学而思网校初一数学暑期复习计划---乘法公式和因式分解初一数学教研组1、（1）解：原式(21)(21)(221)(241)……(2321)1(221)(241)(2321)1(241)(2321)126411264（2）解：原式2(11)(11)(11)……(11)22422n2(11)(11)……(11)4422n2(121)24n124n1（3）解：设S3132134138131613321,两边乘以31，得31S3131321341381316133212S3213213413813161332136412S13641,24816323641即31313131313122、解：2961(2481)(2481)(261)(261)(2121)(2241)(2481)6365(2121)(2241)(2481)，这两个数是63和65。3、解：若把M视为2ab这一项，9x2M4(3x)2M22，此时M可以为12x；若把9x2视为2ab这一项，9x2M4M22229x2，此时M为81x4；416若把4视为2ab这一项，9x2M4(3x)2M23x2，此时M为4；3x9x2M还可以是9x2,4。4、解：(1)由a2b2c2abbcca，1ab2ac2bc202可得abc；则abc4。a3b2c84。(2)已知关系式可以转化为a2b2c2abbcca0,1ab2ac2bc202abc。选A5、解：（1）a15a2a123,a2a4a21a2a21124a211121（2）a27,a229,即a9,a3a2a2aa6、解：(1)x22x7x117x10,x0,0,即x7xx(2)x17,x21249,x2147xx2x2(3)x21x247,x41x422209,x41x422077、解：原式(1x)1xx(1x)x(1x)2x(1x)3x(1x)2005(1x)21xx(1x)x(1x)2x(1x)3x(1x)2004(1x)31xx(1x)x(1x)2x(1x)3x(1x)2003(1x)20051xx(1x)(1x)20078、解：原式10079、解：原式a3b3a2b2ab(ab)(a2abb2)(ab)(ab)(ab)(ab)(a2abb2ab1)10、解：原式(a33a23a1)(b33b23b1)(a1)3(b1)3(a1b1)(a1)2(a1)(b1)(b1)2(ab2)(a22a1abab1b22b1)(ab2)(a2b2abab1)