Question

Hãy so sánh:\(A=\frac{17^{201}+1}{17^{202}+1}vàB=\frac{17^{200}+1}{17^{201}+1}\)

Answer 1

\(A=\frac{17^{201}+1}{17^{202}+1}< 1\)

\(\rightarrow A=\frac{17^{201}+1}{17^{202}+1}< \frac{17^{201}+1+16}{17^{202}+1+16}\)

\(\rightarrow A=\frac{17^{201}+1}{17^{202}+1}< \frac{17^{201}+17}{17^{202}+17}\)

\(\rightarrow A=\frac{17^{201}+1}{17^{202}+1}< \frac{17\left(17^{200}+1\right)}{17\left(17^{201}+1\right)}\)

\(\rightarrow A=\frac{17^{201}+1}{17^{202}+1}< \frac{17^{200}+1}{17^{201}+1}\)

\(\rightarrow A=\frac{17^{201}+1}{17^{202}+1}< B\)

\(\rightarrow A< B\)

Answer 2

GIẢI GIÙM MÌNH NHA:THANKS:)