First, we write $1\,\dfrac 2 {3}$ as an improper fraction:

\[ 1\,\dfrac 2 {3}=\dfrac{(1\times 3)+2}{3}= \dfrac {5} {3}\]

We also write $2\,\dfrac 3 {4} $ as an improper fraction:

\[ 2\,\dfrac 3 {4}=\dfrac{(2\times 4)+3}{4}= \dfrac {11} {4}\]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[\dfrac{5}{3}\times\dfrac {11} {4}= \dfrac {5\times 11} {3\times 4} = \dfrac{\color{blue}55}{12} \]

Therefore, the missing number is ${\color{blue}{55}}.$

First, we write $1\,\dfrac 3 {5}$ as an improper fraction:

\[ 1\,\dfrac 3 {5}=\dfrac{(1\times 5)+3}{5}= \dfrac {8} {5}\]

We also write $2\,\dfrac 3 {4} $ as an improper fraction:

\[ 2\,\dfrac 3 {4}=\dfrac{(2\times 4)+3}{4}= \dfrac {11} {4}\]

Now, we can multiply the two numbers.

\begin{align*} \dfrac{8}{\color{red}5}\times\dfrac {11} {\color{blue}4} \end{align*}

We can simplify this product by swapping the denominators:

\begin{align*} \dfrac{8}{\color{red}5}\times\dfrac {11} {\color{blue}4} &= \dfrac{8}{\color{blue}4}\times\dfrac {11} {\color{red}5} \\[5pt] &= 2\times\dfrac {11} {5} \\[5pt] &=\dfrac{22}{5} \end{align*}

Therefore, the missing number is $22.$

First, we write $2\,\dfrac 45$ as an improper fraction: \[ 2\,\dfrac 45=\dfrac{(2\times 5)+4}{5}= \dfrac {14} {5}\]

We also write $2\,\dfrac 12 $ as an improper fraction: \[ 2\,\dfrac 12=\dfrac{(2\times 2)+1}{2}= \dfrac {5}{2}\]

Now, we can multiply the two numbers. \begin{align*} \dfrac{14}{\color{red}5}\times\dfrac {5} {\color{blue}2} \end{align*}

We can simplify this product by swapping the denominators: \begin{align*} \dfrac{14}{\color{red}5}\times\dfrac {5} {\color{blue}2} &= \dfrac{14}{\color{blue}2}\times\dfrac {5} {\color{red}5} \\[5pt] &= 7\times1 \\[5pt] &=7 \end{align*}

Therefore, we conclude that \[ 2\,\dfrac 4 {5} \times 2\,\dfrac 1 {2} = \bbox[4pt,border: 1px solid lightgray]{7}. \]

First, we write $4\,\dfrac 12$ as an improper fraction: \[ 4\,\dfrac 12=\dfrac{(4\times 2)+1}{2}= \dfrac {9} {2}\]

We also write $1\,\dfrac 13 $ as an improper fraction: \[ 1\,\dfrac 13=\dfrac{(1\times 3)+1}{3}= \dfrac {4}{3}\]

Now, we can multiply the two numbers. \begin{align*} \dfrac{9}{\color{red}2}\times\dfrac {4} {\color{blue}3} \end{align*}

We can simplify this product by swapping the denominators: \begin{align*} \dfrac{9}{\color{red}2}\times\dfrac {4} {\color{blue}3} &= \dfrac{9}{\color{blue}3}\times\dfrac {4} {\color{red}2} \\[5pt] &= 3\times 2 \\[5pt] &=6 \end{align*}

Therefore, we conclude that \[ 4\,\dfrac 1 {2} \times 1\,\dfrac 1 {3} = \bbox[4pt,border: 1px solid lightgray]{6}. \]

First, we write $3\,\dfrac 12 $ as an improper fraction: \[ 3\,\dfrac 12=\dfrac{(3\times 2)+1}{2}= \dfrac{7}{2}\]

We also write $1\,\dfrac 2 {3} $ as an improper fraction: \[ 1\,\dfrac 2{3}=\dfrac{(1\times 3)+2}{3}= \dfrac{5}{3}\]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac {7}2 \times \dfrac{5}{3} = \dfrac {7\times 5} {2\times 3} = \dfrac{35}{6} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{35}{6} =5\,\textrm{R} 5 = \bbox[4pt,border: 1px solid lightgray]{5\,\dfrac 5{6}}\]

First, we write $2\,\dfrac 1{3} $ as an improper fraction:

\[ 2\,\dfrac 1 {3}=\dfrac{(2\times 3)+1}{3}= \dfrac{7}{3}\]

We also write $1\,\dfrac 3 {4} $ as an improper fraction:

\[ 1\,\dfrac 3 {4}=\dfrac{(1\times 4)+3}{4}= \dfrac{7}{4}\]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac 7 {3} \times \dfrac{7}{4} = \dfrac {7\times 7} {3\times 4} = \dfrac{49}{12} \]

Finally, we write the resulting improper fraction as a mixed number:

\[ \dfrac{49}{12} =4\,\textrm{R} 1 = 4\,\dfrac 1 {12}\]

First, we write $1\,\dfrac 1{5} $ as an improper fraction:

\[ 1\,\dfrac 1 {5}=\dfrac{(1\times 5)+1}{5}= \dfrac{6}{5}\]

We also write $1\,\dfrac 1 {4} $ as an improper fraction:

\[ 1\,\dfrac 1 {4}=\dfrac{(1\times 4)+1}{4}= \dfrac{5}{4}\]

Now, we can multiply the two numbers.

\begin{align*} \dfrac{6}{\color{red}5}\times\dfrac {5} {\color{blue}4} \end{align*}

We can simplify this product by swapping the denominators:

\begin{align*} \dfrac{6}{\color{red}5}\times\dfrac {5} {\color{blue}4} &= \dfrac{6}{\color{blue}4}\times\dfrac {5} {\color{red}5} \\[5pt] &= \dfrac 64 \times1 \\[5pt] &=\dfrac64 \end{align*}

Next, we write this fraction in its lowest terms:

\begin{align*} \dfrac64 = \dfrac{6\div 2}{4\div 2} = \dfrac32 \end{align*}

Finally, we write the resulting improper fraction as a mixed number:

\[ \dfrac{3}{2} =1\,\textrm{R} 1 = \bbox[3pt,Gainsboro]{\color{blue}1\,\dfrac 12} \]

First, we write $2\,\dfrac 2{11}$ as an improper fraction: \[ 2\,\dfrac2{11}=\dfrac{(2\times 11)+2}{11}= \dfrac{24}{11}\]

We also write $1\,\dfrac 1 {10} $ as an improper fraction: \[ 1\,\dfrac 1 {10}=\dfrac{(1\times 10)+1}{10}= \dfrac{11}{10}\]

Now, we can multiply the two numbers. \begin{align*} \dfrac{24}{\color{red}11}\times\dfrac {11} {\color{blue}10} \end{align*}

We can simplify this product by swapping the denominators: \begin{align*} \dfrac{24}{\color{red}11}\times\dfrac {11} {\color{blue}10} &= \dfrac{24}{\color{blue}10}\times\dfrac {11} {\color{red}11} \\[5pt] &= \dfrac{24}{10} \times 1 \\[5pt] &=\dfrac{24}{10} \end{align*}

Next, we write this fraction in its lowest terms: \[ \dfrac{24}{10} = \dfrac{24 \div 2}{10\div 2} = \dfrac{12}{5} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{12}{5} =2\,\textrm{R} 2 = \bbox[4pt,border: 1px solid lightgray]{2\,\dfrac 25} \]

First, we write $2\,\dfrac 47$ as an improper fraction: \[ 2\,\dfrac47=\dfrac{(2\times 7)+4}{7}= \dfrac{18}{7}\]

We also write $1\,\dfrac 5{6} $ as an improper fraction: \[ 1\,\dfrac 5 {6}=\dfrac{(1\times 6)+5}{6}= \dfrac{11}{6}\]

Now, we can multiply the two numbers. \begin{align*} \dfrac{18}{\color{red}7}\times\dfrac {11} {\color{blue}6} \end{align*}

We can simplify this product by swapping the denominators: \begin{align*} \dfrac{18}{\color{red}7}\times\dfrac {11} {\color{blue}6} &= \dfrac{18}{\color{blue}6}\times\dfrac {11} {\color{red}7} \\[5pt] &= 3 \times \dfrac{11}{7} \\[5pt] &=\dfrac{33}{7} \end{align*}

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{33}{7} =4\,\textrm{R} 5 = \bbox[4pt,border: 1px solid lightgray]{4\,\dfrac 57} \]