đề :

= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 ) 

=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)

=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)

=1/100 - ( 1- 1/100)

=1/100 - 99 /100

= -98/100

= -49 /50

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

=1.100/99.98/99.....2/1

=1.100

=100

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

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

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

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

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

\(=0\)

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

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

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

Chưa tính

ra nhé bạn

tử có 98 số 1 = 98.1

97 số 2= 98.2

.....

1 số 98 = 1.98 cộng tất cả lại bằng mẫu nên kết quả =1

\(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+98\right)}{1.98+2.97+3.96+...+97.2+98.1}\)
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.98+2.97+3.96+...+98.1}=1\)