❶ 新疆版原创新课堂 数学答案
http://jpkc.ncwu.e.cn/gdds/UploadFiles/200882322143238.pdf 自己进上面那个网站看吧因为答案有分数显示不了，就像我发的一样你看不懂所以你自己进去看下 多项式习题解答
P44.1 1)
1 7 26 2
( ) ( )( ) ( )
3 9 9 9
f x = g x x - + - x -
2)
f (x) = g(x)(x2 + x -1) + (-5x + 7)
P44.2 1) x2 + mx -1| x3 + 9x + q.余式
( p +1+ m2 )x + (q -m) = 0
2 1
m q
p q
. =
\.
. = -
方法二，
设
3 2 0
( 1)( )
1
m q
x px q x m x q
mq p
. - =
+ + = + - + ..
.- - = 同样。
2 ) x2 + mx +1| x4 + px2 + q.余式
m( p + 2 -m2 )x - (q - p +1+ m2 ) = 0
\m(m2 + p - 2) = 0.
m2 + p =1+ q, (x2 =1- p + q)
P44.3.1用g(x) = x + 3除f (x) = 2x5 - 5x3 -8x
解：
\ f (x) = 2(x + 3)5 - 30(x + 3)4 +175(x + 3)3 - 495(x + 3)2 + 667(x + 3) - 327
P44.3 .2)
3 2
3 2
( )
( 1 2 ) (2 8 )( 1 2 )
x x
x i i x i
\ - -
= - + + - - +
x
-(12 + 8i)(x -1+ 2i) - (9 - 8i)
即余式-9 +8i
商
x2 - 2ix - (5 + 2i)
P44. 4.1). 5
0 f (x) = x , x =1：即
\ f (x) = (x -1)5 + 5(x -1)4 +10(x -1)3 +10(x -1)2 + 5(x -1) +1
当然也可以f (x) = x5 = [(x -1) +1]5
= (x -1)5 + 5(x -1)4 + ×××+1
P44.4 2) 结果
f (x) = x4 - 2x2 + 3 = (x + 2)4 -8(x + 2)3 + 22(x + 2)2 - 24(x + 2) +11
3)
f (x) = x4 + 2ix3 - (1+ i)x2 + 3x + 7 + i
4 3 2
4 3 2
( ) 2 ( ) (1 )( ) 3( ) 7
( ) 2 ( ) (1 )( ) 5( ) 7 5
x i i i x i i i x i i x i i i
x i i x i i x i x i i
= + - + + - - + + - - + - + +
= + - + + + + - + + +
P45.5
(1)
g(x) = (x -1)(x2 + 2x +1) = (x -1)(x +1)2
( ) ( 1)( 3 1) f x = x + x3 - x -
∴( f (x), g(x)) = x +1
(2)
g(x) = x3 -3x2 +1不可约
( ) 4 1 f x = x4 - x3 + 不可约
∴( f (x), g(x)) =1
(3) ( ) 10 1 ( 2 2 1)( 2 2 1) f x = x4 - x2 + = x2 + x - x2 - x -
g(x) = x4 - 4 2x3 + 6x2 + 6 2x +1, f (x) = 4 2(-x3 + 2 2x2 + x) = (x2 - 2 2x -1)2
∴( f (x), g(x)) = x2 - 2 2x -1
P45.6
(1) ( ) ( 1) ( 2) f x = x + 2 x2 -
g(x) = (x2 - 2)(x2 + x +1)
∵ (x +1)2 [-(x +1)]+ (x2 + x +1)(x + 2) =1
∴
(x2 - 2) = -(x +1) f (x) + (x + 2)g(x)
(2) ( ) ( 1)(4 2 14 ) f x = x - x3 + x2 - x - y ，
g(x) = (x -1)(2x2 + x - 4)
( 1) ( ) 1 = x - f x 1 = (x -1)g (x)
而
1 1
1
( ) ( ) 2 -3(2 3)
( ) (2 3) ( 1)
f x g x x x
g x x x
= × +
= + × -
∴ 1 1 1 1
2 1
1 (2 3)( 1) ( )( 1)
3 3
x
= x + x - - g = y - f x - - g
∵
2 1 2 2
1 ( 1) ( ) ( 1) ( )
3 3 3
x - = - x - f x + x - x - g x
(3) ( ) 4 4 1 f x = x4 - x3 - x2 + x + ,
g(x) = x2 - x -1
∴
f (x) = g(x)(x2 -3) + (x - 2) , g(x) = (x - 2)(x +1) +1
∵
1 = -( f - g(x2 - 3))(x +1) + g
= -(x +1) f (x) + (x3 + x2 - 3x - 2)g(x) .
P45.7
f (x) = g(x)1+ (1+ t)x2 + (2 -t)x + u = r(x)
2 2
2 2 2
1 2 ( ) ( 2) 2
( ) ( )( ) (1 )
1 (1 ) (1 ) ( 1)
t t t u t t
g x r x x x u
t t t t
= + - + + + + - + - -
+ + + +
由题意r(x)与g(x)的公因式应为二次所以r(x) | g(x)
∴
. .
.
. .
.
.
. .
.
. .
.
.
=
+
+ +
=
+
+ - + + -
0
(1 )
3
0
(1 )
3 ( 3) (4 )
2
2
2
3 2
u
t
t t
t
t t u t u
∴ ..
.. .
..
.. .
+ + =
+ - + + - =
.
1 -
( 3) 0
3 ( 3) (4 ) 0
1 ( )
2
3 2
t t u
t t u t u
t 否则r x 为一次的
解出(ⅰ)当0 3 3 4 0( 4)( 1) u = 时t 3 + t 2 - t + = t + t 2 - t +
∴
.
3
2
1 3.
4
±p t = - 或t = ± = e
(ⅱ) 1 3
1
0 , 2 3 0, t
t
u t t = -
+
当1 时只有+ + =
( 3 3 4)
1 3
3 3 4
3 ( 3) (4 ) 3 2
3
3 2
2
= - + - +
+
+ - + + - . = + - + t t t
t
t
t t t
t t u t u u
∴
[( 3)( 2 8) 6 24] 2( 4)
3
= - 1 2 + + 2 + - + + = - + u t t t t t t
即 .
.
.
. . .
+ + =
= - +
3 0
2( 4)
t 2 t
u t
2
= -1± -11 t
P45、8d(x) | f (x), d(x) | g(x)表明d (x)是公因式
又已知：d(x)是f (x)与g(x)的组合 表明任何公因式整除d (x)
所以 d (x) 是一个最大的公因式。
P45，9. 证明( f (x)h(x), g(x)h(x)) = ( f (x), g(x)h(x))（h(x)的首系=1）
证：设( f (x)h(x), g(x)h(x)) = m(x)由
( f (x), g(x))h(x) | f (x)h(x) ( f (x), g(x))h(x) | g(x)h(x).
\( f (x), g(x))h(x) | m(x) \( f (x), g(x))h(x)是一个公因式。
设d(x) = ( f (x), g(x)) = u(x) f (x) + v(x)g(x).
\d(x)h(x) = ( f (x), g(x))h(x) = u(x) f (x)h(x) + v(x)g(x)h(x).
而首项系数=1，又是公因式得（由P45、8），它是最大公因式，
且
( f (x), g(x))h(x) = ( f (x)h(x), g(x)h(x)).
P45、10 已知f (x), g(x)不全为0。证明
( ) ( )
( , ) 1.
( ( ), ( )) ( ( ), ( ))
f x g x
f x g x f x g x
=
证：设d(x) = ( f (x), g(x)).则d(x) 1 0.
设 1
( )
( ),
( )
f x
f x
d x
=
1
( )
( ),
( )
g x
g x
d x
=
及d(x) = u(x) f (x) + v(x)g(x).
所以1 1 d(x) = u(x) f (x)d(x) + v(x)g (x)d(x).
消去d (x) 1 0得1 1 1 = u(x) f (x) + v(x)g (x)
P45.11 证：设1 1 ( f (x), g(x)) = d(x) 1 0, f (x) = f (x)d(x), g(x) = g (x)d(x)
∴ 1 1 1 1 u(x) f (x)d(x) + u(x)g (x)d(x) = d(x),u(x) f (x) + u(x)g (x) =1
P45.12
设
2
1 1 1 1 1 1 uf + vg =1, u f + v h =1 .uu f + ufv h + vgu f + vu gh =1
∴ 1 1 1 1 (uu f + uv h + vgu ) f + (v u)gh =1.( f , gh) =1.
P45.13
∵ ( ,g ) 1 i i f = ，
固定1 2 : ( , ) 1 i i f g g =
1 2 ( , . . ) 1 i n f g g g =
P45.14
( f , g) =1.uf + vg =1.(u - v) f + v(g + f ) = 1.( f , g + f ) =1
同理(g, g + f ) =1.
由12 题( fg, f + g) = 1
令 1 2 n g = g g .g
, ( , ) 1 i \每个i f g =
1 1 .( f f , g) =1,
. 1 2 3 ( f f f , g) =1,
. 1 2 1 2 ( , ) 1 m n f f . f g g .g = （注反复归纳用12 题）。
推广
若( f (x), g(x)) =1,则"m，n 有( ( ) , ( ) ) 1 f x m g x n =
P45,15
f(x)=x3+2x2+2x+1, g(x)=x4+x3+2x2+x+1
解：g(x)=f(x)(x-1)+2(x2+x+1),
f(x)=(x2+x+1)(x+1)
即(f(x)，g(x)) = x2+x+1.
令(x2+x+1)=0 得2
1 3
,
2
1 3
1 2
e = - + i e = - - i
∴f(x)与g(x)的公共根为1 2 e ,e .
P45.16 判断有无重因式
①
f (x) = x5 - 5 x4 + 7x3 + 2x2 + 4x -8 ② ( ) 4 4 3 f x = x4 + x2 - x -
解① '( ) 5 20 21 4 4 f x = x4 - x3 + x2 - x +
5 f (x) = f '(x)(x -1) - 3(2x3 - 5x24x +12)
2
3 2 2 15 49
'( ) (2 5 4 12)(5 ) ( 4 4)
2 2
f x = x - x - x + x - + x - x +
(2x3 - 5x2 - 4x +12) = (x2 - 4x + 4)(2x + 3)
故f (x)有重因式
(x - 2)3
② '( ) 4 8 4 f x = x3 + x -
f (x) = (x3 + 2x -1)x + (2x2 - 3x + 3)
f '(x) = (2x2 - 3x + 3)(2x + 3) + (11x -13)
2 6 6 13
11(2 3 3) (11 13)(2 ) (33 )
11 11
x x x x
- + = - - + + ′
\( f (x). f '(x)) =1
P45.17 ? ( ) 3 1 t = 时f x = x3 - x2 + tx - 有重因式（有重根）
解. f '(x) = 3x - 6x + t 2
3 f (x) = f '(x)(x -1) + (2t - 6)x + (t - 3)
如t = 3则有重因式：3重因式( 1) ( ) x - 3 = f x
如t 1 3.则
)
2
15
) (2
2
15
2 f '(x) = (2x + 2)(3′ - + t +
此时必须4
15 t = -
有重因式
) ( 4)
2
1
f (x) = (x + 2 x -
P45.18 求多项式f x = x + px + q ( ) 3 有重根因式的条件
证 f x = x + p ( ) 3 2
3 f (x) = (3x2 + p)x + 2 px + 3q ( p 1 0)
2
2
2 2
3 3 27
(3 ) (2 3 )( ) ( )
2 4 4
a q
x p px q x p
p p p
+ = + - + +
\4 p3 + 27q2 = 0
p45.19 令
f (x) = Ax4 + Bx2 +1,因为(x -1)2 | f (x),所以(x -1) f '(x))
即 '( ) 4 2 ( 1)( ) f x = Ax3 + Bx = x - ax2 + bx + c
4
0
2
0
a A
b a
c b B
c
= .
. - = .
.
- = .
. . - = 2 0
4 2
\ + =
\ = = = -
A B
A a b B
又( 1) ( ) ( ) ( 1)( ' ' ' ') x - f x . f x = x - a x3 + b x2 + c x + d
'
' ' 0
' ' 0
' 1
a A
b a
c b
d
= .
. - = .
.
- = .
. .- = b B
b a
a A
- - =
=
\ =
1 '
' '
'
\A + B +1 = 0
\A = 1,B = -2
P46，20 证
2
( ) 1
2! !
x xn
f x x
n
= + + +.
无重因式（重根）
证：
'( ) ( )
!
xn
f x f x
n
= -
( ', ) ( , )
!
xy
f f f
n
\ = ∵( f , x) =1) ( f , xn ) =1. f (x)无重因式.
P46，21
g′(x)=
1
2 [ f′(x)+ f′(a)]+
( ) ( )
2
x a
f x f x
- ￠ - ￠ .
g′(a)=0
又g(a)=0
// 1 1 1
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0
2 2 2 2
x a
g x f x f x f x f x x a f x g a
￠ = ￠￠ + ￠ + - ￠￠ - ￠ = - . ￠ =
/// (4)
/ // ///
1
( ) ( )
2 2
( ), ( ), ( ), ( )
x a
g x f x f x
a x g x g x g x
= ￠￠ + -
\
\
// 是g 根,且使g (x)的k+1重根
a是g(x)的k+3重根.
P46，22
“ü”必要性显然（见定理6 推论1）
“.”若x0 是f(x)的t 重根，t＞k，
由定理.f(k)( x0)=0
若t＜k. ( 1)
0 f k- (x ) 1 0，所以矛盾.
P46.23
例如f (x) = xm+1,则x = 0是f ￠(x) = (m+1)xm的m重根
但x = 0不是f (x)的根.
P46.24 若( 1) ( ) ( 1) | ( ) x - f x n则xn - f xn
证若f (x) = (x -1)g(x) + r(由上节课命题2)
f (xn ) = (xn -1)g(xn ) + r = g(x) + r .r = 0
所以1| ( ) xn - f xn
P46,25
证明设 x2+x+1 的两个根3
1 2 , , 1 i e e e =
2
1 2
3 3
1 1 1 2 1
3 3
2 2 2 2 2
1 ( )( )
( ) ( ) 0
( ) ( ) 0
x x x x
f f
f f
e e
e e e
e e e
+ + = - -
.. + = \.
+ = ..
1 1 2
1 2 2
(1) (1) 0
(1) (1) 0
f f
f f
e
e
+ = .
.
. + =
即
2 1 . f (1) = 0 f (1)= 0
1 2 3 4 \ f (x) = l (x) + 2l (x) + 5l (x) +10l (x) = x +1
P49.补14）， f (x)..[x], f (0), f (1)奇,则f(x)无整数根.
证:反设f(x)有整数数根m,则x-m|f(x),
f(0)奇.-m奇f(1)奇.1-m奇矛盾!
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一根圆柱形柱子，底面周长是12.56m，高3m，体积是多少立方米
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已知线段ab=cd,且彼此重合各自的三分之一，m,n分别为ab,cd的中点，若mn=14,求ab的长。
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在快对作业上查，就可以找到原创新课堂六年级数学答案了。
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这样可以不？其实甲乙丙三者高相等，只要比较底边就可以了，乙底边等于甲的加上丙的 请你告诉我哪里是BE 你知道答案吗
❻ 人教版初一数学上册原创新课堂45.46面的答案。给的话，悬赏分送给你
计算（2x的三次方-3x的二次方y-2xy的二次方）-（x的三次方-2xy的二次方回+y的三次方）答+（-x的三次方+3x的三次方y-y的三次方）的值，其中x=0.5，y=-1时，甲同学把x=0.5错抄成x=-0.5,但他计算的结果也是正确的，是说明理由，并求这个结果。。 要全部过程 完整的！！！！！！
❼ 试卷数学原创新课堂五年级上册47,48页答案
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