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

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

A<B

nhanh và gọn, lẹ

tính B

\(1+\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}\)

\(=1+1+\frac{1}{98}-\left(1+\frac{1}{97}\right)+\frac{1}{97}-\frac{1}{98}\)

\(=1+1+\frac{1}{98}-1-\frac{1}{97}+\frac{1}{97}-\frac{1}{98}\)

\(=1+1-1\)

\(=1\)

thank

a)=7x(1/4+1/28+1/70+1/130)

  =7x(1/1x4+1/4x7+1/7x10+1/10x13)

  =7x(1-1/4+1/4-1/7+1/7-1/10+1/10-1/13)

  =7x(1+1/13)

  =7x14/13=98/13

Các câu còn lại p dựa theo cách trên làm nốt nhé!!!

\(\frac{99}{98}-\frac{98}{97}+\frac{1}{97\times98}\)

\(=\left(1+\frac{1}{98}\right)-\left(1+\frac{1}{97}\right)+\frac{1}{97\times98}\)

\(=\frac{1}{98}-\frac{1}{97}+\frac{1}{97\times98}\)

\(=\frac{-1}{97\times98}+\frac{1}{97\times98}\)

\(=0\)

\(\frac{99}{98}-\frac{98}{97}+\frac{1}{97X98}=\left(1+\frac{1}{98}\right)-\left(1+\frac{1}{97}\right)+\frac{1}{97X98}\)

\(=\frac{1}{97X98}-\frac{1}{97X98}=0\)

Nguyễn tuấn thảo ơi người ta mới lớp 5 bạn thêm -1 vào chi ? :)

Nhanh nhanh nha

nhanh nhanh nha

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