Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1)\\
\dfrac{{17}}{5} > \dfrac{{15}}{5} = 3 = \dfrac{{12}}{4} > \dfrac{9}{4} \Rightarrow \dfrac{{17}}{5} > \dfrac{9}{4}\\
\dfrac{{26}}{6} > \dfrac{{24}}{6} = 4 = \dfrac{{36}}{9} > \dfrac{{29}}{9} \Rightarrow \dfrac{{26}}{6} > \dfrac{{29}}{9}\\
\dfrac{{63}}{7} = 9 = \dfrac{{99}}{{11}} > \dfrac{{93}}{{11}} \Rightarrow \dfrac{{63}}{7} > \dfrac{{93}}{{11}}\\
2)\\
*)\\
\dfrac{{46}}{9} - 5 = \dfrac{{46}}{9} - \dfrac{{45}}{9} = \dfrac{1}{9}\\
\dfrac{{36}}{7} - 5 = \dfrac{{36}}{7} - \dfrac{{35}}{7} = \dfrac{1}{7}\\
\dfrac{1}{9} < \dfrac{1}{7} \Rightarrow \dfrac{{46}}{9} - 5 < \dfrac{{36}}{7} - 5 \Rightarrow \dfrac{{46}}{9} < \dfrac{{36}}{7}\\
*)\\
\dfrac{{10}}{3} - 3 = \dfrac{{10}}{3} - \dfrac{9}{3} = \dfrac{1}{3}\\
\dfrac{{13}}{4} - 3 = \dfrac{{13}}{4} - \dfrac{{12}}{4} = \dfrac{1}{4}\\
\dfrac{1}{3} > \dfrac{1}{4} \Rightarrow \dfrac{{10}}{3} - 3 > \dfrac{{13}}{4} - 3 \Rightarrow \dfrac{{10}}{3} > \dfrac{{13}}{4}\\
*)\\
\dfrac{{22}}{5} - 4 = \dfrac{{22}}{5} - \dfrac{{20}}{5} = \dfrac{2}{5}\\
\dfrac{{26}}{6} - 4 = \dfrac{{26}}{6} - \dfrac{{24}}{6} = \dfrac{2}{6}\\
\dfrac{2}{5} > \dfrac{2}{6} \Rightarrow \dfrac{{22}}{5} - 4 > \dfrac{{26}}{6} - 4 \Rightarrow \dfrac{{22}}{5} > \dfrac{{26}}{6}\\
3)\\
*)\\
2 - \dfrac{{11}}{9} = \dfrac{{18}}{9} - \dfrac{{11}}{9} = \dfrac{7}{9}\\
2 - \dfrac{{13}}{{10}} = \dfrac{{20}}{{10}} - \dfrac{{13}}{{10}} = \dfrac{7}{{10}}\\
\dfrac{7}{9} > \dfrac{7}{{10}} \Rightarrow 2 - \dfrac{{11}}{9} > 2 - \dfrac{{13}}{{10}} \Rightarrow \dfrac{{11}}{9} < \dfrac{{13}}{{10}}\\
*)\\
\dfrac{{13}}{3} - 4 = \dfrac{{13}}{3} - \dfrac{{12}}{3} = \dfrac{1}{3}\\
\dfrac{{18}}{4} - 4 = \dfrac{{18}}{4} - \dfrac{{16}}{4} = \dfrac{2}{4} = \dfrac{1}{2}\\
\dfrac{1}{3} < \dfrac{1}{2} \Rightarrow \dfrac{{13}}{3} - 4 < \dfrac{{18}}{4} - 4 \Rightarrow \dfrac{{13}}{3} < \dfrac{{18}}{4}\\
*)\\
\dfrac{{17}}{5} - 3 = \dfrac{{17}}{5} - \dfrac{{15}}{5} = \dfrac{2}{5}\\
\dfrac{{21}}{6} - 3 = \dfrac{{21}}{6} - \dfrac{{18}}{6} = \dfrac{3}{6} = \dfrac{1}{2} = \dfrac{2}{4}\\
\dfrac{2}{5} < \dfrac{2}{4} \Rightarrow \dfrac{{17}}{5} - 3 < \dfrac{{21}}{6} - 3 \Rightarrow \dfrac{{17}}{5} < \dfrac{{21}}{6}\\
4)\\
A = \dfrac{{{{100}^{90}} + 1}}{{{{100}^{80}} + 1}}\\
B = \dfrac{{{{100}^{89}} + 1}}{{{{100}^{79}} + 1}} = \dfrac{{100.\left( {{{100}^{89}} + 1} \right)}}{{100.\left( {{{100}^{79}} + 1} \right)}} = \dfrac{{{{100}^{90}} + 100}}{{{{100}^{80}} + 100}}\\
A - B = \dfrac{{{{100}^{90}} + 1}}{{{{100}^{80}} + 1}} - \dfrac{{{{100}^{90}} + 100}}{{{{100}^{80}} + 100}}\\
 = \dfrac{{\left( {{{100}^{90}} + 1} \right).\left( {{{100}^{80}} + 100} \right) - \left( {{{100}^{90}} + 100} \right).\left( {{{100}^{80}} + 1} \right)}}{{\left( {{{100}^{80}} + 1} \right).\left( {{{100}^{80}} + 100} \right)}}\\
 = \dfrac{{\left( {{{100}^{90}}{{.100}^{80}} + {{100.100}^{90}} + {{100}^{80}} + 100} \right) - \left( {{{100}^{90}}{{.100}^{80}} + {{100.100}^{80}} + {{100}^{90}} + 100} \right)}}{{\left( {{{100}^{80}} + 1} \right).\left( {{{100}^{80}} + 100} \right)}}\\
 = \dfrac{{{{100.100}^{90}} + {{100}^{80}} - {{100.100}^{80}} - {{100}^{90}}}}{{\left( {{{100}^{80}} + 1} \right).\left( {{{100}^{80}} + 100} \right)}}\\
 = \dfrac{{99.\left( {{{100}^{90}} - {{100}^{80}}} \right)}}{{\left( {{{100}^{80}} + 1} \right).\left( {{{100}^{80}} + 100} \right)}} > 0\\
 \Rightarrow A - B > 0 \Rightarrow A > B
\end{array}\)