题目内容
递等式计算.(能巧算的要巧算) (1)(80-4)×125 |
(2)(29+29+29+49)×250 | ||||||||||
(3)96×(718-94) | (4)32×88-23×88+88 | ||||||||||
(5)124×25 | (6)125×12×25 | ||||||||||
(7)[332-(32+278)]×186 | (8)
| ||||||||||
(9)
|
(10)1-
|
分析:(1)(4)根据乘法分配律计算;
(2)求出括号里的和,再把分解成100+36再根据乘法分配律进行计算;
(3)先求括号里的差,再算乘;
(5)把124分解成100+20+4再根据乘法分配律进行计算;
(6)把12分解成3×4,再根据乘法结合律进行计算;
(7)根据减法的性质,先算中括号里的,最后算乘;
(8)根据同分母分数加减法的计算方法进行计算;
(9)从左到右依次计算;
(10)根据减法的性质进行计算.
(2)求出括号里的和,再把分解成100+36再根据乘法分配律进行计算;
(3)先求括号里的差,再算乘;
(5)把124分解成100+20+4再根据乘法分配律进行计算;
(6)把12分解成3×4,再根据乘法结合律进行计算;
(7)根据减法的性质,先算中括号里的,最后算乘;
(8)根据同分母分数加减法的计算方法进行计算;
(9)从左到右依次计算;
(10)根据减法的性质进行计算.
解答:解:(1)(80-4)×125,
=80×125-4×125,
=10000-500,
=9500;
(2)(29+29+29+49)×250,
=136×250,
=(100+36)×250,
=100×250+9×4×250,
=25000+9000,
=34000;
(3)96×(718-94),
=96×624,
=59904;
(4)32×88-23×88+88,
=(32-23+1)×88,
=10×88,
=880;
(5)124×25,
=(100+20+4)×25,
=100×25+20×25+4×25,
=2500+500+100,
=3100;
(6)125×12×25,
=125×3×4×25,
=(125×3)×(4×25),
=375×100,
=37500;
(7)[332-(32+278)]×186,
=(332-32-278)×186,
=22×186,
=4092;
(8)
+
+
=1;
(9)
+
-
,
=1-
,
=
;
(10)1-
-
,
=1-((
+
),
=1-1,
=0.
=80×125-4×125,
=10000-500,
=9500;
(2)(29+29+29+49)×250,
=136×250,
=(100+36)×250,
=100×250+9×4×250,
=25000+9000,
=34000;
(3)96×(718-94),
=96×624,
=59904;
(4)32×88-23×88+88,
=(32-23+1)×88,
=10×88,
=880;
(5)124×25,
=(100+20+4)×25,
=100×25+20×25+4×25,
=2500+500+100,
=3100;
(6)125×12×25,
=125×3×4×25,
=(125×3)×(4×25),
=375×100,
=37500;
(7)[332-(32+278)]×186,
=(332-32-278)×186,
=22×186,
=4092;
(8)
2 |
9 |
3 |
9 |
4 |
9 |
(9)
17 |
26 |
9 |
26 |
3 |
8 |
=1-
3 |
8 |
=
5 |
8 |
(10)1-
5 |
7 |
2 |
7 |
=1-((
5 |
7 |
2 |
7 |
=1-1,
=0.
点评:本题综全考查了学生的计算能力,在计算中要灵活运用简便算法.
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