考點:正、負(fù)數(shù)的運算
專題:運算順序及法則
分析:根據(jù)正、負(fù)數(shù)的加減運算方法,依次進(jìn)行解答即可;其中最后一個:把相鄰的兩個數(shù)相減,可得2500個-1.即可求解.
解答:
解:(1)(-16)+(-8)
=-(16+8)
=-24
(2)(-72)+(+63)
=-72+63
=-(72-63)
=-9
(3)(-
)+
+0
=-(
-
)
=-
(4)(-8)-(+4)+(-6)-(-1)
=-8-4-6+1
=-17
(5)
-
-
+
=
+
-
-
=1-
-
=-
(6)0-21
+(+3
)-(-
)-(+
)
=-21
+3
+
-
=-21
+
+3
-
=-21+3
=-18
(7)[1.4-(-3.6+5.2)-4.3]-(-1.5)
=[1.4-1.6-4.3]+1.5
=[-0.2-4.3]+1.5
=-4.5+1.5
=-3
(8)1-2+3-4+5-6+7-8+…+4999-5000
=(1-2)+(3-4)+(5-6)+(7-8)+…+(4999-5000)
=(-1)+(-1)+(-1)+(-1)+…+(-1)
=-2500.
點評:本題考查了正、負(fù)數(shù)的加減運算,正確理解運算法則是關(guān)鍵.