đề có pải là A=\(\frac{19^{30}+5}{19^{31}+5}\)  ; B=\(\frac{19^{31}+5}{19^{32}+5}\) PẢI KO BẠN

\(A=\frac{19^{30}+5}{19^{31}+5}=>19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\left(1\right)\)

\(B=\frac{19^{31}+5}{19^{32}+5}=>19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\left(2\right)\)

từ (1) and (2)

=>19A>19B

=>A>B

C = 19³⁰+5/19³¹+5

=>19C = 19³¹+95/19³¹+5 = 1+ [90/19³¹+5]

D = 19³¹+5/19³²+5

=>19D = 19³²+95/19³²+5 = 1 + [90/19³²+5]

ma 90/19³¹+5 > 90/19³²+5

=>19C > 19D

=>C > D

Ta có: \(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19.\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

    \(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19.\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Nên  \(19A< 19B\Rightarrow A< B\)

Nhầm: Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow A>B\)

Ta có 19³⁰<19³¹ 
         \(\left(\frac{5}{19}\right)^{31}< \left(\frac{5}{19}\right)^{32}\)
          5=5
công vế theo vế ta có
\(19^{30}+\left(\frac{5}{19}\right)^{31}+5< 19^{31}+\left(\frac{5}{19}\right)^{32}+5\)
Vậy A<B

=))

\(a,\frac{-8}{15}=\frac{-8.12}{15.12}=\frac{-96}{180}\left(1\right)\)

\(\frac{7}{12}=\frac{7.15}{12.15}=\frac{105}{180}\left(2\right)\)

Từ 1 và 2 \(\Rightarrow\frac{-8}{15}< \frac{7}{12}\)

\(b,\frac{13}{19}=\frac{13.53}{19.53}=\frac{689}{1007}\left(1\right)\)

\(\frac{47}{53}=\frac{47.19}{53.19}=\frac{893}{1007}\left(2\right)\)

Từ 1 và 2 \(\Rightarrow\frac{13}{19}< \frac{47}{53}\)

\(\frac{19^{31}}{19^{30}}=\frac{19^{30}.19}{19^{30}}=19\)

\(\frac{19^{32}}{19^{31}}=\frac{19^{31}.19}{19^{31}}=19\)

\(\Rightarrow\frac{19^{31}}{19^{30}}=\frac{19^{32}}{19^{31}}\)