Question

[Lê Hoàng Tiến Đạt]

so sánh     A = \(\frac{17^{18}+1}{17^{19}+1}\)   B = \(\frac{17^{17}+1}{17^{18}+1}\)

Answer 1

[Bùi Thế Hào]

\(A=\frac{17^{18}+1}{17^{19}+1}\) <=> \(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)<=> \(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Nhận thấy: 17¹⁹+1 > 17¹⁸+1  => \(\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

=> 17B > 17A

=> B > A