1) A = 1997¹⁹⁹⁹ - 1997¹⁹⁹⁸

=> A = 1997¹⁹⁹⁸.(1997-1)

=> A = 1997¹⁹⁹⁸ . 1996

Vậy a chia hết cho 4 (vì 1996 chia hết cho 4)

2) B = 1997¹⁹⁹⁸ - 1998¹⁹⁹⁹ 

Mà 1997¹⁹⁹⁸ là số lẻ; 1998¹⁹⁹⁹

=> 1997¹⁹⁹⁸ - 1998¹⁹⁹⁹  là số lẻ

Vậy đề bài sai.

Bạn lưu ý lần sau gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn nhé.

Lời giải:

\(A=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{1998-1997}{1997.1998}\\ =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{1997}-\frac{1}{1998}\\ =(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{1997})-(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{1998})\\ =(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1997}+\frac{1}{1998})-2(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{1998})\\ =(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1997}+\frac{1}{1998})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{999})\\ =\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{1998}\)

\(2998B=\frac{1000+1998}{1000.1998}+\frac{1001+1997}{1001.1997}+...+\frac{1998+1000}{1998.1000}\\ =\frac{1}{1998}+\frac{1}{1000}+\frac{1}{1997}+\frac{1}{1001}+....+\frac{1}{1000}+\frac{1}{1998}\\ =(\frac{1}{1998}+\frac{1}{1997}+...+\frac{1}{1000})+(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{1998})\\ =2(\frac{1}{1000}+\frac{1}{1001}+...+\frac{1}{1998})\\ \Rightarrow B=\frac{1}{1499}(\frac{1}{1000}+\frac{1}{1001}+....+\frac{1}{1998})=\frac{1}{1499}A\)

$\Rightarrow A:B=1499$ là số nguyên.

1998^2-1997(1998+1)

=1998^2-(1998-1)(1998+1)

=1998^2-(1998^2-1)

=1998^2-1998^2+1

=1

bạn theo link: http://h.vn/hoi-dap/question/161130.html

-1/2000

1997, 1997 + 1998, 1998 + 1999, 1999

= 1997 x 1,0001 + 1998 x 1,0001 + 1999 x 1,0001

= (1997 + 1998 + 1999) x 1,0001

= 5994 x 1,0001 ( nhân 1 tổng với 1 số)

(1999×1998+1998+1997)×(1/1+1/2:3/2-4/3)

(1999×1998+1998+1997)×(1/1+1/2×2/3-4/3)

1999×1998+1998+1997)×(1/1+1/3-4/3)

(1999×1998+1998+1997)×(4/3-4/3)

(1999×1998+1998+1997)×0

0

bằng 0 nha bn

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(\dfrac{4}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times0\)

\(=0\)