Nếu nghĩ kĩ thì thấy bài này cũng đơn giản thôi.Thử xem cách giải của mk nè:
Giải: Ta có: A=\(\frac{17^{18}+1}{17^{19}+1}\) B=\(\frac{17^{17}+1}{17^{18}+1}\)
17A=\(\frac{17^{19}+17}{17^{19}+1}\) 17B=\(\frac{17^{18}+17}{17^{18}+1}\)
17A=\(\frac{\left(17^{19}+1\right)+16}{17^{19}+1}\) 17B=\(\frac{\left(17^{18}+1\right)+16}{17^{18}+1}\)
17A=\(\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\) 17B=\(\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)
17A=\(1+\frac{16}{17^{19}+1}\) 17B= \(1+\frac{16}{17^{18}+1}\)
Lại có: 1719+1>1718+1
Suy ra:\(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)
17A<17B
A<B
Vậy A<B
\(\text{Ta có:}\frac{17^{18}+1}{17^{19}+1}\)
\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)
\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)
\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)
\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)
\(\text{Vì }\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)
\(\Rightarrow17A< 17B\)
\(\Rightarrow A< B\)
