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习题二 1. 2. 3.解: 设4 x 4 ? 4 px 3 + 4qx 2 ? 2 p (m + 1)x + (m + 1) = 2 x 2 + ax + b .则有 2 2 ( ) 4 x 4 ? 4 px 3 + 4qx 2 + 2 p (m + 1) x + (m + 1) 2 = 4 x 4 + 4ax 3 + (4b + a 2 ) x 2 + 2abx + b 2 ? - 4 p = 4a ? 4q = 4b + a 2 ? ∴? ?2 p (m + 1) = 2ab ? (m + 1) 2 = b 2 ? ? p = ?a ? 1 ∴ ?q = a 2 + b 4 ? m = ?n ? 1 ? 4.证明: (1)因1, λ , λ2 , λ3 , λ4 是方程x 5 ? 1 = 0的5个互异的根 又x5?1=(x?1) x 4 + x 3 + x 2 + x + 1 = ( x ? 1) F ( x) 所以 λ , λ2 , λ3 , λ4 是方程F(x) = 0的根,依据因式定理, ( ) F( x) = ( x ? λ ) x ? λ2 x ? λ3 x ? λ4 1) (2)设G(x) = F ( x 5 ) + xQ( x 5) x 2 R( x 5 ) = F ( x) S ( x)2) + 由(1)知,G(λ ) = G λ2 = G λ3 = G (λ4 ) = 0, 而p(1) + λQ(1) + λ2 R(1) = 0 ? 2 4 ? p(1) + λ Q(1) + λ R(1) = 0 ? 2 ? p (1) + λ Q(1) + λR(1) = 0 ? p (1) + λ4 Q(1) + λ3 R (1) = 0 ? 因为 T 1 = 0,由以上方程组易得: 4 3 2 P(1) = 0, Q (1) = 0, R (1) = 0 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.由题设a + b + c = 0,推出a + b + c = -(ab + bc + ca), a + b + c = 3abc, 即2222333-(ab + bc + ca) = 1 2 (a + b 2 + c 2 ) 2 1 abc = (a 3 + b 3 + c 3 ) 3 因此(a 3 + b 3 + c 3 )(a 2 + b 2 + c 2 ) = a 5 + b 5 + c 5 + a 2 b 2 (a + b) + b 2 c 2 (b + c) + a 2 c 2 (a + c) = a 5 + b 5 + c 5 + a 2 b 2 (?c) + b 2 c 2 (? a ) + a 2 c 2 (?b) = a 5 + b 5 + c 5 + abc(ab + bc + ac) = a 5 + b5 + c5 + a3 + b3 + c3 a 2 + b2 + c 2 . 3 2 5 ∴ a 5 + b 5 + c 5 = (a 3 + b 3 + c 3 ).(a 2 + b 2 + c 2 ) 6 ∴ a5 + b5 + c5 a3 + b3 + c3 a 2 + b2 + c 2 . = 5 3 2 6.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, f ( x) = ( x 2 ? x + 1)( x 2 ? 2) 设4-x3?kx 2 + akx ? a = x 2 + mx + n)( x 2 + px + q ) ( = x 4 + ( p + m) x 3 + (q + mp + n) x 2 + (mq + np ) x + nq p + m = ?1.1) ? ?q + mp + n = ?k 2) ? 则有: ? ( ? mq + np = 2k 3) ? nq = ?2.4) ? 由(4) ,有??n=1?n=?1 ? n = 2 ? n = ?2 q = -2 ? q = 2 ?q = ?1? q = 1 ? n =1 把? 代入(3), 有-2m + p = 2k 5) ? q = ?2 ?m = ?1 2 ? ? m = ?1 + (1 ? k ) ? 3 6)故当K = 1? p = 0 由(1)(5)得: ? ? 2 ? p = (k ? 1) ? n =1 3 ? ? q=2 ? ? n =1 代入(2)不满足,故? 不合 ? q = ?2 4 7.解: (1)原式= x 4 + y 4 + x + y) + x 2 y 2 ? x 2 y 2 ( = x 4 + y 4 + ( x + y ) 2 ? xy . ( x + y ) 2 + xy + x 2 y 2 x 2 + y 2 ? x 2 y 2 + ( x + y ) ? xy . (x + y ) + xy 2 2 x 2 + y 2 + xy . x 2 + y 2 xy )] + [( x + y ) 2 = x 2 + y 2 + xy x 2 + y 2 ? xy + (x + y ) = x 2 + xy + y 2 2 ? x 2 + y 2 + xy = 2 x 2 + xy + y 2 )[ ] + xy ](x + xy ] 2 2 + y 2 + xy ) ) ( ) 2 ) ( (2)原式= [x( x + 1)] + 2 x( x + 1) + 1 2 = [x( x + 1) + 1] = x2 + x +1 2 ( ) 2 (3)此多项式是对称多项式.当x=-(y+z)时, f ( x, y, z ) = ( y + z )[ z ? ( y + z )][?( y + z ) + y ] + [?( y + z )] yz = ( y + z )(? y )(? z )( y + z ) yz = 0 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, y + z)z + x )( x + y ) + xyz = ( x + y + z )[ m( x 2 + y 2 + z 2 ) + n( xy + yz + xz )] ( ( 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, 当x = y时,原式 = ( y ? z )( y + z ) 3 + ( z ? y )( z + y ) 3 = 0 因式有因式(x-y) (y-z) (z-x) ,设(x?y)( x + y ) 3 + ( y ? z )( y + z ) 3 + ( z ? x)( z + x) 3 = k ( x + y + z )( x ? y )( y ? z )( z ? x) 令x=1.y=2,z=0 得: (?1).33 + 2.2 3 + ( ?1)13 = 3k ( ?1).2.( ?1) 所以 k=-2 ∴ ( x ? y )( x ? y ) 3 + ( y ? z )( y + z ) 3 + ( z ? x)( z + x) 3 = ?2( x + y + z )( x ? y )( y ? z )( z ? x) 8.解: (1)先用综合除法,可能的试除数是 ± 1, 3, 5, 15 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x4?x3+6x2?x+15 = ( x 2 + mx + k )( x 2 + nx + l ) = x 4 + m + n)x 3 + k + mn + l ) x 2 + ml + nk ) + kl ( ( (比较等式两端对应次的系数,得方程组: ( ?m + n = -1.1) ? k + mn + l = 6.2) ? ? ? ml + nk = ?1.3) ?kl = 15.4) ? 由(4)知,k 和l的值可能有下面八组: ? k = 1 ? k = ?1 ?k = 3?k = ?3 ?k = 15 ?k = ?15?k = 5 ?k = ?5 l = 15 ?l = ?15 ? l = 5 ? l = ?5 ? l = 1 ? l = ?1 ? l = 3 ? l = ?3 ? m =1 ? n = ?2 ? 经检验得.方程组的解为 ? ? k =3 ? l=5 ? ∴ x 4 ? x 3 + 6 x 2 ? x + 5 = ( x 2 + x + 3)( x 2 ? 2 x + 5) (2)先用综合除法可能的试除数是 ± 1,±3,±7,±21 , 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x4+7x3+20 x 2 + 29 x + 21 = ( x 2 = mx + k)x 2 + nx + l ( ) = x 4 + (m + n) x 3 + (k + mn + l ) x 2 + (ml + nk ) x + kl 比较各式两端对应次的系数,得方程组: ?m + n = 7.1) ? k + mn + l = 20.2) ? ? ? ml + nk = 29.3) ? kl = 21.4) ? 由(4)知,k 和l的值可能有 8 组?k=1?k=?1 ?k = 21?k = ?21 ?k = 3 ?k = ?3 ?k = 7 ?k = ?7 l = 21 ?l = ?21 ? l = 1 ? l = ?1 ? l = 7 ? l = ?7 ? l = 3 ? l = ?3 ?m = 2 ?n = 5 ? 经检验得方程组解为: ? ?k = 3 ?l =6 ? ∴ x 4 + 7 x 3 + 20 x 2 + 29 x + 21 = ( x 2 + 2 x + 3)( x 2 + 5x + 7) 9.解: (1)原式= x ? x ? 6 x ? 15 x = 45 3 2 = x x 2 ? x ? 6 ? 15( x ? 3) = x( x ? 3)( x ? 2) ? 15( x ? 3) =(x-3)[x(x-2)-15] ( ) = x - 3) x 2 + 2 x ? 15) ( ( =(x-3)(x+5)(x-3) = ( x ? 3) 2 ( x + 5) (2)原式 = 2 x + 7 x ? 4 x + 2 x 4 3 2 2? ? 13 x + 6 = x 2 (2 x 2 + 7 x ? 4) + (2 x ? 1)( x ? 6) = x 2 (2 x ? 1)( x + 4) + (2 x ? 1)( x ? 6) 2 = (2 x ? 1)[ x(x + 4) + x - 6) ] ( = 2 x - 1)x 3 + 4 x 2 + x ? 6 ( ( ) = 2 x - 1) x 3 + 4 x 2 ? 5 x + 6 x ? 6)2 x ? 1)[ x( x 2 + 4 x ? 5) + (2 x - 1)] 6 = (2 x ? 1)[ x(x - 1) x + 5) + (x - 1)] ( 6 = 2 x ? 1)( x ? 1)[ x 2 + 5 x + 6] ( =(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22x(y + z) + y ( z + x) 2 + z ( x + y) - 4 xyz = k ( x + y )( y + z )( z + x) 2 令x = 1. y = 2, z = 0, 得: 2 + 0) 2 + 2(0 + 1) 2 + (1 + 2) = k .3.2.1 1. ( 0 即6k=6,所以 k=1 ∴ x( y + z ) 2 + y ( z + x) 2 + z(x + y) 2 - 4 xyz = x + y) y + z) z + x)4)原式 = [( x 2 + 24 + 11x )][( x 2 + 24) + 14] ? 4 x 2 = x 2 + 24) 2 + 14 x. x 2 + 24) 11x.( x 2 + 24) + 154 x 2 ? 4 x 2 x 2 + 24) 2 + ( x 2 + 24).25 x + 150 x = [( x 2 + 24) + 15 x].[( x 2 + 24) + 10 x] = ( x 2 + 15 x + 24)( x 2 + 10 x + 24) = ( x 2 + 15 x + 24)( x + 4)( x + 6) 第10,11 题,无答案 12.解:Q x + x + 1 = 0 2 ∴x+ 1 = ?1和x 3 = 1 x ∴ x14 + 1 1 1 = x 2 .( x 3 ) 4 + 2 3 4 = x 2 + 2 14 x x(x ) x 1 = x + ) 2 ? 2 = 1 ? 2 = ?1 ( x 13 证明: Q 1 1 1 1 + + = a b c a+b+c 1 1 1 1 0 a b c a+b+c bc + ac + ab 1 ∴ ? =0 abc a+b+c ∴ (bc + ac + ab)(a + b + c)a + b + c) - abc ( =0 abc(a + b + c) abc + b 2 c + bc 2 + a 2 c + abc + ac 2 + a 2 b + ab 2 + abc ? abc ∴ =0 abc(a + b + c) ∴ 2 bc(a + b) + ac(a + b) + c(a + b) + ab(a + b) =0 abc(a + b + c) ∴ (a + b)(b + c)(c + a ) =0 abc(a + b + c) 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 则b k = ?a k (因k为奇数 )因而得证等式必然成立 14.证明: (分析法) 要证明原式成立,即证明: 1 1 ?1 1? ?1 1? a ( + ) + 1 + b? + ? + 1 + c ? + ? + 1 = 0 b c ?a c? ?a b? 即要证: 1 1 1 1 1 1 ?1 1 1? a b?c(0 a b c a b c ?a b c? 即要证: (a + b + c)? 1 + 1 + 1 ? = 0 ? ? ?a b c? 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) 当n = 1时,左边 = 1 + a +1 1 1 , 右边 = 1 = 1 + ,,左边 = 右边 a1 a1 a1 所以,当n=1 时,等式成立 (2)假设当 n=k 时,等式成立,即1+ (a + 1)(a 2 + 1)K (a k +1 + 1) 1 a1 + 1 + +K+ 1 a1 a1 a 2 a1 .a 2 K a k +1a k + 2 = (a1 + 1)( a2 + 1) (a K +1 + 1) (a1 + 1)( a2 + 1)K ak + 1)( ak +1 + 1) K ( + a1a2 K ak +1 a1a2 K ak +1ak + 2 (a1 + 1)a 2 + 1) K (a k +1 + 1) k + 2 + (a1 + 1)a 2 + 1) L (a k + 1)(a k +1 + 1) ( .a ( a1 a 2 L a k + 2 = (a + 1)a 2 + 1) L (a k +1 + 1)(a k + 2 + 1) ( = 1 a1 a 2 L a k + 2 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: ∴ 2 x 4 ? x 3 + 2 x ? 6 = 2( x ? 2) 4 + 15( x ? 2) 3 + 42( x ? 2) 2 + 54( x ? 2) + 22 2 x 4 ? x 3 + 2 x ? 6 2( x ? 2) 4 + 15( x ? 2) 3 + 42( x ? 2) 2 + 52( x ? 2) + 22 = ( x ? 2) 5 ( x ? 2) 5 = 2 15 42 54 22 + + + + 2 3 4 ( x ? 2) (x - 2) ( x ? 2) ( x ? 2) ( x ? 2) 5 (2)设5x2?4x+16 A Bx + C Dx + E = + 2 + 2 ,则有 2 2 x ? 3 x ? x + 1 ( x ? x + 1) 2 ( x ? 3)( x ? x + 1) 5 x 2 ? 4 x + 16 = A( x 2 ? x + 1) 2 + ( Bx + C(x - 3) x 2 ? x + 1) + ( Dx + E )( x ? 3) ) (令x=3,有:49A=49,所以A=1 令x=0,有:-3C-3E=15,所以 C+E=-5…1) 令x=1,有:17=1+(B+C) (-2)+(D+E) (-2) ,所以 B+C+D+E=-8…2) 令x=2,有:28=9+(2B+C) (-1).3+(2D+E)(-1).所以 6B+3C+2D+E=-19…………….(3) 令x=4,有:80=169+13(4B+C)+4D+E,所以 52B+13C+4D+E=-89…4) ? B = -1 ? C = -2 ? 解(1)(2)(3)(4)得:D = -2 ? E = -3 ? 5x 2 1 x+2 2x + 3 ∴ = ? 2 ? 2 2 ( x ? 3)( x ? x + 1) ( x ? 3) x ? x + 1 ( x ? x + 1) 2 17. (1)解:设411+8=4+2 2 = x + y 2 2 1 ? ?x = 4 ? ?x + y = 4 则有? , 解得: ?y = 1 2 ? ? xy = 2 2 ? ? 1 1 1 ∴ 4 + 8 = 4+ = 2+ 2 2 2 2 (2 )原式= a ? 2 + 2 a 2 ( a 2 ? 4) = a 2 ? 2 ( a 2 ? 2) 2 ? a 2 ( a 1 ? 4) a2 ? 2 ? + 2 2 a2 ? 4 2 = a + a2 ? 4 2 2 3 3 2 ?2 3 ? ? 5 2 3? 2 33 2 +2 3 (a 2 ? 2 ? a2 a2 ? 4 2 ) 2 (3 ) = a 2 ? 2+ 2 3? 2 (3) = ( 2 3+ 2 3? 2 ( )( 3+ 2 ) ) ( 5(2 3 + 2) ? 3 2 ? 2 3 3 2 + 2 3 (2 3 - 2) 3 + 2 2 ( 3 5 = 2 3 + 2 2 - (3 2 + 2 3) 2 3+ 2 6 10 3 1 =2 3+2 22 -2 3+ 2 =0 2 2 (4)原式= 3 a+ =( a? ( )( a ? a ? 1) + ( a ? 1) + ( a + a ? 1 ) a ?1 3 2 2 a ? a ?1 ) 6 a ? a ?1 ( a + a ?1 )( a + a ?1 ) 6 ) = a ? 2 a(a ? 1) + (a ? 1) + a + 2 a . a ? 1 + (a ? 1) = 4a ? 2 18.解: (1)原式= ( ( 2+ 3? 5 2+ 3+ 5 2+ 3? 5 2 + 3 -5 )( 2+ 3? 5 ) = ) 2 = (2)略 19.解: (1)原式= 2+ 3- 5 2 6 1 6- 5 - 3 5- 2 - 4 6+ 2 = ( 6+ 5 6- 5 )( 6+ 5 ) ( -3 5+ 2 5? 2 )( 5+ 2 ) ( ? 4 6? 2 6+ 2 ( )( 6? 2 ) ) = 6+ 5? 5? 2? 6+ 2 =0 (2)原式= (1 + 3 )( 3 = 5 )( 3 - 1)( 5 - 3 ) ( 3 - 1 + 2 * 3 - 2 3 + 15 - 5 )( 5 - 3 ) = 2* 2 5- 3 4 (1 + 2 3+ 5 )( 3 -1 )( 5- 3 ) = ( 15 + 5 = )( 5- 3 ) 5 3 - 3 5 + 5 5 - 5 5 - 5 + 15 - 15 + 3 4 = 2 5?2 5 ?1 = 4 2 1 ? 1 6 +1?1 1 7 + 2 6 +1 =0 (3)原式= 7 ? 2 6 +1 = 1 6 ?1+1 ? (4)原式= 5 ? 3 ? 29 ? 2 180 = 5 ? 3 ? ( 20 ? 3 = = 5 ? 6 ? 20 5 ? 6?2 5 = =1 5? ( ( 5 ?1 20.解:原式= x2 ?1 1+ 1? x 1? x ? 1? x 1? 1+ x 1+ x + 1+ x [ + ]2 + 1 2 1? x + 1? x 1? x ? 1? x 1+ x ? 1? x 1+ x + 1+ x 2 ( ? x2 ?1? ? x 1? x ? x 1? x ? ? +1 = + 2 2 2 ? (1 ? x ) ? (1 ? x ) (1 + x ) ? (1 + x _ ? ? ? x2 ?1? x 1? x x 1+ x ? ? ? +1 = ? 2 ? x(1 ? x ) x 1 + x ? ? ? x2 ?1? 1? x 1+ x ? ? ? +1 = ? 2 ? 1? x 1+ x ? ? ? = x2 ?1? 1 1 1 ? ? 2 ? 1? x 1+ x ? 2 2 2 2 x2 ?1? 1+ x ? 1? x ? ? ? +1 = ? 2 ? 1? x2 本文来源：https://www.dywdw.cn/b16b5067f4ec4afe04a1b0717fd5360cba1a8d8a.html
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2022-04-24
