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Ta có
100 x 100 = 100 x (99 + 1) = 100 x 99 + 100
99 x 101 = 99 x (100 + 1) = 99 x 100 + 99
Vì 100 > 99 nên 100 x 100 > 99 x 101
101/99=1-2/99
81/79=1-2/79
mà -2/99>-2/79
nên 101/99>81/79
Tất nhiên là \(\frac{102}{97}>\frac{99}{101}\) rùi vì \(\frac{102}{97}>1\)và\(\frac{99}{101}< 1\) nên \(\frac{102}{97}>\frac{99}{101}\)
\(\frac{102}{97}>1\\ \frac{99}{101}< 1\\ \Rightarrow\frac{102}{97}>1>\frac{99}{101}\)
Chúc bạn học tốt!
\(\dfrac{1}{2022}\cdot A=\dfrac{2022^{100}+1}{2022^{100}+100}=1-\dfrac{99}{2022^{100}+100}\)
\(\dfrac{1}{2022}B=\dfrac{2022^{101}+1}{2022^{101}+100}=1-\dfrac{9}{2022^{101}+100}\)
2022^100+100<2022^101+100
=>-99/2022^100+100<-99/2022^101+100
=>A<B
Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
\(\Rightarrow2017A>2017B\Rightarrow A>B\)
Vậy...
Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
Hay \(2017A>2017B\)nên \(A>B\)
Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
Theo bài ra , ta có :
100 . 100 = 1002
99 . 101 = (99+1) . (101-1) < 100 . 100 = 1002
Vậy 100.100 > 99.101
100x100>99x101 bạn nhé