Breuken herleiden

\({2 \over a} : {8 \over b} = {2 \over a} ⋅ {b \over 8} = {2 b \over 8 a} = {b \over 4 a}\)

\(-{1 \over 6} : p = -{1 \over 6} : {p \over 1} = -{1 \over 6} ⋅ {1 \over p} = -{1 \over 6 p}\)

\(-{8 \over 7} : {p - 6 q \over q} = -{8 \over 7} ⋅ {q \over p - 6 q} = -{8 q \over 7 (p - 6 q)} = -{8 q \over 7 p - 42 q}\)

\({3 \over 9 x} - {8 \over 9 x} = -{5 \over 9 x}\)

\({3 \over p} + {2 \over 7 p} = {21 \over 7 p} + {2 \over 7 p} = {23 \over 7 p}\)

\({3 \over 5 a} + {7 \over 2 b} = {6 b \over 10 a b} + {35 a \over 10 a b} = {6 b + 35 a \over 10 a b}\)

\(2 + {9 \over 8 x} = {2 \over 1} + {9 \over 8 x} = {16 x \over 8 x} + {9 \over 8 x} = {16 x + 9 \over 8 x}\)

\(9 p - {5 \over 6 p} = {9 p \over 1} ⋅ {6 p \over 6 p} - {5 \over 6 p} = {54 p^{2} \over 6 p} - {5 \over 6 p} = {54 p^{2} - 5 \over 6 p}\)

\({7 a \over b} - {2 \over 5 b} = {35 a \over 5 b} - {2 \over 5 b} = {35 a - 2 \over 5 b}\)

\({2 b \over 4 a} + {5 a \over 9 b} = {18 b^{2} \over 36 a b} + {20 a^{2} \over 36 a b} = {20 a^{2} + 18 b^{2} \over 36 a b} = {10 a^{2} + 9 b^{2} \over 18 a b}\)

\({4 x \over 5} + {x - 3 \over 2} = {8 x \over 10} + {5 (x - 3) \over 10} = {8 x + 5 (x - 3) \over 10} = {13 x - 15 \over 10}\)

\({-8 x + 7 \over -6 x + 2} - 3 = {-8 x + 7 \over -6 x + 2} + {-3 (-6 x + 2) \over -6 x + 2} = {-8 x + 7 - 3 (-6 x + 2) \over -6 x + 2} = {-8 x + 7 + 18 x - 6 \over -6 x + 2} = {10 x + 1 \over -6 x + 2}\)

\({4 a^{2} + 2 a + 60 \over 2 a} = {4 a^{2} \over 2 a} + {2 a \over 2 a} + {60 \over 2 a} = 2 a + 1 + {30 \over a}\)

\({a^{2} - 4 a - 6 \over 2 a^{2}} = {a^{2} \over 2 a^{2}} - {4 a \over 2 a^{2}} - {6 \over 2 a^{2}} = \frac{1}{2} - {2 \over a} - {3 \over a^{2}}\)

\({5 p \over p} = {5 \over 1} = 5\)

\({x \over 7 x} = {1 \over 7}\)

\({35 x \over -45 x} = -\frac{7}{9}\)

\({-12 p \over -2 p} = 6\)

\({-8 p q \over -14 p r} = {4 q \over 7 r}\)

\({-6 b \over -21 a b} = {2 \over 7 a}\)

\({-15 p q r \over 3 q r} = -5 p\)

\({7 a b \over b} + {3 a c \over c} = 7 a + 3 a = 10 a\)

\({2 \over x} ⋅ {8 \over y} = {16 \over x y}\)

\({p \over 2} ⋅ -{9 \over q} = -{9 p \over 2 q}\)

\(-{1 \over 2} ⋅ x = -{x \over 2}\)

\({4 b \over a} ⋅ {a - 5 \over 9} = {4 b (a - 5) \over 9 a} = {4 a b - 20 b \over 9 a}\)