Szczegółowe wyjaśnienie:
[tex]a)\ \dfrac{2}{3}:x=\dfrac{10}{21}\to x=\dfrac{2}{3}:\dfrac{10}{21}\\\\x=\dfrac{2\!\!\!\!\diagup^1}{3\!\!\!\!\diagup_1}\cdot\dfrac{21\!\!\!\!\!\diagup^7}{10\!\!\!\!\!\diagup_5}\\\\\boxed{x=\dfrac{7}{5}}\\\\spr:\\\\L=\dfrac{2}{3}:\dfrac{7}{5}=\dfrac{2}{3}\cdot\dfrac{5}{7}=\dfrac{10}{21}=P[/tex]
[tex]b)\ \dfrac{4}{7}:x=\dfrac{4}{35}\to x=\dfrac{4}{7}:\dfrac{4}{35}\\\\x=\dfrac{4\!\!\!\!\diagup^1}{7\!\!\!\!\diagup_1}\cdot\dfrac{35\!\!\!\!\!\diagup^5}{4\!\!\!\!\diagup_1}\\\\\boxed{x=5}\\\\spr:\\\\L=\dfrac{4}{7}:5=\dfrac{4}{7}\cdot\dfrac{1}{5}=\dfrac{4}{35}=P[/tex]
[tex]c)\ x:\dfrac{1}{4}=12\qquad|\cdot\dfrac{1}{4}\\\\x=12\!\!\!\!\!\diagup^3\cdot\dfrac{1}{4\!\!\!\!\diagup_1}\\\\\boxed{x=3}\\\\spr:\\\\L=3:\dfrac{1}{4}=3\cdot\dfrac{4}{1}=12=P[/tex]
[tex]d)\ \dfrac{2}{5}:x=2\to x=\dfrac{2}{5}:2\\\\x=\dfrac{2\!\!\!\!\diagup^1}{5}\cdot\dfrac{1}{2\!\!\!\!\diagup_1}\\\\\boxed{x=\dfrac{1}{5}}\\\\spr:\\\\L=\dfrac{2}{5}:\dfrac{1}{5}=\dfrac{2}{5\!\!\!\!\diagup_1}\cdot\dfrac{5\!\!\!\!\diagup^1}{1}=2=P[/tex]
[tex]e)\ x:\dfrac{2}{3}=\dfrac{3}{8}\qquad|\cdot\dfrac{2}{3}\\\\x=\dfrac{3\!\!\!\!\diagup^1}{8\!\!\!\!\diagup_4}\cdot\dfrac{2\!\!\!\!\diagup^1}{3\!\!\!\!\diagup_1}\\\\\boxed{x=\dfrac{1}{4}}\\\\spr:\\\\L=\dfrac{1}{4}:\dfrac{2}{3}=\dfrac{1}{4}\cdot\dfrac{3}{2}=\dfrac{3}{8}=P[/tex]
[tex]f)\ \dfrac{3}{7}:x=1\to\boxed{x=\dfrac{3}{7}}\\\\spr:\\\\L=\dfrac{3}{7}:\dfrac{3}{7}=1=P[/tex]
