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

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

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

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

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

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

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

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 

to lam ko biết là đúng hay sai đây đấy

bỏ hai số 5 nằm ở  2 mẫu số 

 ta có biểu thức 1

(19^30+5).(19^32)/19^31.19^32  

= (19^30+5).(19^31.19)/19^31.19^32

biểu thức 2

(19^31+5).19^31/19^31.19^32

=(19^30+5).(19.19^31)/19^31.19^32

suy ra  bằng nhau

M= \(\frac{1}{19}\)

N= \(\frac{1}{19}\)

=> M=N

Tính 19M và 19N rồi so sánh

M>N nhé

k trước mới trả lời :D

= nhau

mk chắc chắn cần gải ra bảo mk 

:3

Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}.19}{19^{32}.19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)Vậy A > B

Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+19}{19^{32}+19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)

Vậy A > B

vậy thì phải là <

bằng nhau khong tin bấm máy tính

bằng nhau

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

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

Ta thấy \(19A>19B\) nên A > B

Ta có \(A=\frac{19^{30}+5}{19^{31}+5}\)

Suy ra \(19A=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

Ta có \(B=\frac{19^{31}+5}{19^{32}+5}\)

Suy ra  \(19B=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

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

Do đó \(19A>19B\Rightarrow A>B\)

Vậy A > B

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

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

Do \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\)

Nên \(1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\)

Hay 19A>19B 

Suy ra A>B 

Vậy A>B