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当前位置:首页 > 金融/证券 > 综合/其它 > 初等数学研究答案李长明周焕山编习题二1至20题
习题二1.2.3.解:则有设.2112444222234baxxmxmpqxpxx2223422342)4(44)1()1(2444babxxabaxxmxmpqxpxx222)1(2)1(24444-bmabmpabqap1412nmbaqap4.证明:(1)因个互异的根的是方程501,,,,15432x又)()1(1112345xFxxxxxxx所以的根,依据因式定理,(是方程0)F,,,432x)1.....(..........)()F(432xxxxx(2)设)2.......().........()()(()()(G5255xSxFxRxxxQxFx)而)知,由(,0)()G(1432GGG0)1(R)1()1(0)1()1()1(0)1()1()1(0)1()1()1(342422QpRQpRQpRQp因为由以上方程组易得:,01)(234T0)1(R,0)1(,0)1(PQ故由因式定理可知,x-1是P(x),Q(x)和R(x)的因式,又根据(2),x-1也是F(x),S(x)的因式,但x-1不是F(x)的因式,所以x-1是S(x)的因式5.即(,推出由题设,3),2-0cba333222abccbacabcabcba)(21(-222cbacabcab))(31333cbaabc))(222333cbacba因此()cacacbcbbabacba()()(222222555)()()(222222555bcaacbcbacba)(555acbcababccba2.3222333555cbacbacba)).(65222333555cbacbacba(2.35222333555cbacbacba6.解:由试除法知,当k=2时,有一次因式,为了探求二次因式,可用待定系数法,求得当k=1时,)2)(1()(22xxxxf))(-22234qpxxnmxxaakxkxx(设nqxnpmqxnmpqxmpx)()()(234)4.......(....................23.....................2)2.(....................)1...(..........1nqknpmqknmpqmp)(则有:由(4),有1212,212-1qnqnqnqn,)5.....(....................22-),321kpmqn有代入(把21011K)6(........................................)1(3232)1(151qnpmkpkm故当)得:)(由(不合)不满足,故代入(212qn………………….7.解:(1)原式=2222444yxyxyxyx)(=222244)(.)(yxxyyxxyyxyxxyyxxyyxyxyx222222.xyyxxyyxxyyxxyyx2222222.xyyxxyyxxyyx22222xyyxyxyx222222222yxyx(2)原式=11212xxxx211xx=221xx(3)此多项式是对称多项式。当x=-(y+z)时,yzzyyzyzyzzyzyxf)]([])()][()[(),,(0))()()((yzzyzyzy所以f(x,y,z)有因式(x+y+z),因原式为三次式,故还有另一个二次对称式的因式,设,)]()()[())((zy222xzyzxynzyxmzyxxyzyxxz)(令x=1,y=1,z=0.得,2=(2m+n)即2m+n=1…………………………(1)令x=1,y=1,z=1.得,9=3(3m+3n)即m+n=1…………………………….(2)由(1)(2)得,m=0,n=1,所以,(y+z)(z+x)(x+y)+xyz=(x+y+z)(xy+yz+zx)(4)原式是轮换多项式,0))(())((33yzyzzyzyyx时,原式当因式有因式(x-y)(y-z)(z-x),设))()()(())(())(())((333xzzyyxzyxkxzxzzyzyyxyx令x=1.y=2,z=0得:)1.(2).1(31)1(2.23).1(333k所以k=-2))()()((2))(())(())((333xzzyyxzyxxzxzzyzyyxyx8.解:(1)先用综合除法,可能的试除数是1553,1,,,,由于多项式偶次次系数都是正数,奇次次系数都是负数,就只选择正的试除数,试除结果都被排除,因此原式在Q上没有一次因式,假设因式含有x的二次因式。设))((15622234lnxxkmxxxxxxklnkxlmnkx)ml)n)xm234(((比较等式两端对应次的系数,得方程组:)4.....(........................................15)3(..............................1)2.........(....................61........................................-1nmklnkmllmnk)(由(4)知,k和l的值可能有下面八组:35,35115,115,5353,151,151lklklklklklklklk经检验得。方程组的解为5321lknm)52)(3(5622234xxxxxxxx(2)先用综合除法可能的试除数是21,7,3,1,由于多次式系数均为正数,因此只能选择负的试除数,试除结果都被排除,因此原式在Q上没有一次因式,假设原式含有x的二次因式,设lnxxxxxxx22234kmx(2129207)klxnkmlxlmnkxnmx)()()(234比较各式两端对应次的系数,得方程组:)4......(..............................21)3.......(....................29)2...(....................20)1.....(..............................7klnkmllmnknm由(4)知,k和l的值可能有8组3737,73,73,121121,211,211lklklklklklklklk经检验得方程组解为:6352lknm7)x5)(32(212920722234xxxxxxx9.解:(1)原式=4515623xxxx)3(1562xxxx)3(15)2)(3(xxxx=(x-3)[x(x-2)-15])152x3)-x2x((=(x-3)(x+5)(x-3))5()3(2xx(2)原式61324722234xxxxx)6)(12()472(22xxxxx)6)(12()4)(12(2xxxxx6)-x4)x)[12(2((xx]641-x223xxx)()66541)-x223xxxx((1)]-x26)54()[12(2(xxxx1)]-x65)x1)-x)[12((((xx]65)[1)(122xxxx(=(2x-1)(x-1)(x+2)(x+3)(3)此多项式是轮换多次试,当x=-y时。原式=0所以有因式(x+y)(y+z)(z+x)设))()((4-()(zyx222xzzyyxkxyzyxzxzy))(1.2.3.210)10(2)02.1,0,2.1222kzyx)((得:令即6k=6,所以k=1x)zz)yy)x4-y)x)()(222((((xyzzxzyzyx(4)原式2224]14)24)][(1124[(xxxx2222224154)24.(1124.14)24xxxxxxx)((xxxx15025).24()24(222]10)24].[(15)24[(22xxxx)2410)(2415(22xxxx)6)(4)(2415(2xxxx第10,11题,无答案12.解:012xx1113xxx和2243243214141)1).(1xxxxxxxx(1212)12xx(13证明:cbacba111101111cbacba01cbaabcabacbc0)(abc-c)ba(cb)((cbaabcaabacbc)0)(222222cbaabcabcabcabbaacabccabccbabc0c)babc(ab)ab(ab)a)()a(2(cbaacbbc0)())()((cbaabcaccbba所以(a+b)(b+c)(c+a)=0所以a,b,c中必有两数之和为零,不妨设a+b=0,b=-a因而得证等式必然成立为奇数因则kabkk14.证明:(分析法)要证明原式成立,即证明:01111111)11(baccabcba即要证:0)111(111)111(cbaccbabcbaa即要证:0111cbacba因为a+b+c=0,末式成立,各部皆可逆,故原命题成立15.证明:(运用数学归纳法)(1)右边,左边右边时,左边当,111,1111111aaaan所以,当n=1时,等式成立(2)假设当n=k时,等式成立,即21211212111.1)1(1111kkkaaaaaaaaaaa21211211211211)(1)1(111)(1(kkkkkKaaaaaaaaaaaaaa)()22112121211)(1()1(1(.1()1(1(kkkkkaaaaaaaaaaa))))2212121)1)(1()1(1kkkaaaaaaa)(所以当n=k+1时,等式成立16.解:(1)先将分子表达成(x-2)的幂形式:22)2(54)2
本文标题:初等数学研究答案李长明周焕山编习题二1至20题
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