Breuken herleiden

\({5 \over a}:{2 \over b}={5 \over a}⋅{b \over 2}={5b \over 2a}\)

\({4 \over 7}:x={4 \over 7}:{x \over 1}={4 \over 7}⋅{1 \over x}={4 \over 7x}\)

\(-{4 \over 7}:{x-5y \over y}=-{4 \over 7}⋅{y \over x-5y}=-{4y \over 7(x-5y)}=-{4y \over 7x-35y}\)

\({9 \over 4a}+{7 \over 4a}={16 \over 4a}={4 \over a}\)

\({5 \over p}+{2 \over 3p}={15 \over 3p}+{2 \over 3p}={17 \over 3p}\)

\({9 \over 8a}-{7 \over 4b}={9b \over 8ab}-{14a \over 8ab}={9b-14a \over 8ab}\)

\(2-{4 \over 3x}={2 \over 1}-{4 \over 3x}={6x \over 3x}-{4 \over 3x}={6x-4 \over 3x}\)

\(8p+{7 \over 4p}={8p \over 1}⋅{4p \over 4p}+{7 \over 4p}={32p^2 \over 4p}+{7 \over 4p}={32p^2+7 \over 4p}\)

\({7x \over y}+{8 \over 3y}={21x \over 3y}+{8 \over 3y}={21x+8 \over 3y}\)

\({2b \over 6a}-{4a \over 7b}={14b^2 \over 42ab}-{24a^2 \over 42ab}={-24a^2+14b^2 \over 42ab}={-12a^2+7b^2 \over 21ab}\)

\({5x \over 9}+{x+3 \over 8}={40x \over 72}+{9(x+3) \over 72}={40x+9(x+3) \over 72}={49x+27 \over 72}\)

\({8a+5 \over -6a-2}-9={8a+5 \over -6a-2}+{-9(-6a-2) \over -6a-2}={8a+5-9(-6a-2) \over -6a-2}={8a+5+54a+18 \over -6a-2}={62a+23 \over -6a-2}\)

\({2 \over a}⋅{7 \over b}={14 \over ab}\)

\({a \over 3}⋅{9 \over b}={9a \over 3b}={3a \over b}\)

\({7 \over 9}⋅x={7x \over 9}\)

\({4y \over x}⋅{x-7 \over 2}={4y(x-7) \over 2x}={2y(x-7) \over x}={2xy-14y \over x}\)

\({3p^2-9p-60 \over 3p}={3p^2 \over 3p}-{9p \over 3p}-{60 \over 3p}=p-3-{20 \over p}\)

\({a^2+8a+9 \over 3a^2}={a^2 \over 3a^2}+{8a \over 3a^2}+{9 \over 3a^2}=\frac{1}{3}+{8 \over 3a}+{3 \over a^2}\)

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

\({x \over 9x}={1 \over 9}\)

\({-8a \over -10a}=\frac{4}{5}\)

\({12x \over -3x}=-4\)

\({-15ab \over -35ac}={3b \over 7c}\)

\({6q \over -14pq}=-{3 \over 7p}\)

\({-12xyz \over 3yz}=-4x\)

\({4pq \over q}-{7pr \over r}=4p-7p=-3p\)