Question

So sánh

A=97^98+1/ 97^99+1 va B=97^97+1/ 97^98+1

Answer 1

Ta có: \(A=\frac{97^{98}+1}{97^{99}+1}\Rightarrow97A=\frac{97^{99}+97}{97^{99}+1}=\frac{97^{99}+1+96}{97^{99}+1}=1+\frac{96}{97^{99}+1}\)

\(B=\frac{97^{97}+1}{97^{98}+1}\Rightarrow97B=\frac{97^{98}+97}{97^{98}+1}=\frac{97^{98}+1+96}{97^{98}+1}=1+\frac{96}{97^{98}+1}\)

Vì \(\frac{96}{97^{99}+1}< \frac{96}{97^{98}+1}\Rightarrow1+\frac{96}{97^{99}+1}< 1+\frac{96}{97^{98}+1}\Rightarrow97A< 97B\Rightarrow A< B\)

Vậy A < B

Answer 2

Ta thấy A < 1 và 96 > 1 nên ta có:

            A <     97⁹⁸ + 1 + 96 / 97⁹⁹ + 1 + 96

      =>  A <     97⁹⁸ + 97 / 97⁹⁹ + 97

      =>  A <     97(97⁹⁷ + 1) / 97(97⁹⁸ + 1)

      =>  A <     97⁹⁷ + 1 / 97⁹⁸ + 1 = B

      => A < B

Answer 3

A<B

nhanh và gọn, lẹ