a. Có: \(\frac{100^{101}+1}{100^{100}+1}>1\Rightarrow\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+\left(1+99\right)}{100^{100}+\left(1+99\right)}\)
\(\Rightarrow B>\frac{100^{101}+100}{100^{100}+100}\\ \Rightarrow B>\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\\ \Rightarrow B>\frac{100^{100}+1}{100^{99}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
b. Có: \(\frac{13^{16}+1}{13^{17}+1}< 0\Rightarrow\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+\left(1+12\right)}{13^{17}+\left(1+12\right)}\)
\(\Rightarrow B< \frac{13^{16}+13}{13^{17}+13}\\ \Rightarrow B< \frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\\ \Rightarrow B< \frac{13^{15}+1}{13^{16}+1}=A\\ \Leftrightarrow A>B\)
Vậy A > B
c. Có: \(\frac{1999^{2000}+1}{1999^{1999}+1}>1\Rightarrow\frac{1999^{2000}+1}{1999^{1999}+1}>\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\)
\(\Rightarrow B>\frac{1999^{2000}+1999}{1999^{1999}+1999}\\ \Rightarrow B>\frac{1999\left(1999^{1999}+1\right)}{1999\left(1999^{1998}+1\right)}\\ \Rightarrow B>\frac{1999^{1999}+1}{1999^{1998}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
