Quy đồng mẫu số A và B

$A=\frac{\left( {{100}^{100}}+1 \right)\left( {{100}^{68}}+1 \right)}{\left( {{100}^{99}}+1 \right)\left( {{100}^{68}}+1 \right)}$

$B=\frac{\left( {{100}^{69}}+1 \right)\left( {{100}^{99}}+1 \right)}{\left( {{100}^{99}}+1 \right)\left( {{100}^{68}}+1 \right)}$

Xét hiệu :

$\left( {{100}^{100}}+1 \right)\left( {{100}^{68}}+1 \right)$ - $\left( {{100}^{69}}+1 \right)\left( {{100}^{99}}+1 \right)$

= $\left( {{100}^{168}}+{{100}^{68}}+{{100}^{100}}+1 \right)$ - $\left( {{100}^{168}}+{{100}^{99}}+{{100}^{69}}+1 \right)$

= ${{100}^{100}}+{{100}^{68}}-{{100}^{99}}-{{100}^{69}}$

=$\left( {{100}^{100}}-{{100}^{99}} \right)-\left( {{100}^{69}}-{{100}^{68}} \right)$

= ${{100}^{99}}(100-1)-{{100}^{68}}(100-1)$ >0

$\Rightarrow $ A > B