Đáp án:
`A>B`
Giải thích các bước giải:
`A=\frac{19^{30}+5}{19^{31}+5}`
`=>19A=\frac{19(19^{30}+5)}{19^{31}+5}`
`=>19A=\frac{19^{31}+95}{19^{31}+5}`
`=>19A=\frac{(19^{31}+5)+90}{19^{31}+5}`
`=>19A=1+\frac{90}{19^{31}+5}(1)`
`B=\frac{19^{31}+5}{19^{32}+5}`
`=>19B=\frac{19(19^{31}+5)}{19^{32}+5}`
`=>19B=\frac{19^{32}+95}{19^{32}+5}`
`=>19B=\frac{(19^{32}+5)+90}{19^{32}+5}`
`=>19B=1+\frac{90}{19^{32}+5}(2)`
Vì `19^{32}+5>19^{31}+5`
`=>\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}(3)`
Từ `(1),(2),(3)`
`=>19A>19B=>A>B`
Vậy `A>B`
