9.2.3分式的混合运算

马鞍山市金瑞中学初一数学备课组一、复习：1.分式的乘法、除法、加法、减法法则。acbdcdabadbcdcabcdabababcccacadbcadbcbdbdbdad2.练习：32)3(4422xxxxx2)3)(2(2xx除法转化为乘法之后可以运用乘法的交换律和结合律3231)2(22××xxxxxmnnmaa）（③nnnbaab)(④nmnmaaa①3.整数指数幂的运算性质：若m,n为整数，且a≠0,b≠0，则有②nmnmaaa23223)()2(abbaaba（2）221232)()2()()2(yxyxyxyx（3）例1.(1)4232)()(abcabccba）（4232)()(abcabccba）（解：(1)原式4422332)()()()(abcabccba444222336acbbaccba35cb分子、分母分别乘方例1.(1)4232)()(abcabccba）（4232)()(abcabccba）（2226233)(8)(babaaba226233)()(8)(bababaaba26)(8)(baabab23223)()2(abbaaba（2）221232)()2()()2(yxyxyxyx4264)()2()()2(yxyxyxyx把负整数指数写成正整数指数的形式积的乘方221232)()2()()2(yxyxyxyx（3）46)2(4)()2(yxyx22)()2(yxyx22)()2(yxyx同底数幂相乘，底数不变指数相加结果化为只含有正整数指数的形式4264)()2()()2(yxyxyxyx例2.计算：1.2.3.4.aaaaaaaaa2444122222)225(423xxxxxxxxxxxx4244222111128422aaaaaaaa1.解法一：aaaaaaaa42)2()1(4222aaaaaa4)2()2(4221aaaaaaaaaa24441222221.解法二：aaaaaaaaaaaa4244142222222)4)(2()1()2)(4()2)(2(aaaaaaaaaaaaaaaaa2444122222aaaaaa42142)2)(4()(422aaaaa21a2.解：2)2)(2(5423xxxxx292423xxxx)3(21x)225(423xxxxxxxxx)2)(2(2121x)2x)(2x()2x(1x)2x)(2x()2x(1xxxx22x43.解：xxxxxxxx42442224.解：111128422aaaaaaaa)1)(1(4)1)(2()2(4aaaaaaaaaaaa4)1)(1()1(41a仔细观察题目的结构特点，灵活运用运算律，适当运用计算技巧，可简化运算，提高速度，优化解题。例3.计算：xyxyxxyxyxx3232分析与解：原式yxxyxxyxyxx)(3232yxx2yxx2巧用分配律yxxxx1312322.ba1ba1)ba(1)ba(122把和看成整体，题目的实质是平方差公式的应用。ba1ba1分析与解：原式babababababa111111baba11222baa巧用公式繁分式的化简：1.把繁分式些成分子除以分母的形式，利用除法法则化简；2.利用分式的基本性质化简。例4.111111aa原式)111()111(aa11aaaa11aa练习1.2.xxxxxx24222122412232aaaa