Question

[Thiên Thần Bảo Bình]

\(\frac{2001x2002+2003x21+1981}{2002x2003-2001x2002}\)

Đề bài : Tính 

Answer 1

[Nguyễn Hưng Phát]

Ta có:

\(\frac{2001.2002+2003.21+1981}{2002.2003-2001.2002}=\frac{2001.2002+2002.21+21+1981}{2002.\left(2003-2001\right)}\)

=\(\frac{2002.\left(2001+21\right)+2002}{2002.2}=\frac{2002.2022+2002}{2002.2}\)

=\(\frac{2002.\left(2022+1\right)}{2002.2}=\frac{2002.2023}{2002.2}\)

=\(\frac{2023}{2}\)