题目内容
(1)1.25×16
(2)3.15+2.64+1.36+6.85
(3)14.28-1.5×0.3-3.55
(4)1-2+3-4+5-6+7-8…+2001-2002+2003-2004+2005.
(2)3.15+2.64+1.36+6.85
(3)14.28-1.5×0.3-3.55
(4)1-2+3-4+5-6+7-8…+2001-2002+2003-2004+2005.
分析:(1)将1.25×16变形为1.25×8×2,运用乘法结合律,再计算即可求解.
(2)运用加法交换律和结合律来解决问题;
(3)首先计算乘法,然后运用减法的简便方法,减去两个数的和,即可得解;
(4)运用加法结合律化成1+(-2+3)+(-4+5)+(-6+7)+…+(-2002+2003)+(-2004+2005)=1+1+1+…+1,有(2005-1)÷2+1个1,据此得解.
(2)运用加法交换律和结合律来解决问题;
(3)首先计算乘法,然后运用减法的简便方法,减去两个数的和,即可得解;
(4)运用加法结合律化成1+(-2+3)+(-4+5)+(-6+7)+…+(-2002+2003)+(-2004+2005)=1+1+1+…+1,有(2005-1)÷2+1个1,据此得解.
解答:解:(1)1.25×16
=1.25×8×2
=10×2
=20
(2)3.15+2.64+1.36+6.85
=(3.15+6.85)+(2.64+1.36)
=10+4
=14
(3)14.28-1.5×0.3-3.55
=14.28-0.45-3.55
=14.28-(0.45+3.55)
=14.28-4
=10.28
(4)1-2+3-4+5-6+7-8…+2001-2002+2003-2004+2005
=1+(-2+3)+(-4+5)+(-6+7)+…+(-2002+2003)+(-2004+2005)
=1+1+1+…+1
=1+(2005-1)÷2
=1003.
=1.25×8×2
=10×2
=20
(2)3.15+2.64+1.36+6.85
=(3.15+6.85)+(2.64+1.36)
=10+4
=14
(3)14.28-1.5×0.3-3.55
=14.28-0.45-3.55
=14.28-(0.45+3.55)
=14.28-4
=10.28
(4)1-2+3-4+5-6+7-8…+2001-2002+2003-2004+2005
=1+(-2+3)+(-4+5)+(-6+7)+…+(-2002+2003)+(-2004+2005)
=1+1+1+…+1
=1+(2005-1)÷2
=1003.
点评:考查了小数乘法,注意灵活应用运算律简便计算.
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