Vì \(\frac{19^{30}+5}{19^{31}+5}< 1\); \(\frac{19^{31}+5}{19^{30}+5}>1\)

Nên A  < B

1,  Vì A, B < 1

  \(\Rightarrow B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)

2, Đề là thế này?? \(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)

\(\Rightarrow C=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)

\(\Rightarrow C=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)

\(\Rightarrow C=\frac{\left(2+201\right).200}{4}=10150\)

A = \(\frac{5^{30}-2}{5^{31-2}}\) = 5

B = \(\frac{5^{31}-2}{5^{32}-2}\) = \(\frac{1}{5}\) = 0.2

Mà 5 > 0.2

Nên: A > B

\(5A=\frac{5^{31}-10}{5^{31}-2}=\frac{5^{31}-2-8}{5^{31}-2}=\frac{5^{31}-2}{5^{31}-2}-\frac{8}{5^{31}-2}=1-\frac{8}{5^{31}-2}\left(1\right)\)

\(5B=\frac{5^{32}-10}{5^{32}-2}=\frac{5^{32}-8}{5^{32}-2}=\frac{5^{32}-2}{5^{32}-2}-\frac{8}{5^{32}-2}=1-\frac{8}{5^{32}-2}\left(2\right)\)

từ (1) và (2)

=>A>B

k trước mới trả lời :D

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

\(\frac{19^{30}+5}{19^{31}+5}\)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 \(A>B\)