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a) ta có: \(1-\frac{2012}{2013}=\frac{1}{2013}\)
\(1-\frac{2013}{2014}=\frac{1}{2014}\)
mà \(\frac{1}{2013}>\frac{1}{2014}\) nên \(\frac{2013}{2014}>\frac{2012}{2013}\)
1) \(16^{2020}+\dfrac{1}{16^{2021}}+1\)
\(=16^{2021}\div16^{2020}+1\)
\(=16+1\)
\(=17\)
2) \(16^{2021}+\dfrac{1}{16^{2022}}+1\)
\(=16^{2022}\div16^{2021}+1\)
\(=16+1\)
= 17
Vì 17=17 nên \(16^{2020}+\dfrac{1}{16^{2021}}+1=16^{2021}+\dfrac{1}{16^{2022}}+1\)
Đặt \(A=\frac{2^{15}+1}{2^{16}+1}\)
\(\Rightarrow2A=\frac{2^{16}+2}{2^{16}+1}=\frac{2^{16}+1+1}{2^{16}+1}=1+\frac{1}{2^{16}+1}\)
Đặt \(B=\frac{2^{14}+1}{2^{15}+1}\)
\(\Rightarrow2B=\frac{2^{15}+2}{2^{15}+1}=\frac{2^{15}+1+1}{2^{15}+1}=1+\frac{1}{2^{15}+1}\)
Vì 216+1>215+1
\(\Rightarrow\frac{1}{2^{16}+1}< \frac{1}{2^{15}+1}\)
\(\Rightarrow1+\frac{1}{2^{16}+1}< 1+\frac{1}{2^{15}+1}\)
\(\Rightarrow2A< 2B\Rightarrow A< B\)
Vậy...
Lời giải:
\(A=\frac{98^{12}+1}{98^{13}+1}\\ 98A=\frac{98^{13}+98}{98^{13}+1}=1+\frac{97}{98^{13}+1}> 1+\frac{97}{98^{14}+1}=\frac{98^{14}+98}{98^{14}+1}=98.\frac{98^{13}+1}{98^{14}+1}=98B\)
$\Rightarrow A>B$