题目内容
(1)3
×2345+5555÷
+654.3×36
(2)
(3)1+
+
+
+
+
+
.
3 |
5 |
25 |
256 |
(2)
1×3×24+2×6×48+3×9×72 |
1×2×4+2×4×8+3×6×12 |
(3)1+
1 |
2 |
1 |
4 |
1 |
8 |
1 |
16 |
1 |
32 |
1 |
64 |
分析:(1)根据数字特点,把原式变为3.6×2345+5555×
+6543×3.6,运用乘法分配律简算,再把5555×
变为1111×8×
=8888×6.4,再次运用乘法分配律简算;
(2)通过观察,分数的分子与分母有一定特点:都含有相同的因式,把原式变为
,约分即可;
(3)通过观察,每个分数都可以拆成两个分数相减的形式,然后通过加减相抵消的方法,求得结果.
256 |
25 |
256 |
25 |
32 |
5 |
(2)通过观察,分数的分子与分母有一定特点:都含有相同的因式,把原式变为
1×3×24×(1+8+27) |
1×2×4×(1+8+9) |
(3)通过观察,每个分数都可以拆成两个分数相减的形式,然后通过加减相抵消的方法,求得结果.
解答:解:(1)3
×2345+5555÷
+654.3×36,
=3.6×2345+5555×
+6543×3.6,
=3.6×(2345+6543)+1111×8×
,
=3.6×8888+8888×6.4,
=(3.6+6.4)×8888,
=10×8888,
=88880;
(2)
,
=
,
=9;
(3)1+
+
+
+
+
+
,
=1+(1-
)+(
-
)+(
-
)+(
-
)+(
-
)+(
-
),
=1+(1-
),
=1+
,
=1
.
3 |
5 |
25 |
256 |
=3.6×2345+5555×
256 |
25 |
=3.6×(2345+6543)+1111×8×
32 |
5 |
=3.6×8888+8888×6.4,
=(3.6+6.4)×8888,
=10×8888,
=88880;
(2)
1×3×24+2×6×48+3×9×72 |
1×2×4+2×4×8+3×6×12 |
=
1×3×24×(1+8+27) |
1×2×4×(1+8+27) |
=9;
(3)1+
1 |
2 |
1 |
4 |
1 |
8 |
1 |
16 |
1 |
32 |
1 |
64 |
=1+(1-
1 |
2 |
1 |
2 |
1 |
4 |
1 |
4 |
1 |
8 |
1 |
8 |
1 |
16 |
1 |
16 |
1 |
32 |
1 |
32 |
1 |
64 |
=1+(1-
1 |
64 |
=1+
63 |
64 |
=1
63 |
64 |
点评:解答此题应仔细观察,根据题目特点,运用运算定律或运算技巧,灵活解答.
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