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a) Ta có: 266 . 734 = 232 . 234 . 734 < (2.2.7)34 = 2834
Vậy 2834 > 266 . 734
Tương tự
a) \(A=7+7^2+...+7^{99}\)
\(7A=7^2+7^3+...+7^{100}\)
\(7A-A=7^2+7^3+...+7^{100}-7-7^2-...-7^{99}\)
\(6A=7^{100}-7\)
\(A=\frac{7^{100}-7}{6}\)
Mà 7100 > 7100 - 7 => A < \(\frac{7^{100}}{6}\)
b) \(A=7+7^2+...+7^{99}\)
\(A=\left(7+7^2+7^3\right)+...+\left(7^{97}+7^{98}+7^{99}\right)\)
\(A=\left(7+7^2+7^3\right)+...+7^{96}.\left(7+7^2+7^3\right)\)
\(A=399+...+7^{96}.399\)
\(A=399.\left(1+...+7^{96}\right)⋮19\left(đpcm\right)\)
\(\frac{A}{B}=\frac{7^{2013}+1}{7^{2014}+1}.\frac{7^{2015}+1}{7^{2014}+1}=\frac{7^{4028}+7^{2013}+7^{2015}+1}{7^{4028}+2.7^{2014}+1}=\)
\(=\frac{7^{4028}+7^{2013}\left(1+7^2\right)+1}{7^{4028}+2.7.7^{2013}+1}=\frac{7^{4028}+50.7^{2013}+1}{7^{4028}+14.7^{2013}+1}>1\)
\(\Rightarrow A>B\)
\(A=1+7+7^2+7^3+...+7^{100}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{101}\)
\(\Rightarrow7A-A=7^{101}-1\)
\(\Rightarrow6A=7^{101}-1< 7^{101}\)
Vậy : \(A< B\)
\(A=1+7+7^2+7^3+...+7^{100}\)
\(7A=7+7^2+7^3+7^4+....+7^{101}\)
\(7A-A=\left(7+7^2+7^3+...+7^{101}\right)-\left(1+7+7^2+....+7^{100}\right)\) \(6A=7^{101}-1\)
\(A=\dfrac{7^{101}-1}{6}< 7^{101}\)
\(\Rightarrow A< B\)
Vậy....................................................