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 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

=))

\(\frac{19^{30}+5}{19^{31}+5}\)và \(\frac{19^{31}+5}{19^{32}+5}\)

Xét biểu thức \(\frac{19^{30}+5}{19^{31}+5}\)là A và biểu thức \(\frac{19^{31}+5}{19^{32}+5}\)là B

\(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}\)

\(=\)\(\frac{19^{31}.19}{19^{32}.19}\)\(=\)\(\frac{19.\left(19^{30}+1\right)}{19.\left(19^{31}+1\right)}\)

\(=\)\(\frac{19^{30}+1}{19^{31}+1}\)

Mà \(\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}\)

Nên \(A>B\)

= nhau

mk chắc chắn cần gải ra bảo mk 

:3

\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Vì \(19^{31}+5< 19^{32}+5\) nên \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\) \(\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\)

Do đó \(M>N\)

Ta có :

\(N=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19.\left(19^{30}+5\right)}{19.\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=M\)

=> N < M 

\(M=\frac{19^{30}+5}{19^{31}+5}\)

\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\)

\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

chung tử rồi so sánh mẫu đi

#)Giải :

\(M=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19M=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19N=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19M>19N\Rightarrow M>N\)

              #~Will~be~Pens~#

Ta có : \(N=\frac{19^{31}+5}{19^{32}+5}< 1\)

Áp dụng công thức \(\forall a,b,m\in N;b,m\inℕ^∗\)

\(\Rightarrow\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

Ta có :

\(N=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\cdot\left(19^{30}+5\right)}{19\cdot\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=M\)

Vậy N < M