Tính hợp lý \(A = \left( {4\dfrac{5}{{17}} - 3\dfrac{4}{5} + 8\dfrac{{15}}{{29}}} \right) - \left( {3\dfrac{5}{{17}} - 6\dfrac{{14}}{{29}}} \right)\) ta được

$ - \dfrac{{21}}{4}$   

\( - \dfrac{{11}}{4}\)  

\( - \dfrac{{10}}{4}\)

\( - \dfrac{5}{4}\)

$\dfrac{{19.20}}{{19 + 20}} = \dfrac{1}{{19}} + \dfrac{1}{{20}}$             

\(6\dfrac{{23}}{{11}} = \dfrac{{6.23 + 11}}{{11}}\)  

\(a\dfrac{a}{{99}} = \dfrac{{100a}}{{99}}\left( {a \in {N^*}} \right)\)

\(1\dfrac{{15}}{{23}} = \dfrac{{1.23}}{{15}}\)

\( - \dfrac{{31}}{{24}} <  - 1\dfrac{1}{{12}} < \dfrac{{12}}{{100}} < 5\dfrac{1}{2} < 23\% \)           

\( - \dfrac{{31}}{{24}} <  - 1\dfrac{1}{{12}} < 23\%  < \dfrac{{12}}{{100}} < 5\dfrac{1}{2}\)

\( - \dfrac{{31}}{{24}} <  - 1\dfrac{1}{{12}} < \dfrac{{12}}{{100}} < 23\%  < 5\dfrac{1}{2}\)             

\( - 1\dfrac{1}{{12}} <  - \dfrac{{31}}{{24}} < \dfrac{{12}}{{100}} < 23\%  < 5\dfrac{1}{2}\)

$\dfrac{{11}}{6}$ 

\(\dfrac{7}{6}\)  

\(\dfrac{{13}}{6}\)     

\( - \dfrac{5}{6}\)

$1\dfrac{1}{7}$ 

\(\dfrac{2}{7}\)  

\(\dfrac{6}{7}\)

\(\dfrac{7}{8}\)

$ - \dfrac{5}{6}$                  

\(0\)          

\( - \dfrac{6}{5}\)

\(1\)