so sánh 100^2015 4/ 100^2005 4

Lời giải:

Đặt \(A = \frac{{{{100}^{2015}} + 1}}{{{{100}^{2005}} + 1}}\)

Nên \(\frac{A}{{{{100}^{10}}}} = \frac{{{{100}^{2015}} + 1}}{{{{100}^{2015}} + {{100}^{10}}}} = \frac{{{{100}^{2015}} + {{100}^{10}} - 999}}{{{{100}^{2015}} + {{100}^{10}}}} = 1 - \frac{{999}}{{{{100}^{2015}} + {{100}^{10}}}}\)

Đặt \(B = \frac{{{{100}^{2016}} + 1}}{{{{100}^{2006}} + 1}}\) nên \(\frac{B}{{{{100}^{10}}}} = \frac{{{{100}^{2016}} + {{100}^{10}} - 999}}{{{{100}^{2016}} + {{100}^{10}}}} = 1 - \frac{{999}}{{{{100}^{2016}} + {{100}^{10}}}}\)

Ta thấy \(1 - \frac{{999}}{{{{100}^{2015}} + {{100}^{10}}}} < 1 - \frac{{999}}{{{{100}^{2016}} + {{100}^{10}}}}\)

Do đó A < B