分式的四则运算

中小学个性化辅导专家教学内容——分式的四则运算★知识精要1.分式的乘法法则：将分子相乘的积作为分子，分母相乘的积作为分母bdacdcba；2.分式的除法法则：将除式的分子和分母颠倒位置后，再与被除式相乘bcadcdbadcba注意：当分子、分母是多项式时，先进行因式分解再约分例：3x2x2x2x1x3x1x2xx4x23x4x2x3x223.分式乘方的法则：nnnba)ba(（n为正整数）4.分式的加减法则：（1）同分母的分式加减法法则：cbacbca（2）异分母的分式加减法法则是先通分，变为同分母的分式，然后再加减通分通分先要确定公分母，如果各分母的系数是整数，通常各取分母系数的最小公倍数与字母因式的最高次幂的积作公分分母.这样的公分母叫做最简公分母.例：a1019a1015a104a23a52222x12x332x3x4x314x348x41x382a1a1a22a1a1a2a1a2a2a11a1注意：分式的运算结果一般化简成最简分式或整式中小学个性化辅导专家【热身练习】1.快速计算：(1)yxyx_______________(2)32ab______________(3)322yx3_______________(4)axbab222______________(5)xx1x12_______________(6)x6yxyx22332____________(7)bab2ba2_______________(8)y1x2___________(9)2x21x1______________(10)a41ba232__________2.已知3a1a,则22a1a_____________3.若02xy,则分式2222y2xyxyxyx_____________4.分式xy51,y21,x212的最简公分母为_____________5.已知3y1x1,则22yxy2y2xyyx_____________6.已知1ab,3ba,则abba的值是_____________【精解名题】1.化简ababa222的结果为()A.a2baB.abaC.abaD.baba中小学个性化辅导专家2.若0y2x3,则1yx等于()A.32B.23C.35D.353.计算yxxxyxyx222的结果()A.yxx22B.yx2C.y1D.y114.下列分式：(1)3a2a2(2)22baba(3)ba12a4(4)2x1中,最简分式有()A.1个B.2个C.3个D.4个5.分式2nm21,22nm31,nm41的最简公分母是()A.22nmnm24B.nmnm12C.nmnm242D.nmnm1226.计算yxyxyy2xyxx2的结果是()A.1B.3C.yxyxD.yxyx37.化简x31x21x1等于()A.x21B.x23C.x611D.x658.计算1aa1a的结果是()A.1a1B.1a1C.1-a1aa2D.1a9.已知实数a,b满足1ab,设1bb1aaM,1b11a1N,则M，N的大小关系为()A.NMB.NMC.NMD.不确定10.下列分式的运算中，其中结果正确的是()中小学个性化辅导专家A.ba2b1a1B.323aaaC.bababa22D.3a19a6a3a211.计算下列各题：(1)4222xxxxxx(2)226336xxxxxx(3)2a1a4a4a1a22(4)1x2x3x1x4x4x1x222(5)244222y2x12y16xy4xxy6x3(6)ab2ab2baababab2a222(7)222221x4x8x42x21x2x1x(8)3m19mm22中小学个性化辅导专家(9))50x15x3050x15xx2(50x15xx222(10)2a4a22a2a3a2aa3a2222222(11)2222222222y9xxy4y5xy9y2xy9xyx(12)22yxx2yx1yx1中小学个性化辅导专家(13)2a2a4(14)3232xyxyyxyx(15)2x3x33x4x16x5x2222(16)20x9x520x55x212.化简求值：(1)1xx1x112,其中2x(2)1x3x1x1x1x3x2x22,其中2x中小学个性化辅导专家(3)当1x时,226336xxxxxx的值为多少？(4)化简2011x2010x1.......2x1x11xx1,并求当x=1时该代数式的值【巩固练习】1.已知3x1x,求1xxx242的值中小学个性化辅导专家2.化简:1x81x41x21x11x18423.设n为自然数,计算:)1n(n1....431321211【自我测试】1.若0yxxy,则分式x1y1_____________中小学个性化辅导专家2.若3xB1xA3x1x3x，则2008BA_____________3.1x21x2x42=____________4.已知05y4x2yx22，则x1y4____________5.计算下列各题：(1)x2x4x412(2)22xx2x2x2x2x2x(3)2a2a54a23a(4)x4x4x4x1xx2x2x22(5)x11x2xxxx1x22