Question

So sánh:

\(A=\frac{10^8+1}{10^9+1}\)với\(B=\frac{10^9+1}{10^{10}+1}\)

Answer 1

Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)

Ta có:

\(B=\frac{10^9+1}{10^{10}+1}< \frac{10^9+1+9}{10^{10}+1+9}\)

\(B< \frac{10^9+10}{10^{10}+10}\)

\(B< \frac{10.\left(10^8+1\right)}{10.\left(10^9+1\right)}\)

\(B< \frac{10^8+1}{10^9+1}=A\)

=> B < A

Answer 2

Ta có:

\(10A=\frac{10\left(10^8+1\right)}{10^9+1}=\frac{10^9+10}{10^9+1}=\frac{10^9+1+9}{10^9+1}=\frac{10^9+1}{10^9+1}+\frac{9}{10^9+1}=1+\frac{9}{10^9+1}\)

tương tự với B ta có:\(10B=1+\frac{9}{10^{10}+1}\)

Vì 10⁹+1<10¹⁰+1 \(\Rightarrow\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\)

\(\Rightarrow1+\frac{9}{10^9+1}>1+\frac{9}{10^{10}+1}\)

\(\Rightarrow10A>10B\Leftrightarrow A>B\)