Ta có :
\(\displaystyle {{A}} = {{{{10}^8} + 2} \over {{{10}^8} - 1}} = {{{{10}^8} - 1 + 3} \over {{{10}^8} - 1}} \)\(\displaystyle = {{{{10}^8} - 1} \over {{{10}^8} - 1}} + {3 \over {{{10}^8} - 1}} = 1{3 \over {{{10}^8} - 1}}\)
\(\displaystyle B = {{{{10}^8}} \over {{{10}^8} - 3}} = {{{{10}^8} - 3 + 3} \over {{{10}^8} - 3}} \)\(\displaystyle = {{{{10}^8} - 3} \over {{{10}^8} - 3}} + {3 \over {{{10}^8} - 3}} = 1{3 \over {{{10}^8} - 3}}\)
Ta có : \(\displaystyle {3 \over {{{10}^8} - 1}} < {3 \over {{{10}^8} - 3}}\) nên \(\displaystyle 1{3 \over {{{10}^8} - 1}} < 1{3 \over {{{10}^8} - 3}}\)
Vậy \(A < B.\)
