計算題:
(1)-24÷(-14)×(-7);
(2)3×(-9)-(-15)×(-3)-3×18;
(3)(5a2-3b2)-[(a2-2ab-b2)-(5a2-2ab-3b2)];
分析:(1)先算乘方,再把除法變?yōu)槌朔ㄓ嬎慵纯桑?br />(2)根據乘法的分配律的逆運算,可以簡化計算;
(3)先去小括號,再去中括號,最后合并同類項.
解答:解:(1)-2
4÷(-14)×(-7)=-16×(-
)×(-7)=-8;
(2)3×(-9)-(-15)×(-3)-3×18
=3×(-9-15-18)
=3×(-42)
=-126;
(3)(5a
2-3b
2)-[(a
2-2ab-b
2)-(5a
2-2ab-3b
2)]
=5a
2-3b
2-[a
2-2ab-b
2-5a
2+2ab+3b
2]
=5a
2-3b
2-a
2+2ab+b
2+5a
2-2ab-3b
2
=9a
2-5b
2.
點評:(1)解決整式加減混合運算的關鍵是熟記去括號法則,熟練運用合并同類項的法則,這是各地中考的常考點.
(2)有理數混合運算的順序,有括號先算括號里面的,再按照先乘方,后乘除,最后算加減的順序進行計算.注意運用運算律簡化計算.