Question

Cho:

\(A=\frac{10^{19}+1}{10^{20}+1}\)

\(B=\frac{10^{20}+1}{10^{21}+1}\)

So sánh A và B

Ai làm nhanh mà mik tick cho!!!

Answer 1

Ta có: \(A=\frac{10^{19}+1}{10^{20}+1}\Rightarrow10A=\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)

\(B=\frac{10^{20}+1}{10^{21}+1}\Rightarrow10B=\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)

Vì \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\Rightarrow1+\frac{9}{10^{201}+1}>1+\frac{9}{10^{21}+1}\Rightarrow10A>10B\)

\(\Rightarrow A>B\)

Vậy A > B

Answer 2

Ta có: nếu \(\frac{a}{b}< 1\) thì \(\frac{a}{b}< \frac{a+n}{b+n}\left(n\in N\right)\)

\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

Vậy A > B