We have $\dfrac{1}{2}$ on the left and $\dfrac{3}{8}$ on the right.

Therefore, the multiplication problem shown is:

\[ \dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8} \]

We have $\dfrac{1}{4}$ on the left and $\dfrac{3}{20}$ on the right.

Therefore, the multiplication problem shown is

\[ \bbox[3pt,Gainsboro]{\color{blue}\dfrac{1}{4}} \times \dfrac{3}{5} = \dfrac{3}{20} . \]

We have $\dfrac{5}{6}$ on the left and $\dfrac{10}{18}$ on the right.

Therefore, the multiplication problem shown is

\[ \bbox[3pt,Gainsboro]{\color{blue}\dfrac{5}{6}} \times \dfrac{2}{3} =\bbox[3pt,Gainsboro]{\color{blue} \dfrac{10}{18}} \,. \]

We have $\dfrac{2}{3}$ on the left and $\dfrac{6}{12}$ on the right.

The shape on the right-hand side is split into $4$ equal parts by horizontal lines. So, the shaded part on the right is $\dfrac{\color{blue}3}{4}$ of the shaded part on the left.

Therefore, the multiplication problem shown is

\[ \dfrac{2}{3} \times \dfrac{\color{blue}3}{4} = \dfrac{6}{12}. \]

So, the missing number is $\color{blue}3.$

We have $\dfrac{1}{4}$ on the left and $\dfrac{3}{20}$ on the right.

The shape on the right-hand side is split into $5$ equal parts by horizontal lines. So, the shaded part on the right is $\dfrac{\color{blue}3}{5}$ of the shaded part on the left.

Therefore, the multiplication problem shown is \[ \dfrac{1}{4} \times \dfrac{\color{blue}3}{5} = \dfrac{3}{20}. \]

So, the missing number is $\bbox[3pt,Gainsboro]{\color{blue}3}.$

We have $\dfrac{1}{6}$ on the left and $\dfrac{1}{18}$ on the right.

The shape on the right-hand side is split into $3$ equal parts by horizontal lines. So, the shaded part on the right is $\dfrac{\color{blue}1}{3}$ of the shaded part on the left.

Therefore, the multiplication problem shown is \[ \dfrac{1}{6} \times \dfrac{\color{blue}1}{3} = \dfrac{1}{18}. \]

So, the missing number is $\bbox[3pt,Gainsboro]{\color{blue}1}.$

In the model, we have $\dfrac{3}{4}$ on the left.

Since we need to multiply by $\dfrac{\color{red}2}{\color{blue}3}$, we split the whole into $\color{blue}3$ equal parts horizontally and ensure that $\color{red}2$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{6}{12}$ of the whole.

Therefore, the missing picture is as follows:

In the model, we have $\dfrac{2}{4}$ on the left.

Since we need to multiply by $\dfrac{\color{red}3}{\color{blue}4}$, we split the whole into $\color{blue}4$ equal parts horizontally and ensure that $\color{red}3$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{6}{16}$ of the whole.

Therefore, the missing picture is as follows:

In the model, we have $\dfrac{3}{5}$ on the left.

Since we need to multiply by $\dfrac{\color{red}4}{\color{blue}5}$, we split the whole into $\color{blue}5$ equal parts horizontally and ensure that $\color{red}4$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{12}{25}$ of the whole.

Therefore, the missing picture is as follows:

Since we need to multiply by $\dfrac{\color{red}2}{\color{blue}3}$, we split the whole into $\color{blue}3$ equal parts horizontally and ensure that $\color{red}2$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{2}{9}$ of the whole.

Therefore, \[ \dfrac{1}{3} \times \dfrac{2}{3} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{2}{9}}. \]

Since we need to multiply by $\dfrac{\color{red}5}{\color{blue}7}$, we split the whole into $\color{blue}7$ equal parts horizontally and ensure that $\color{red}5$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{5}{14}$ of the whole.

Therefore, \[ \dfrac{1}{2} \times \dfrac{5}{7} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{5}{14}}. \]

Since we need to multiply by $\dfrac{\color{red}3}{\color{blue}4}$, we split the whole into $\color{blue}4$ equal parts horizontally and ensure that $\color{red}3$ of the parts remain shaded.

From the picture on the right, we obtain that the shaded part represents $\dfrac{15}{24}$ of the whole.

Therefore, \[ \dfrac{5}{6} \times \dfrac{3}{4} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{15}{24}}. \]