Odpowiedź:
[tex]a)\ \ 3x\cdot12=36x\\\\b)\ \ 4\cdot0,7x=2,8x\\\\c)\ \ \frac{1}{3}\cdot6x=\frac{6}{3}x=2x\\\\d)\ \ \frac{1}{2}\cdot\frac{1}{3}x=\frac{1}{6}x\\\\e)\ \ \not2^1y\cdot(-\frac{3}{\not4_{2}})=y\cdot(-\frac{3}{2})=-\frac{3}{2}y\\\\f)\ \ \frac{x}{\not2_{1}}\cdot\not6^3=x\cdot3=3x\\\\g)\ \ 6x:3=2x\\\\h)\ \ \frac{3}{4}x:2=\frac{3}{4}x\cdot\frac{1}{2}=\frac{3}{8}x\\\\i)\ \ \frac{15x}{3}=5x\\\\j)\ \ \not6^2\cdot\frac{x}{\not9_{3}}=2\cdot\frac{x}{3}=\frac{2x}{3}=\frac{2}{3}x[/tex]
[tex]k)\ \ \frac{2,5x}{0,5}=\frac{25x}{5}=5x\\\\l)\ \ \frac{x}{\not3_{1}}\cdot(-\not9^3)=x\cdot(-3)=-3x[/tex]
