Schrijf zo eenvoudig mogelijk.
\(\ds \frac {1}{6}+\frac {1}{2} =\answer {\frac {2}{3}}\)

\(\ds \frac {1}{6}+\frac {1}{2}= \frac {1}{6}+\frac {3}{6}= \frac {4}{6}= \frac {2}{3} \)
\(\ds \frac {2}{9}-\frac {1}{6} = \answer {\frac {1}{18}}\)

\(\ds \frac {2}{9}-\frac {1}{6}= \frac {4}{18}-\frac {3}{18}= \frac {1}{18}\)
\(\ds \frac {1}{3}-\frac {1}{9}+\frac {2}{27} = \answer {\frac {8}{27}}\)

\(\ds \frac {1}{3}-\frac {1}{9}+\frac {2}{27}= \frac {9}{27}-\frac {3}{27}+ \frac {2}{27}= \frac {8}{27}\)
\(\ds \frac {2}{3} \cdot \frac {5}{7} = \answer {\frac {10}{21}}\)

\(\ds \frac {2}{3} \cdot \frac {5}{7} =\frac {10}{21}\)
\(\ds \frac {15}{6} \cdot \frac {3}{2} = \answer {\frac {15}{4}}\)

\(\ds \frac {15}{6} \cdot \frac {3}{2} = \frac {5}{2} \cdot \frac {3}{2}= \frac {15}{4}\)
\(\ds \frac {2}{5} \cdot \frac {9}{22} \cdot \frac {4}{18} = \answer {\frac {2}{55}}\)

\(\ds \frac {2}{5} \cdot \frac {9}{22} \cdot \frac {4}{18} = \frac {\cancel {2}^1 \cdot \cancel {9}^1 \cdot 2}{5 \cdot \cancel {22}_{11} \cdot \cancel {9}_1} = \frac {2}{55}\)
\(\ds \frac {6}{5} : \frac {2}{15} = \answer {9}\)

\(\ds \frac {6}{5} : \frac {2}{15} = \frac {6}{5} \cdot \frac {15}{2}= \frac {\cancel {6}^3 \cdot \cancel {15}^3}{\cancel {5}_1\cdot \cancel {2}_1}= \frac {9}{1}=9\)
\(\ds \frac {12}{25} : \frac {18}{35} = \answer {\frac {14}{15}}\)

\(\ds \frac {12}{25} : \frac {18}{35} = \frac {12}{25} \cdot \frac {35}{18}=\frac {\cancel {12}^2\cdot \cancel {35}^7}{\cancel {25}_5 \cdot \cancel {18}_3} = \frac {14}{15}\)
\(\ds \frac {\frac {1}{4}}{\frac {3}{2}} = \answer {\frac {1}{6}}\)

\(\ds \frac {\frac {1}{4}}{\frac {3}{2}}= \frac {1}{4} : \frac {3}{2}= \frac {1}{4} \cdot \frac {2}{3} = \frac {1}{6}\)
\(\ds \frac {\frac {1}{2}+\frac {1}{4}}{\frac {1}{6}+\frac {1}{3}} = \answer {\frac {3}{2}}\)

\(\ds \frac {\frac {1}{2}+\frac {1}{4}}{\frac {1}{6}+\frac {1}{3}}= \frac {\frac {2+1}{4}}{\frac {1+2}{6}}=\frac {\frac {3}{4}}{\frac {3}{6}}= \frac {3}{4} \cdot \frac {6}{3}= \frac {3}{2} \)
Schrijf zo eenvoudig mogelijk.
\( 20\cdot (\frac {5}{4}-\frac {4}{5}) =\answer [format=integer,onlineshowanswerbutton]{9} \)
\( -\frac {6}{27}+\frac {27}{1} + \frac {16+14}{9} -\frac {3}{14+13} - 3\cdot 9 =\answer [format=integer,onlineshowanswerbutton]{3} \)
\( -\frac {(1-a)-2}{a+1} =\answer [format=integer,onlineshowanswerbutton]{1} \) (\(a\neq -1\))
Schrijf als een zo eenvoudig mogelijke breuk. Veronderstel dat alle uitdrukkingen bestaan.
\( \frac {a-b}{c}-\frac {a-2b}{2c } =\answer [onlineshowanswerbutton]{\frac {a}{2c} } \)
\( \frac {\frac {a-b}{b}}{1-\frac {a}{b} } =\answer [onlineshowanswerbutton]{-1 } \)
\( \frac {1-\frac {a+b}{b}}{\frac {a^{2}}{b} } =\answer [onlineshowanswerbutton]{-\frac {1}{a} } \)
\( \ds \frac {a}{b}+\frac {b}{c } =\answer [onlineshowanswerbutton]{\frac {ac+b^2}{bc} } \)
\( a+\cfrac {a}{1+a } =\answer [onlineshowanswerbutton]{\frac {2a+a^2}{1+a} } \)
\( 1+\cfrac {a}{1+a } =\answer [onlineshowanswerbutton]{\frac {1+2a}{1+a} } \)
\( 1+\cfrac {1}{1+a } =\answer [onlineshowanswerbutton]{\frac {2+a}{1+a} } \)
\( a+\cfrac {1}{1+a } =\answer [onlineshowanswerbutton]{\frac {1+a+a^2}{1+a} } \)
\( \cfrac {1}{1+\cfrac {1}{1+a} } =\answer [onlineshowanswerbutton]{\frac {1+a}{2+a} } \)
\( \cfrac {1}{1+\cfrac {1}{1+\cfrac {1}{1+a}} } =\answer [onlineshowanswerbutton]{\frac {2+a}{3+2a} } \)