题目内容
计算:
32×25×125 | 484-286-14 | 1+2+3+4+5+…+99+100 |
9+99+999+9999 | 2356-(1356-721) | 6×(120-200÷25) |
分析:(1)先把32分解成8×4,再运用乘法结合律简算;
(2)根据减法的性质简算;
(3)根据高斯求和的方法计算;
(4)先把9,99,999,9999,化成一个数减1,然后再运用加法结合律简算;
(5)先去括号,再计算,注意运算符号的变化;
(6)先算小括号里面的除法,再算小括号里面的减法,最后算括号外的乘法.
(2)根据减法的性质简算;
(3)根据高斯求和的方法计算;
(4)先把9,99,999,9999,化成一个数减1,然后再运用加法结合律简算;
(5)先去括号,再计算,注意运算符号的变化;
(6)先算小括号里面的除法,再算小括号里面的减法,最后算括号外的乘法.
解答:解:(1)32×25×125,
=8×4×25×125,
=(8×125)×(4×25),
=1000×100,
=100000;
(2)484-286-14,
=484-(286+14),
=484-300,
=184;
(3)1+2+3+4+5+…+99+100,
=(1+100)+(2+99)+…+(49+52)+(50+51),
=101×50,
=5050;
(4)9+99+999+9999,
=(10-1)+(100-1)+(1000-1)+(10000-1),
=(10+100+1000+10000)-(1+1+1+1),
=11110-4,
=11106;
(5)2356-(1356-721),
=2356-1356+721,
=1000+721,
=1721;
(6)6×(120-200÷25),
=6×(120-8),
=6×112,
=672.
=8×4×25×125,
=(8×125)×(4×25),
=1000×100,
=100000;
(2)484-286-14,
=484-(286+14),
=484-300,
=184;
(3)1+2+3+4+5+…+99+100,
=(1+100)+(2+99)+…+(49+52)+(50+51),
=101×50,
=5050;
(4)9+99+999+9999,
=(10-1)+(100-1)+(1000-1)+(10000-1),
=(10+100+1000+10000)-(1+1+1+1),
=11110-4,
=11106;
(5)2356-(1356-721),
=2356-1356+721,
=1000+721,
=1721;
(6)6×(120-200÷25),
=6×(120-8),
=6×112,
=672.
点评:此题是考查四则混合运算,要仔细观察算式的特点,灵活运用一些定律进行简便计算.
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