frac{12}{10}\\=\frac{-3}{4}+\frac{5}{2}-\frac{6}{5}\\=\frac{-15}{20}+\frac{50}{20}-\frac{24}{20}\\=\frac{-15+50-24}{20}\\=\frac{11}{20}\)
10. \(+\left(\frac{-7}{-3}\right) -\left(\frac{-3}{-3}\right) -\left(\frac{-8}{-4}\right)\\= \frac{7}{3}- \frac{3}{3}-\frac{8}{4}\\=\frac{7}{3}-\frac{1}{1}-\frac{2}{1}\\=\frac{7-1-2}{3}\\=\frac{4}{3}\)
11. \(-\left(\frac{-2}{9}\right) +\left(\frac{6}{4}\right) +\left(\frac{8}{10}\right)\\= \frac{2}{9}+ \frac{6}{4}+\frac{8}{10}\\=\frac{2}{9}+\frac{3}{2}+\frac{4}{5}\\=\frac{20}{90}+\frac{135}{90}+\frac{72}{90}\\=\frac{20+135+72}{90}\\=\frac{227}{90}\)
12. \(+\left(\frac{-8}{-5}\right) +\left(\frac{-7}{6}\right) -\left(\frac{-6}{11}\right)\\= \frac{8}{5}- \frac{7}{6}+\frac{6}{11}\\=\frac{528}{330}-\frac{385}{330}+\frac{180}{330}\\=\frac{528-385+180}{330}\\=\frac{323}{330}\)