Question

Tính bằng cách thuận tiện

\(\frac{2003\cdot2005+200}{2004\cdot2004+199}\)

Answer 1

(2004-1)(2004+1)+200/2004.2004+199

=2004^2-1+200/2004^2+199

=2004^2+199/2004^2+199

=1

Answer 2

\(\frac{2003.2005+200}{2004.2004+199}\)

= \(\frac{2003.\left(2004+1\right)+200}{\left(2003+1\right).2004+199}\)

= \(\frac{2003.2004+2003+200}{2003.2004+2004+199}\)

= \(\frac{2003.2004+2203}{2003.2004+2203}\)

= \(1\)

                 ~~~~Hok tốt~~~~

Answer 3

Dấu ^ có nghĩa là gì vậy bạn

Answer 4

\(\frac{2003.2005+200}{2004.2004+199}\)

\(=\frac{\left(2004-1\right).2005+200}{2004.\left(2005-1\right)+199}\)

\(=\frac{2005.2004-2005+200}{2004.2005-2004+199}\)

\(=\frac{2004.2005-\left(2005-200\right)}{2004.2005-\left(2004-199\right)}=1\)