急需!!!!!!!!!! 六年级下册简便运算!!!!!!!!!!!!!!

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例题1答
计算:1234+2341+3412+4123
【思路导航】整体观察全式,可以发现题中的4个四位数均由数1,2,3,4组成,且4个数字在每个数位上各出现一次,于是有
原式=1×1111+2×1111+3×1111+4×1111
=(1+2+3+4)×1111
=10×1111
=11110
1.23456+34562+45623+56234+62345
= 2×11111+3×11111+4×11111+5×11111+6×11111
=(2+3+4+5+6)×11111
= 222220

2.45678+56784+67845+78456+84567
= 4×11111+5×11111+6×11111+7×11111+8×11111
=(4+5+6+7+8)×11111
= 333330

3.124.68+324.68+524.68+724.68+924.68
= 524.68×5
= 2623.4

例题2答
计算:2.8×23.4+11.1×57.6+6.54×28
【思路导航】我们可以先整体地分析算式的特点,然后进行一定的转化,创造条件运用乘法分配律来简算。所以
原式=2.8×23.4+2.8×65.4+11.1×8×7.2
=2.8×(23.4+65.4)+88.8× 7.2
=2.8×88.8+88.8×7.2
=88.8×(2.8+7.2)
=88.8×10
=888
1.99999×77778+33333×66666
= 99999×(77778+22222)
= 9999900000

2.34.5×76.5-345×6.42-123×1.45
= 34.5×(76.5-64.2)-12.3×14.5
= 34.5×12.3-12.3×14.5
= (34.5+14.5)×12.3
= 246

3.77×13+255×999+510
= 77×13+255×(999+2)
= 1001+255×1001
=(1+255)×1001
= 256256
例题3答
计算(1993×1994-1)/(1993+1992×1994)
【思路导航】仔细观察分子、分母中各数的特点,就会发现分子中1993×1994可变形为1992+1)×1994=1992×1994+1994,同时发现1994-1 = 1993,这样就可以把原式转化成分子与分母相同,从而简化运算。所以
原式=【(1992+1)×1994-1】/(1993+1992×1994)
=(1992×1994+1994-1)/(1993+1992×1994)
=1
练习3
1.(362+548×361)/(362×548-186)
= (362+548×361)/(361×548+548-186)
= (362+548×361)/(361×548+362)
= 1

2.(1988+1989×1987)/(1988×1989-1)
= (1988+1989×1987)/(1987×1989+1989-1)
= (1988+1989×1987)/(1987×1989+1988)
= 1

3.(204+584×1991)/(1992×584―380)―1/143
=(204+584×1991)/(1991×584+584―380)―1/143
= (204+584×1991)/(1991×584+204)―1/143
= 1―1/143
= 142/143
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