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а) \(\raisebox{-1em}{\)\dfrac{1}{2} + \dfrac{3}{8} = \(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{4} ,\ \hline8\ \end{array} \(}}\) [ ] \(\raisebox{-1em}{\)+\(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{3} ,\ \hline8\ \end{array} \(}}\) [ ] \(\raisebox{-1em}{\)= \(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{7} ,\ \hline8\ \end{array} \(}}\) [ ];

б) \(\raisebox{-1em}{\)\dfrac{1}{3} + \dfrac{1}{6} = \(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{2} ,\ \hline6\ \end{array} \(}}\) [ ] \(\raisebox{-1em}{\)+ \(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{2} ,\ \hline6\ \end{array} \(}}\) [ ] \(\raisebox{-1em}{\)=\(}\) \(\mathrlap{\,\raisebox{-1em}{\)\begin{array}{c}, \phantom{3} ,\ \hline6\ \end{array} \(}}\) [ ];

в) \(\dfrac{1}{4} + \dfrac{5}{12} = \) [ ] \(+\) [ ] \(=\) [ ];

г) \(\dfrac{2}{5} + \dfrac{3}{10} = \) [ ] \(+\) [ ] \(=\) [ ];

д) \(\dfrac{1}{2} + \dfrac{2}{5} = \) [ ] \(+\) [ ] \(=\) [ ];

е) \(\dfrac{1}{3} + \dfrac{1}{4} = \) [ ] \(+\) [ ] \(=\) [ ].