1949×(1/51−1/2000)+51×(1/1949−1/2000)−2000×(1/1949+1/51)+3.

问题描述:

1949×(

1
51
1
2000
)+51×(
1
1949
1
2000
)−2000×(
1
1949
+
1
51
)+3.

1949×(

1
51
-
1
2000
)+51×(
1
1949
-
1
2000
)-2000×(
1
1949
+
1
51
)+3,
=
949
51
-
1949
2000
+
51
1949
-
51
2000
-
2000
1949
-
2000
51

=(
1949
51
-
2000
51
)+(
51
1949
-
2000
1949
)-(
1919
2000
+
51
1949
)+3,
=-1-1-1+3,
=0.