Question

[trần quang duy]

so sánh A và B biết:

A=\(\frac{10^{15}+1}{10^{16}+1}\)

B=\(\frac{10^{16}+1}{10^{17}+1}\)

Answer 1

[Đinh Đức Hùng]

Ta có :

\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)

Vì \(10^{16}+1< 10^{17}+1\) nên \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) \(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)

=> 10A > 10B Do đó A > B

Vậy A > B

Answer 2

[le thi minh thu]

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

Ta có:

\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{\left(10^{15}+1\right).10}{\left(10^{16}+1\right).10}=\frac{10^{16}+10}{10^{17}+10}=\frac{10^{16}+1+9}{10^{17}+1+9}\)

Vì \(B=\frac{10^{16}+1}{10^{17}+1}< 1\)

\(\Rightarrow B=\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=A\)

Vậy B < A