-2000 and 5 / 6) plus (- 1999 and 2 / 3) plus 4000 and 1 / 6 plus 8 and 1 / 3 =?

-2000 and 5 / 6) plus (- 1999 and 2 / 3) plus 4000 and 1 / 6 plus 8 and 1 / 3 =?

Original formula = (- 2000 5 / 6) + (- 1999 2 / 3) + 4000 1 / 6 + 8 1 / 3
=(4000 and 1 / 6 + 8 and 1 / 3) - (2000 and 5 / 6 + 1999 and 2 / 3)
=4008 and 1 / 2-4000 and 1 / 2
=8

(- 2000 and 5 / 6) + (- 1999 and 2 / 3) + 4000 and 2 / 3 + (- 1 and 1 / 2), using the demolition method, the formula is also written down

(- 2000 and 5 / 6) + (- 1999 and 2 / 3) + 4000 and 2 / 3 + (- 1 and 1 / 2),
=（-2001-2000+4001-1）+1/6+1/3-1/3-1/2
=-1+1/6-1/2
=-1 and 1 / 3

(2000 5 / 6) + (- 1999 2 / 3) + 4000 2 / 3 + (- 1 1 / 2)
How to do this problem?

(2000 5 / 6) + (- 1999 2 / 3) + 4000 1 2 / 3 + (- 1 / 2)
=2000 + 5 / 6 + (- 1999-2 / 3) + 4000 + 2 / 3-1 / 2
=1 + 5 / 6-2 / 3 + 4000 + 2 / 3-
=4000 and one in two

(- 2000 5 / 6) + (- 1999 2 / 3) + 4000 1 / 6 + 8 1 / 3 = how much

(- 2000 5 / 6) + (- 1999 2 / 3) + 4000 1 / 6 + 8 1 / 3
=-2000 5 / 6 + 4000 1 / 6 - 1999 2 / 3 + 8 1 / 3
=2000-2 / 3-1991-1 / 3
=2000-1992
=8