递等式计算,能简便的用简便方法.
×+×××1649××12+(+)×12×35(++)×48.
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解:(1)×+×,
=()×,
=1×,
=;
(2)××16,
=×16×,
=10×,
=2;
(3)49×,
=48×+1×,
=11+,
=11;
(4)×12+,
=12×(),
=12×1,
=12;
(5)(+)×12×35,
=×12×35+×12×35,
=9×35+12×7,
=315+84,
=499;
(6)(++)×48,
=×48+×48+×48,
=42+8+32,
=82.
解析分析:(1)(3)(4)(5)(6)运用乘法分配律解答,(2)运用乘法交换律解答.
点评:完成本题要注意分析式中数据,运用合适的简便方法计算.