22.1.1★化簡(jiǎn)1x1xx2x322.1.1★化簡(jiǎn)1x1xx2x3xn1，其中n1原式1xx2x3 xn1xx2x3xn1xn1xnaban1an2babn2bn1anbn（1）abcdcadb（2）x2yx2yx416y48x2cbd2c2b2d22bd2bc2cda2（2）x2yx2y的結(jié)果是x48x2y216y4x24y2原式x24y2x24y22x24y2x223x224y23x24y224y2x612x4y248x2y464y62.1.3gxfxx33x21，求gxfx所得的商qx及余rx2x1x 3x22x1x33x2xx32x21 7x24x1 7x214x 26x 因此，所求的商q1x 3x22x1x33x2xx32x21 7x24x1 7x214x 26x 因此，所求的商qx1x7，余式rx26x2 x為3次多項(xiàng)式，首項(xiàng)系數(shù)gx2次，首項(xiàng)系數(shù)為3，故商qx1次，首項(xiàng)系1，112，于是可設(shè)商式q xa，余式rxbxc數(shù)必為，而余式次數(shù)33根據(jù)fxqxgxrxx33x2x 3x2x xabx2 2 1 x3a xb2a xac32 3 3 3a23b2a13ac解得a7b26c2，故商式qx1x7，余式rx26x2999 2.1.4x7時(shí)，代數(shù)式ax5bx84x7ax5bx3 x7ax5bx31ax5bx8 2147922.1.5xyzxyz12，試求2x3y4z 解 y2x，z3x，代xyzx2y4z6，所以2xxyzx2y4z6，所以2x3yz4402.1.6★★試確定a和bx4ax2bx2x23x23x2x1x2，因此fxx4ax2bx2f10x1x2x11ab20f20x2164a2b20ab2.1.7★若x1x1x5x3bx2cxd，求bd解析x1x1x5x21x5x35x2x5，所以bd5．bd02.1.8★將3x25x7表示成ax22bx2c3x25x 2 x25x23x2217x2152.1.9★已知a2a10，求a32a22的值．解析1 由a2a1，有a32a22a3a2a2aa2aa2aa22122由a21a．a(chǎn)32a22a2a221aa22aa224a4a1a4a1a2.1.10xymx3y3nm0x2y2的值．解析因?yàn)閤ym，所以m3xy3x3y33xyx2.1.10xymx3y3nm0x2y2的值．解析因?yàn)閤ym，所以m3xy3x3y33xyxyn3mxy所以xy nx2y2xy2mn2m2 ． 2.1.11★★若x2xyy14y2xyx28，求xy把兩個(gè)方程相加，得xy2xy42xy6xy70xy6xy72.1.12xy1x2y22x7y72，所以1xy2x2y22xy22xyxy1x22x3y3xy33xyxy1513 1 2 2x4y4x2y222x217222 2 1故x7y7x3y3x4y4x3y3xy 2 82.1.13★★已知a1999x2000，b1999x2001，c1999x2002a2b2c2abbcca解 由a2b2c2abbc多項(xiàng)式1ab2bc2ca22又因?yàn)閍b1bc1ca2，原式1121222322.1.14★★已知實(shí)數(shù)a、b、x、y滿足abx原式1121222322.1.14★★已知實(shí)數(shù)a、b、x、y滿足abxy2，axby5，求a2b2xyabx2y2的abxy2，得abxyaxbyaybx4．因?yàn)閍xby5aybxa2b2xyabx22.1.15★★已知3xax7ax6ax5 axa，試求aaa aa的值765 fx的系數(shù)和，就是f1 多項(xiàng)3127aaaa ．x，它x12；被x2除8x12.1.16★★求一個(gè)關(guān)于xffxax2bxc則f1abc2①②③f24a2bc8f1abc04ac6④⑤aca53c235x2x233x2y2m22yxnmy2n20m2x2m2y22mxym2x2m2y22mxy2mnyy2n20即mxy2myn20所以mxy0,mynnn因?yàn)閙0，所以y ，x mxyzxyzxyzx1y21z2y1x21z2z1x21y24xyz解 因?yàn)閤yzxyz，所左邊x1z2y2y2z2y1z2x2x2z2z1y2x2x2y2xyzxz2xy2xy2z2yz2yx2yx2z2zy2zx2zx2xyzxyyxxzxzyzyzxyzxyyzxyzxyxyzzxzxyzyyzxyzxxyzxyyzxyzxyzxyz4xyz★已知a2b2c2abbcca，證明abc．解析因?yàn)閍2b2c2abbcca，所以2a2b2c22abbcca0，即ab2bc2ca20，因此abbcca0，abcyz2x3zx2y3xy2z3yz2xzx2yxy2zyz2xazx2ybxy2zc①②③則要證的等式變?yōu)閍3b3則要證的等式變?yōu)閍3b3c3a3b3c3abca2b2c2abbc，abcyz2xzx2yxy2z0，所以a3b3c33abc0，所以yz2x3zx2y3xy2z3yz2xzx2yxy2z2.1.21★★已知a4b4c4d44abcda、b、c、dabcda4b4c4d44abcd0a2b22c2d222a2b22c2d24abcd0，a2b22c2d222abcd20a2b220c2d220abcd20，所a2b2c2d2abcd0所以ababcdcd0又因?yàn)閍、b、c、d都為正數(shù)，所以ab0cd0ab，cdabcda2c2acac0，所以ac．故abcd成立．2.1.22★★已知abc02a4b4c4a2b2c22 用作差法，注意利用abc0的條件．左右2a4b4c4a2b2c2a4b4c42a2b22b2c2a2b2c22a2b2c22bca2b2c2a2bc2a2bc2abcaabcabcabcab0（1）2x5n1yn4x3n1yn22xn1yn4（2）x38y3z36xyz（3）a2b2c22bc2ca2ab（4）a7a5b2a2b5b7解析（1）原式2xn1ynx4n2x2ny2y42xn1ynx2n22x2ny2y222xn1ynx2ny22xn1ynxny2xny2（2）原式x32y3z33x2yzx2yzx24y2z22xyxz2yza22abb22bc2caab22cabc2ab．a(chǎn)2b2c22bc2ca2aab．（4）原式a7a5b2a2b5b7a5a2b2b5a2b2a2b2a5b5abababa4a3ba2b2ab3b4ab2aba4a3ba2b2ab3b4原式x32y3x3y3x3y3xyx2xyy2x3y3x3y3xyx2xyy2xyx2xyy2原式x23y2．x2y2x22x2y2y22xyxyx2y22x2y2xyxyx2y2xyx2y22x4x2y2y4x2y2xyx2y2222222x2y2 z2x2y2z2原式中與a3b3ab33abab原式x2y2z2x233x2y2z2x2x2y2z2x2y2z2y2z233x2y2z2x2y2z2y2z23x2y2zxzxy2z2原式ab33ababc3ab3c33abababcab2cabc23abababca2b2c2abbcca評(píng) a3b3c31abc2a22b22c22ab2bc2ca21abcab2bc2ca22顯然，當(dāng)abc0時(shí)，則a3b3c33abc；abc0時(shí)，則a3b3c33abc0a3b3c3≥3abc，而且，當(dāng)且僅當(dāng)abcxa3b3c3≥3abc，而且，當(dāng)且僅當(dāng)abcxa20yb3≥0zc30xy≥3xyz3x15x14x13x2x解 公式anbnx161x1x15x14x13x2x1x1x15x14 x2x原式x xx81x41x21x1xxx81x41x21x1評(píng)注在本題分解過(guò)程中，用到先乘以x1，再除以x的技巧，這一技巧在等式變形中很常用 方法 將常數(shù)項(xiàng)8拆成19．原式x39x1x319xx1x2x19xx1x2x8方法 將一次項(xiàng)9x拆成x8x．原式x3x8xx3x8xxx1x18xx1x2x8方法 將三次項(xiàng)x3拆成9x38x3原式9x3原式9x38x39x9x39x8x39xx1x18x1x2xx1x2x 添加兩項(xiàng)x2x2x39xx3x2x29x原式x2x1x8x1x1x2x（1）x9x6x33（2）m21n214mn（3）x14x212x14（4）a3bab3a21（1）將3拆成111．原式x9x6111x91x61x3x31x6x31x31x31x3x31x62x3x1x2x1x62x33（2）將4mn2mn原式m21n212mn2mnm2n2m2n212mnmn12m原式x142x212x21原式x142x212x212x 2 4 224 x 2xx x x212 222 x1x x22x222x2123x21x23（4）添加兩項(xiàng)abab原式a3bab3a2b21aba3bab3a2ababb2abababaababb2aabbab1abb2aab1abb2a2ab1b2ab評(píng)注（4）是一道較難的題目，由于分解后的因式結(jié)構(gòu)較復(fù)雜，所以不易想到添加abab，而且添加2.2.8x42x32x21．解析原式x42x212x3x2122xx2x21x21x21xbcbccacaabab．解原式bcbccabcabababcbccabccaababacbcabaabcabbccax3x3yzy3zxz3xy．解原式x3yy3xy3zx3zz3xz3xyx2y2zx3y3z3xxyxyxyzx2xyy2z3xyx2yzxyyzzy2z2xyyzx2xyzyz2xyyzzxxyz原式1xx222x31xx2x61xx222x31xx2x3x31xx21xx22x3x3x1xx21xx2x3x4x2x1x2x212母y來(lái)替代，于是原題轉(zhuǎn)化為關(guān)于y的二次三項(xiàng)式的因式分解問(wèn)題了．x2xy原式y(tǒng)1y212y23yy2y5x2x2x2xx1x2x2x5x2x1x2x1u，可以得到同樣的結(jié)果，有興趣的同學(xué)不x23x24x28x390原式x1x22x12x原式x1x22x12x390y2x25x2原式y(tǒng)y190y2yy10y2x25x122x25x2x25x122x7x1對(duì)多項(xiàng)式適當(dāng)?shù)暮愕茸冃问俏覀冋业叫略幕鵤b2abab21ab2．yabxabx22xy2x4yy22yx22xy1y 2xyxy12所以，原式abab1212aa2baa1．解a1x12aa2a122a2x22a2原式x22a2bax2b2x2b2a2b2ab2xx2bax2ab2xababa2ab1272．解析令yx2，則原式y(tǒng)14y14272y22y12y22y12y44y214y32y24yy44y214y32y24y2y412y22y46y22y29y22y3y3y215所以，原式2x5x1x24x19x24x823xx24x82x2解 設(shè)x24x8y，原式y(tǒng)23xy2x2y2xyx26x8x25xx2x4x25x82.2.186x47x336x27x原式6x417xx216x42x212x27xx216x2122x27xx216x2127xx212x213x3x212x23x23x28x2x1x22x213x3x212x23x23x28x2x1x23x1x3 本解法實(shí)際上是將x21看作一個(gè)整體，但并沒(méi)有設(shè)立新元來(lái)代替它，即熟練使用換元法后，76原式 6x7x36 222xx2 1 1x6x 7x 362 2x x令x1t，則x2 t22，于1x原式x26t227tx26t27t24x22t33t2 1 1x2xx33xx ＝2x23x23x28x2x1x23x1x3x2xyy224xyx2y2元對(duì)稱(chēng)式．對(duì)于較難分解的二元對(duì)稱(chēng)式，經(jīng)常令uxyvxy2原式xy2xy4xyxy22xyxyuxyv原式u2v24vu2u2u46u2vx22xyy23xy2x2xyy22 2xy3y22x10y8原式x3yxy2x10y解x3y4xy2xx4x2abxabxx4x2abxababx2acbdxcd十字相乘法把它分解成xaxb或axdbxc的形于原式中的ey，第一、第三列構(gòu)成的十字交叉之積的和等于原式中的dx．6x213xy6y222x23y20解析原式2x3y3x2y22x23y2x3y43x2y5．其十字相乘圖-45-6x27xy3y2xz7yz2z2原式2x3y3xyxz7yz2x3yz3xy2z．其十字相乘圖-zy-x1x2x3x424原式x25x4x25x6x25x4x25xx25x4x25x42＝x25x422x25x4x25x46x25x4xx5x25x10對(duì)于形如exaxbxcxdf（a、b、cd、ef為常數(shù)abcdxaxb與xcxd分別相乘后，構(gòu)成有相同部分：x2abxx2cdx的項(xiàng)x2x3x8x124x2．解原式x214x24x211x24x214x24x214x243xx214x2423xx214x24x214x244xx214x24x210x24x215xx4x6x15129x1512922對(duì)地形如exaxbxcxdfx2（a、b、c、d、e為常數(shù)abcdxaxb與xcxdx2abx2cdx23y28z22xy2xz14yzx22xy3y2x3yxyx23y28z22xy2xz14yzx3ymzxynzx22xy3y2mnxzx22xy3y2mnxz3nmyzmnz2mn2,3nm14mnm2，n4所以，原式x3y2zxy4z2.2.26x4x34x23x5x4x34x23xx2ax1x2bxx4abx3ab6x25abxabab645ab解之，得a1b2所以，原式x2x1x22x式x2ax1x2bx5．2.2.27kx2y23x7kx2y2xyxyx2y23x7yk那么它的兩個(gè)一次因式一定是xym與xyn的形式，其中mnx2y23x7y設(shè)xymxynx2y23x7yx2xymxxyy2mynxnyx2y2mnxnmymnmnnmmnmmnnmmnm5,解之，得nk10x2y23x7yk可以分解成兩個(gè)一次因式的乘積xy5xy2因此，當(dāng)k102.2.28a4ab4b4解 因?yàn)閍4b4ab44a3b6a2b2ab44abab22ab2所以，原式a4b4aab44abab22ab2a2ab42abab2ab222ab22a2b2ab2xyz5x5y5z5 這個(gè)式子是關(guān)于x、y、z的五次齊次對(duì)稱(chēng)式，令xy，則原式0故原式有因式xy．同理，亦有因式y(tǒng)z，zx．這樣原式還有一個(gè)二次齊次對(duì)稱(chēng)式kx2y2z2lxyyzzx．所以，可原式xyyzzxkx2y2z2lxyyzzx．xy1z0時(shí)，得152kl①x2y1z0355k2lk5，l5所以，原式5xyyzzxx2y2z2xyyzzx2.2.30★★分aba2b2bcb2c2cac2a2② 當(dāng)ab時(shí)，易知原式0，所以原式有因式ab．同理，bc與ca也都是原式的因式．但四次多項(xiàng)式應(yīng)有四個(gè)一次因式，由對(duì)稱(chēng)性余下的一個(gè)因式必有為abc 當(dāng)ab時(shí)，易知原式0，所以原式有因式ab．同理，bc與ca也都是原式的因式．但四次多項(xiàng)式應(yīng)有四個(gè)一次因式，由對(duì)稱(chēng)性余下的一個(gè)因式必有為abc，故可設(shè)aba2b2bb2c2cac2a2kabcabbcca233k112．解得ka0b1c21．所以，原式＝abcabbccaa2bc2b2ca2c2ab2abcabca2b2c2abbcca解析所給的式子是一個(gè)四次對(duì)稱(chēng)式．若令ab，則原式b2bc2b2cb2b2c22b2b2bc2cb2c2c22b2b2b2c22bcb22bcc22c22b20所以，原式含有因式ab同理，原式含有因式bcca于是，原式含有因式abbcca由于原式為四次對(duì)稱(chēng)式，故還有因式kabc，其中k為待定系數(shù)．所以，原式kabbccaabc．比較等式兩邊a3b的系數(shù)，得k所以，原式abbccaabc（1）a9ba3b a6．a(chǎn)23a a2aa a 解析 2 a6a23a a2aa6a1a a2aa1a26aa1a2aa29aa1a2a a10a a10．a(chǎn)1a2aa2（1）x；（2）2xyxyxyxyxx1x1 解析（1）xa29aa1a2a a10a a10．a(chǎn)1a2aa2（1）x；（2）2xyxyxyxyxx1x1 解析（1）x1 xxx 1（2）xyxyx2xyx22xyxy xx x 2x2x．x x2a23a2a2a53a24a52a28a．a(chǎn)aaa解 2a22aa11a22a3a63a26a2a4 2a26a2a6原式aaaa2a11a313a212a21aa2a2a2a1a33a22a21111aa a1111 a a a a1a1a a2aa2a3a1aa1a2a2a8aa1a2a2a．11248 1 1 1 1 1 1a2ababa2原式1a1a2481a11 1 1 122481 111248 1 1 1 1 1 1a2ababa2原式1a1a2481a11 1 1 122481 1 1 1 121a221a2481a21a21 1 14481 1 1 1881 1 1 1 1．1 12ab2bc2ca．a(chǎn)2abac b2abbc c2acbc解 本題如果直接通分化為同分母去處較繁而通過(guò)分子拆項(xiàng)分母分解之后利用xy11 原式abacbcbacacaba bcb cac111111a a b b c c02.3.6x23x10x21x21x3x10x0x13x 22xx 11x2 x 27x評(píng) 這里利用x與1互為倒數(shù)的特點(diǎn)．巧妙地運(yùn)用乘法公式加以變形，使問(wèn)題變得較簡(jiǎn)單．同x 11x3 x x12 x2 x37118 x4 x2247x22x3xy2x1x2xy xy 11x3 x x12 x2 x37118 x4 x2247x22x3xy2x1x2xy xy5xy5xy2x3xy2y2xy3xy25xy3xy1xyx2xy5xy xyz．zxz xyxz yxyz 直接通分比較繁，考慮到這里主要涉及xy，yz，zx三個(gè)式子，不妨用換元法．使所求xyayzbzxc，則abc0原式bcacab a3b3ab33abab c33abc32.3.9xyz1xyz2x2y2z216求 xy yz zx2解 因?yàn)閤yz2兩邊平方得x2y2z22xy2yz2zx4已知x2y2z216所以1x2y11xyyzzx6．又z2xy．xyxy42x211y2z11z2x，yzzx2111原式x2y y2z z2xx2y2zx2y2zxyzzyx2xyyzzx4xyz24 1128 2.3.10★★若abc1abcaba bcb cac原式因?yàn)閍bc1a、b、caababcaba abcb abcacaaba abcxyzzyx2xyyzzx4xyz24 1128 2.3.10★★若abc1abcaba bcb cac原式因?yàn)閍bc1a、b、caababcaba abcb abcacaaba abcab abcaabca1aba11ab a1aab11aba 因?yàn)閍bc1，所以a0，b0，c0abbc原式aba bcb bcac1bb1 bcb bcabc1bb1 bcb11bc1由abc1，得a 1bc原式 1b1bcbc1c 1bb1 bcb11bc111．x23x x25x x27x111原式x2x x2x x3x111111x x2x x3x x411 ．x x x25x1xnxn11與x xn2.3.12xyzx14y11z17xyzyz 解 因?yàn)?x1xy1zz11與x xn2.3.12xyzx14y11z17xyzyz 解 因?yàn)?x1xy1zzx1z7x xx7x371 4x44x3x4x37x34x212x902x320x3z5y2xyz2352.3.13xyz3a（a0xyz不全相等xayayazazaxxa2ya2zxa，yaxau，yav，zaw，則分式uvvwwu，且由已知有uvw0．將uvwu2v2u2v2w22uvuwwu0xyz不全相等，所以u(píng)、vw不全為零，所以u(píng)2v2w20uvvwwu1u2v2 即所求分式的值為122.3.14xyz，求xyza b cxyzk設(shè)a b cxabk，ybck，zcak．xyzabkbckcak0a xyz2.3.15★★已知 1， 0， xabk，ybck，zcak．xyzabkbckcak0a xyz2.3.15★★已知 1， 0， 的值 解 令xu，yv，zw，于是條件變uvw