题目内容
计算:
(1)1-3+5-7+9-11+…-1999+2001=
(2)69316.931÷69.31=
(1)1-3+5-7+9-11+…-1999+2001=
1001
1001
(2)69316.931÷69.31=
1000.1
1000.1
.分析:(1)根据题意,一共有1001个数,1后面的每两项进行结合得到(5-3)=2,(9-7)=2,…(2001-1999)=2,共有500个2,然后再进一步计算即可;
(2)把69316.931拆成69310+6.931=69.31×1000+69.31×0.1,然后再根据乘法分配律进一步计算即可.
(2)把69316.931拆成69310+6.931=69.31×1000+69.31×0.1,然后再根据乘法分配律进一步计算即可.
解答:解:
(1)1-3+5-7+9-11+…-1999+2001,
=1+(5-3)+(9-7)+…+(2001-1999).
=1+2×500,
=1+1000,
=1001;
(2)69316.931÷69.31,
=(69310+6.931)÷69.31,
=(69.31×1000+69.31×0.1)÷69.31,
=69.31×(1000+0.1)÷69.31,
=69.31×1000.1÷69.31,
=1000.1.
故答案为:1001,1000.1.
(1)1-3+5-7+9-11+…-1999+2001,
=1+(5-3)+(9-7)+…+(2001-1999).
=1+2×500,
=1+1000,
=1001;
(2)69316.931÷69.31,
=(69310+6.931)÷69.31,
=(69.31×1000+69.31×0.1)÷69.31,
=69.31×(1000+0.1)÷69.31,
=69.31×1000.1÷69.31,
=1000.1.
故答案为:1001,1000.1.
点评:根据题目给出的数据,找准使用的运算定律,然后再进一步计算即可.

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