19A= \(\frac{19^{31}+95}{19^{31}+5}\) = \(\frac{19^{31}+95}{19^{31}}\)+\(\frac{19^{31}+95}{5}\)=95+ \(\frac{19^{31}+95}{5}\)
19B=\(\frac{19^{32}+95}{19^{32}+5}\)=\(\frac{19^{32}+95}{19^{32}}\)+\(\frac{19^{32}+95}{5}\)=95+\(\frac{19^{32}+95}{5}\)
vì \(\frac{19^{31}+95}{5}\)<\(\frac{19^{32}+95}{5}\)=> 19A<19B => A<B
Ta có:
\(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19A>19B\Rightarrow A>B\)