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Question

so sánh: A=$\dfrac{10^{15}+1}{10^{16}+1}$ và B = $\dfrac{10^{16}+1}{10^{17}+1}$

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Phan Công Bằng · Answer

Có:$10A=\dfrac{10^{16}+10}{10^{16}+1}=\dfrac{10^{16}+1+9}{10^{16}+1}=\dfrac{10^{16}+1}{10^{16}+1}+\dfrac{9}{10^{16}+1}=1+\dfrac{9}{10^{16}+1}$

$10B=\dfrac{10^{17}+10}{10^{17}+1}=\dfrac{10^{17}+1+9}{10^{17}+1}=\dfrac{10^{17}+1}{10^{17}+1}+\dfrac{9}{10^{17}+1}=1+\dfrac{9}{10^{17}+1}$

$1+\dfrac{9}{10^{16}+1}>1+\dfrac{9}{10^{17}+1}\Rightarrow A>B$

Vậy $A>B$Câu trả lời: Có:$10A=\dfrac{10^{16}+10}{10^{16}+1}=\dfrac{10^{16}+1+9}{10^{16}+1}=\dfrac{10^{16}+1}{10^{16}+1}+\dfrac{9}{10^{16}+1}=1+\dfrac{9}{10^{16}+1}$

$10B=\dfrac{10^{17}+10}{10^{17}+1}=\dfrac{10^{17}+1+9}{10^{17}+1}=\dfrac{10^{17}+1}{10^{17}+1}+\dfrac{9}{10^{17}+1}=1+\dfrac{9}{10^{17}+1}$

$1+\dfrac{9}{10^{16}+1}>1+\dfrac{9}{10^{17}+1}\Rightarrow A>B$

Vậy $A>B$

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