Bài tập toán cao cấp Tập 1 part 2 - Pdf 20

1.4. Biˆe

udiˆe
˜
nsˆo
´
ph´u
.
cdu
.
´o
.
ida
.
ng lu
.
o
.
.
ng gi´ac 27
T`u
.
d
´othudu
.
o
.
.
c
z
2

ϕ
2

cos


ϕ
2

+ i sin


ϕ
2

= cos ϕ + i sin ϕ. 
V´ı d u
.
3. 1) T´ınh (

3+i)
126
2) T´ınh acgumen cu

asˆo
´
ph´u
.
c sau
w = z

.
c:
(

3+i)
126
=2
126

cos
126π
6
+ i sin
126π
6

=2
126
[cos π + i sin π]=−2
126
.
2) Ta c´o
w = z
4
−z
2
= cos 4ϕ + i sin 4ϕ −[cos 2ϕ −i sin 2ϕ]
= cos 4ϕ − cos 2ϕ + i(sin 4ϕ + sin 2ϕ)
= −2 sin 3ϕ sin ϕ +2i sin 3ϕ cos ϕ
= 2 sin 3ϕ[−sin ϕ + i cos ϕ].

c l`a khi
(2k − 1)π
3
<ϕ<
2kπ
3
, k ∈ Z )th`ı
w =(−2 sin 3ϕ)[sin ϕ −icos ϕ].
28 Chu
.
o
.
ng 1. Sˆo
´
ph´u
.
c
Ta t`ım da
.
ng lu
.
o
.
.
ng gi´ac cu

a v = sin ϕ − i cos ϕ.Hiˆe

n nhiˆen |v| =1.
Ta t´ınh argv

ynˆe
´
u sin 3ϕ<0th`ı
w =(−2 sin 3ϕ)

cos

ϕ −
π
2

+ i sin

ϕ −
π
2

.
(iii) Nˆe
´
u sin 3ϕ =0⇒ ϕ =

3
⇒ w =0.
Nhu
.
vˆa
.
y
argw =

ϕ −
π
2
nˆe
´
u
(2k − 1)π
3
<ϕ<
2kπ
3
· 
V´ı d u
.
4. Ch´u
.
ng minh r˘a
`
ng
1) cos
π
9
+ cos

9
+ cos

9
+ cos


9
+ ···+ cos

9
,
T = sin
π
9
+ sin

9
+ ···+ sin

9
,
z = cos
π
9
+ i sin
π
9
.
1.4. Biˆe

udiˆe
˜
nsˆo
´
ph´u
.

=
z +1
1 −z
2
=
1
1 − z
=
1

1 −cos
π
9

− i sin
π
9
=

1 −cos
π
9

+ i sin
π
9

1 −cos
π
9

.
.
nhu
.
trong 1) ta k´yhiˆe
.
u
S = cos ϕ + cos(ϕ + α)+···+ cos(ϕ + nα),
T = sin ϕ + sin(ϕ + α)+···+ sin(ϕ + nα),
z = cos α + i sin α, c = cos ϕ + i sin ϕ.
Khi d
´o
S + iT = c + cz + ···+ cz
n
=
c(1 − z
n+1
)
1 −z
=
(cos ϕ + i sin ϕ)[1 −cos(n +1)α − i sin(n +1)α]
1 − cos α −i sin α
=
(cos ϕ + i sin ϕ)2 sin
(n +1)α
2

cos
(n +1)α − π
2

(n +1)α
2
sin

ϕ +

2
sin
α
2
i.
T`u
.
d´o so s´anh phˆa
`
n thu
.
.
c v`a phˆa
`
na

o ta thu du
.
o
.
.
ckˆe
´
t qua

+ ···+ a
n
sin b
n
,
a
1
cos b
1
+ a
2
cos b
2
+ ···+ a
n
cos b
n
30 Chu
.
o
.
ng 1. Sˆo
´
ph´u
.
c
nˆe
´
u c´ac acgumen b
1

.
5. T´ınh tˆo

ng
1) S
n
=1+a cos ϕ + a
2
cos 2ϕ + ···+ a
n
cos nϕ;
2) T
n
= a sin ϕ + a
2
sin 2ϕ + ···+ a
n
sin nϕ.
Gia

i. Ta lˆa
.
pbiˆe

uth´u
.
c S
n
+ iT
n

n
=
a
n+1
z
n+1
− 1
az − 1
(nhˆan tu
.

sˆo
´
v`a mˆa
˜
usˆo
´
v´o
.
i
a
z
−1)
=
a
n+2
z
n
− a
n+1

a
2
− 2a cos ϕ +1
=
a
n+2
cos nϕ − a
n+1
cos(n +1)ϕ − a cos ϕ +1
a
2
− 2a cos ϕ +1
+
+ i
a
n+2
sin nϕ −a
n+1
sin(n +1)ϕ + a sin ϕ
a
2
− 2a cos ϕ +1
·
B˘a
`
ng c´ach so s´anh phˆa
`
n thu
.
.

1.4. Biˆe

udiˆe
˜
nsˆo
´
ph´u
.
cdu
.
´o
.
ida
.
ng lu
.
o
.
.
ng gi´ac 31
2) Biˆe

udiˆe
˜
n tuyˆe
´
n t´ınh sin
5
ϕ qua c´ac h`am sin cu


˜
n sin 5ϕ v`a cos 5ϕ
qua sin ϕ v`a cosϕ. Theo cˆong th´u
.
c Moivre ta c´o
cos 5ϕ + i sin 5ϕ = (cos ϕ + i sin ϕ)
5
= sin
5
ϕ +5i cos
4
ϕ sin ϕ
− 10 cos
3
ϕ sin
2
ϕ −10i cos
2
ϕ sin
3
ϕ
+ 5 cos ϕ sin
4
ϕ + i sin
5
ϕ.
T´ach phˆa
`
n thu
.

3
ϕ + sin
5
ϕ
cos
5
ϕ −10 cos
3
ϕ sin
2
ϕ + 5 cos ϕ sin
4
ϕ
(chia tu
.

sˆo
´
v`a mˆa
˜
usˆo
´
cho cos
5
ϕ)
=
5tgϕ −10tg
3
ϕ +tg
5

k
+ z
−k
= 2 cos kϕ, z
k
−z
−k
=2i sin kϕ.
´
Ap du
.
ng c´ac kˆe
´
t qua

n`ay ta c´o
sin
5
ϕ =

z − z
−1
2i

5
=
z
5
− 5z
3

.
ng 1. Sˆo
´
ph´u
.
c
3) Tu
.
o
.
ng tu
.
.
nhu
.
trong phˆa
`
n 2) ho˘a
.
c gia

i theo c´ach sau d
ˆay
1
+
cos
4
ϕ =

e

1
2

e
2ϕi
+ e
−2ϕi
2

+
3
8
=
3
8
+
1
2
cos 2ϕ +
1
8
cos 4ϕ.
2
+
sin
4
ϕ cos
3
ϕ =



=
1
128

e
6ϕi
− 3e
2ϕi
+3e
−2ϕi
− e
−6ϕi

e
ϕi
−e
−ϕi

=
1
128

e
7ϕi
−e
5ϕi
− 3e
3ϕi
+3e

o
.
ng tr`ınh
1
+
(x +1)
n
−(x − 1)
n
=0
2
+
(x + i)
n
+(x − i)
n
=0, n>1.
2) Ch´u
.
ng minh r˘a
`
ng mo
.
i nghiˆe
.
mcu

aphu
.
o

+
Chia hai vˆe
´
cu

aphu
.
o
.
ng tr`ınh cho (x −1)
n
ta du
.
o
.
.
c

x +1
x −1

n
=1⇒
x +1
x − 1
=
n

1=cos
2kπ

d
´o suy r˘a
`
ng
x +1=ε
k
(x −1) ⇒ x(ε
k
− 1)=1+ε
k
.
Khi k =0⇒ ε
0
= 1. Do d´ov´o
.
i k =0phu
.
o
.
ng tr`ınh vˆo nghiˆe
.
m. V´o
.
i
k =
1,n−1 ta c´o
x =
ε
k
+1

− ε
k
− 1
=
−2i sin
2kπ
n
2 −2 cos
2kπ
n
= −i
sin
2kπ
n
1 −cos
2kπ
n
= icotg

n
,k=1, 2, ,n− 1.
2
+
C˜ung nhu
.
trˆen, t`u
.
phu
.
o

n
·
Ta biˆe
´
ndˆo

iphu
.
o
.
ng tr`ınh:
x + i
x −i
− 1=cosψ + i sin ψ − 1

2i
x −i
=2i sin
ψ
2
cos
ψ
2
−2 sin
2
ψ
2

1
x −i

ng 1. Sˆo
´
ph´u
.
c
T`u
.
d
´o suy ra
x −i =
1
sin
ψ
2

cos
ψ
2
+ i sin
ψ
2

=
cos
ψ
2
− i sin
ψ
2
sin

.
ng tr`ınh d
˜a cho. Ta c´o



1+ai
1 −ai



=1⇒
1+ai
1 − ai
= cos α + i sin α
v`a t `u
.
d
´o
1+xi
1 − xi
=
n

1+ai
1 −ai
= cos
α +2kπ
n
+ i sin

.
.
c kh´ac nhau. 
V´ı d u
.
8. Biˆe

udiˆe
˜
n c´ac sˆo
´
ph´u
.
csaudˆay du
.
´o
.
ida
.
ng m˜u:
1) z =
(−

3+i)

cos
π
12
−i sin
π

i. 1) D˘a
.
t z
1
= −

3+i, z
2
= cos
π
12
− i sin
π
12
, z
3
=1− i v`a
biˆe

udiˆe
˜
n c´ac sˆo
´
ph´u
.
cd´odu
.
´o
.
ida


π
12
i
;
z
3
=

2e

π
4
i
.
T`u
.
d´othudu
.
o
.
.
c
z =
2e

6
i
· e


=

3+i du
.
´o
.
ida
.
ng m˜u. Ta c´o
|z
1
| =2; ϕ = arg(

3+i)=
π
6
,
do d
´o

3+i =2e
π
6
i
.T`u
.
d
´othudu
.
o

.
1) c˘an bˆa
.
c3: w =
3

−2+2i
2) c˘an bˆa
.
c4: w =
4

−4
3) c˘an bˆa
.
c5: w =
5


3 −i
8+8i
.
Gia

i. Phu
.
o
.
ng ph´ap tˆo
´

ng lu
.
o
.
.
ng gi´ac (ho˘a
.
cda
.
ng m˜u) rˆo
`
i
´ap du
.
ng c´ac cˆong th´u
.
ctu
.
o
.
ng ´u
.
ng.
1) Biˆe

udiˆe
˜
n z = −2+2i du
.
´o

3


8

cos

4
+2kπ
3
+ i sin

4
+2kπ
3

,k=
0, 2.
T`u
.
d´o
w
0
=

2

cos
π
4

12

.
2) Ta c´o
−4 = 4[cos π + i sin π]
v`a do d
´o
w
k
=
4

4

cos
π +2kπ
4
+ i sin
π +2kπ
4

,k= 0, 3.
T`u
.
d
´o
w
0
=


cos

4
+ i sin

4

= −1 − i,
w
3
=

2

cos

4
+ i sin

4

=1−i.
3) D
˘a
.
t
z =

3 −i
8+8i

´
ph´u
.
cdu
.
´o
.
ida
.
ng lu
.
o
.
.
ng gi´ac 37
Do vˆa
.
y
w
k
=
5


1
4

2



π
12
+
2kπ
5

,k= 0, 4. 
V´ı d u
.
10. 1) T´ınh tˆo

ng mo
.
i c˘an bˆa
.
c n cu

a1.
2) T´ınh tˆo

ng1+2ε +3ε
2
+ ···+ nε
n−1
, trong d´o ε l`a c˘an bˆa
.
c n
cu

ado

t c´ac c˘an bˆa
.
c n cu

a 1. Ta c´o
ε
k
=
n

1=cos
2kπ
n
+ i sin
2kπ
n
,k= 0,n− 1.
T`u
.
d
´o
ε
0
=1,ε
1
= ε = cos

n
+ i sin


.
mcu

a c˘an bˆa
.
c n cu

a1c´othˆe

viˆe
´
tdu
.
´o
.
ida
.
ng
1,ε,ε
2
, ,ε
n−1
.
Bˆay gi`o
.
ta t´ınh
S =1+ε + ε
2
+ ···+ ε
n−1

n t´ınh l`a S.Tax´et biˆe

uth´u
.
c
(1 −ε)S = S − εS =1+2ε +3ε
2
+ ···+ nε
n−1
− ε − 2ε
2
−···−(n −1)ε
n−1
− nε
n
=1+ε + ε
2
+ ···+ ε
n−1
  
0(ε=1)
−nε
n
= −n
v`ı ε
n
=1.
Nhu
.
vˆa

i
α = 0) mo
.
i c˘an bˆa
.
c n cu

a α c´o thˆe

biˆe

udiˆe
˜
ndu
.
´o
.
ida
.
ng t´ıch β
0
ε
k
,
k =1, 2, ,n − 1, trong d
´o ε
k
= cos
2kπ
n

2
)
k
+ ···+(β
0
ε
n−1
)
k
= β
k
0
(1 + ε
k
1
+ ε
k
2
+ ···+ ε
k
n−1
)

ε
k
m
=

cos
2mπ


.
Biˆe

uth´u
.
c trong dˆa
´
u ngo˘a
.
c vuˆong l`a cˆa
´
psˆo
´
nhˆan. Nˆe
´
u ε
k
1
=1,t´u
.
cl`a
k khˆong chia hˆe
´
tchon th`ı
S = β
k
0
1 −ε
nk

.
o
.
.
ng gi´ac 39
Nˆe
´
u ε
k
1
=1t´u
.
cl`ak chia hˆe
´
tchon, k = nq th`ı
S = β
nq
0
[1+1+···+1]=β
nq
0
n = nα
q
(v`ı β
n
0
= α).
Nhu
.
vˆa

.
csaudˆay du
.
´o
.
ida
.
ng lu
.
o
.
.
ng gi´ac
1) −1+i

3(DS. 2

cos

3
+ i sin

3

)
2)

3 −i (DS. 2

cos

π
6
)
5)


3
2
+
1
2
i (DS. cos

6
+ i sin

6
)
6)
1
2
− i

3
2
(DS. cos

3
+ i sin



)
9) 2 −

3 −i (DS. 2

2 −

3

cos
19π
12
+ i sin
19π
12

)
2. Biˆe

udiˆe
˜
n c´ac sˆo
´
ph´u
.
csaud
ˆay du
.
´o

.
c
3) cos ϕ −i sin ϕ (DS. cos(−ϕ)+i sin(−ϕ))
4) −cos ϕ −i sin ϕ (D
S. cos(π + ϕ)+i sin(π + ϕ))
B˘a
`
ng c´ach d˘a
.
t α = θ +2kπ, trong d´o0 θ<2π, ta c´o:
5) 1+cos α+i sin α (DS. 2 cos
θ
2

cos
θ
2
+i sin
θ
2

v´o
.
i0 θ<π;
−2 cos
θ
2

cos
θ +2π

+ i sin
π − θ
2

v´o
.
i0 θ<π;
−2 cos
θ
2

cos
3π − θ
2
+ i sin
3π − θ
2

v´o
.
i π  θ<2π)
8) −sin α + i(1 + cos α)
(DS. 2 cos
θ
2

cos
π + θ
2
+ i sin

100
(DS. −
1
2
− i

3
2
)
2)

4

3+i

12
(DS. 2
12
)
3)
(

3+i)
6
(−1+i)
8
− (1 + i)
4
(DS. −3, 2)
4)


3)(cos ϕ + i sin ϕ)
2(1 − i)(cos ϕ − i sin ϕ)
(DS.

2
2

cos

6ϕ −
π
12

+ i sin

6ϕ −
π
12

)
1.4. Biˆe

udiˆe
˜
nsˆo
´
ph´u
.
cdu

z
n
= 2 cos nϕ.
5. H˜ay biˆe

udiˆe
˜
n c´ac h`am sau d
ˆay qua sin ϕ v`a cos ϕ
1) sin 3ϕ (D
S. 3 cos
2
ϕ sin ϕ −sin
3
ϕ)
2) cos 3ϕ (D
S. cos
3
ϕ −3 cos ϕ sin
2
ϕ)
3) sin 4ϕ (D
S. 4 cos
3
ϕ sin ϕ −4 cos ϕ sin
3
ϕ)
4) cos 4ϕ (D
S. cos
4

1 −15tg
2
ϕ + 15tg
4
ϕ −tg
6
ϕ
)
7. Ch´u
.
ng minh r˘a
`
ng
1 −C
2
n
+ C
4
n
− C
6
n
+ =2
n
2
cos

4
·
C

.
ng cˆong th´u
.
c Moivre v`a
su
.

du
.
ng cˆong th´u
.
c nhi
.
th ´u
.
c Newton rˆo
`
i so s´anh phˆa
`
n thu
.
.
c v`a phˆa
`
n
a

o c´ac sˆo
´
thu d

1
2
3) cos

5
+ cos

5
= −
1
2
4) cos

7
+ cos

7
+ cos

7
= −
1
2
5) cos

9
+ cos

9
+ cos

cotgα + i
cotgα − i
,n∈ N,α∈ R.
(D
S. x =tg
α + kπ
n
, k = 0,n−1)
10. Ch´u
.
ng minh r˘a
`
ng nˆe
´
u A l`a sˆo
´
ph´u
.
cc´omod
un = 1 th`ı mo
.
i nghiˆe
.
m
cu

aphu
.
o
.

a
2
x
n−2
−···−a
n
=0.
(DS. x
k
=
a
ε
k

2 −1
, k = 0,n− 1)
Chı

dˆa
˜
n. D`ung cˆong th´u
.
c nhi
.
th ´u
.
c Newton dˆe

du
.

2
+ x +1=0.
(D
S. x
k
= cos

3
+ i sin

3
, k =1, 2, 3, 4, 5)
13. Gia

iphu
.
o
.
ng tr`ınh
x
5
+ αx
4
+ α
2
x
3
+ α
3
x

´
nhˆan v´o
.
i cˆong bˆo
.
ib˘a
`
ng
α
x
.
14. Gia

su
.

n ∈ N, n>1, c =0,c ∈ R. Gia

i c´ac phu
.
o
.
ng tr`ınh sau
d
ˆa y
1.4. Biˆe

udiˆe
˜
nsˆo

n
, k = 1,n− 1)
3) (x + ci)
n
+ i(x − ci)
n
=0
(DS. x = −cicotg
(3+4k)π
4n
, k = 0,n−1)
4) (x + ci)
n
− (cos α + i sin α)(x −ci)
n
=0,α=2kπ.
(D
S. x = −cicotg
α +2kπ
2n
,k= 0,n− 1)
15. T´ınh
D
n
(x)=
1


1
2

5
x −10 cos
3
x sin
2
x + 5 cos x sin
4
x,
sin 5x = 5 cos
4
x sin x −10 cos
2
x sin
3
x + sin
5
x.
2) sin

5
=

10 + 2

5
4
, cos

5
=

.
ng 2
D
-
ath´u
.
cv`ah`amh˜u
.
uty

2.1 D
-
ath´u
.
c 44
2.1.1 D
-
ath´u
.
c trˆen tru
.
`o
.
ng sˆo
´
ph´u
.
c C 45
2.1.2 D
-

nv´o
.
ihˆe
.
sˆo
´
thuˆo
.
c tru
.
`o
.
ng sˆo
´
P du
.
o
.
.
cbiˆe

udiˆe
˜
ndo
.
n tri
.
du
.
´o

n
l`a c´ac sˆo
´
;v`amˆo
˜
itˆo

ng da
.
ng (2.1) dˆe
`
u
l`a dath´u
.
c.
K´yhiˆe
.
u: Q(z) ∈P[z].
Nˆe
´
u a
0
,a
1
, ,a
n
∈ C th`ı ngu
.
`o
.

thu
.
.
c: Q(z) ∈ R[z].
2.1. D
-
ath´u
.
c 45
Nˆe
´
u Q(z) =0th`ıbˆa
.
ccu

a n´o (k´y hiˆe
.
u degQ(z)) l`a sˆo
´
m˜u cao nhˆa
´
t
cu

amo
.
i lu˜y th`u
.
acu


´
t.
Nˆe
´
u P (z)v`aQ(z) ∈P[z] l`a c˘a
.
pd
ath´u
.
cv`aQ(z) =0th`ıtˆo
`
nta
.
i
c˘a
.
pd
ath´u
.
c h(z)v`ar(z) ∈P[z] sao cho
1
+
P = Qh + r,
2
+
ho˘a
.
c r(z) = 0, ho˘a
.
c degr<degQ.

ph´u
.
c C
Gia

su
.

Q(z) ∈ C[z]. Nˆe
´
u thay z bo
.

isˆo
´
α th`ı ta thu d
u
.
o
.
.
csˆo
´
ph´u
.
c
Q(α)=a
0
α
n

adath´u
.
c Q(z) hay cu

aphu
.
o
.
ng tr`ınh da
.
isˆo
´
Q(z)=0.
D
-
i
.
nh l´y Descate. D
ath´u
.
c Q(z) chia hˆe
´
t cho nhi
.
th´u
.
c z − α khi v`a
chı

khi α l`a nghiˆe

nˆe
´
u Q(z) chia hˆe
´
tcho(z −α)
m
nhu
.
ng khˆong chia hˆe
´
tcho
(z −α)
m+1
.Sˆo
´
m du
.
o
.
.
cgo
.
il`abˆo
.
i cu

a nghiˆe
.
m α. Khi m = 1, sˆo
´

.
clˆa
.
pnˆenb˘a
`
ng
c´ach gh´ep thˆem v`ao cho tˆa
.
pho
.
.
psˆo
´
thu
.
.
c R mˆo
.
t nghiˆe
.
ma

o x = i cu

a
phu
.
o
.
ng tr`ınh x

.
o thˆem c´ac sˆo
´
m´o
.
idˆe

gia

iphu
.
o
.
ng tr`ınh (v`ı thˆe
´
C c`on du
.
o
.
.
cgo
.
i
l`a tru
.
`o
.
ng d
´ong da
.

.
uty

Mo
.
idath´u
.
cd
a
.
isˆo
´
bˆa
.
c n (n  1) trˆen tru
.
`o
.
ng sˆo
´
ph´u
.
cd
ˆe
`
u c´o ´ıt
nhˆa
´
tmˆo
.

.
cd
ˆe
`
uc´od´ung n
nghiˆe
.
mnˆe
´
umˆo
˜
i nghiˆe
.
mdu
.
o
.
.
c t´ınh mˆo
.
tsˆo
´
lˆa
`
nb˘a
`
ng bˆo
.
icu


+ ···+ m
k
= n.
D
ath´u
.
c (2.1) v´o
.
ihˆe
.
sˆo
´
cao nhˆa
´
t a
0
=1du
.
o
.
.
cgo
.
il`adath´u
.
c thu
go
.
n.
2


adath´u
.
c liˆen ho
.
.
p Q(z), trong d´o d a
th ´u
.
c Q(z)du
.
o
.
.
c x´ac di
.
nh bo
.

i
Q(z)
def
= a
0
z
n
+ a
1
z
n−1

+ ···+ a
n−1
z + a
n
(2.4)
l`a dath´u
.
c quy go
.
nv´o
.
ihˆe
.
sˆo
´
thu
.
.
c a
1
,a
2
, ,a
n
.
D
ath´u
.
c n`ay c´o t´ınh chˆa
´

´
thu
.
.
c th`ı sˆo
´
ph´u
.
c liˆen ho
.
.
pv´o
.
in´oα c˜ung l`a nghiˆe
.
mbˆo
.
i m cu

a
dath´u
.
cd´o.
Su
.

du
.
ng di
.

cv´o
.
ihˆe
.
sˆo
´
thu
.
.
cv´o
.
ibiˆe
´
nchı

nhˆa
.
n gi´a tri
.
thu
.
.
cnˆen biˆe
´
nd
´o t a k ´y
hiˆe
.
ul`ax thay cho z.
2.1. D

.
itu
.
o
.
ng ´u
.
ng β
1

2
, ,β
m
v`a c´ac c˘a
.
p nghiˆe
.
mph´u
.
cliˆen ho
.
.
p a
1
v`a a
1
, a
2
v`a a
2

2
···(x −b
m
)
β
m
(x
2
+ p
1
x + q
1
)
λ
1
×
× (x
2
+ p
2
x + q
2
)
λ
2
···(x
2
+ p
n
x + q

.
ihˆe
.
sˆo
´
cao nhˆa
´
tb˘a
`
ng 1 c´o nghiˆe
.
mh˜u
.
uty

th`ı
nghiˆe
.
md
´o l`a sˆo
´
nguyˆen.
D
ˆo
´
iv´o
.
idath´u
.
cv´o

uty

cu

a phu
.
o
.
ng tr`ınh v´o
.
ihˆe
.
sˆo
´
h˜u
.
uty

a
0
x
n
+a
1
x
n−1
+···+a
n−1
x+
a

0
.
C
´
AC V
´
IDU
.
V´ı d u
.
1. Gia

su
.

P (z)=a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n
.Ch´u
.
ng

du
.
o
.
.
c
p(Z)=a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n
= a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n

.

P (z) ∈ R[z]. Khi d
´o
P (z)=a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n
= a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n
= a
0

c P(z)=P (z)v`ı P (z)=P (z). 
V´ı d u
.
2. Ch´u
.
ng minh r˘a
`
ng nˆe
´
u a l`a nghiˆe
.
mbˆo
.
i m cu

adath´u
.
c
P (z)=a
0
z
n
+ a
1
z
n−1
+ ···+ a
n−1
z + a
n

z + a
n
(go
.
il`adath´u
.
c liˆen ho
.
.
pph´u
.
cv´o
.
idath´u
.
c P(z)).
Gia

i. T`u
.
v´ıdu
.
1 ta c´o
P (z)=P (z). (2.6)
V`ı a l`a nghiˆe
.
mbˆo
.
i m cu


`
ng c´ach lˆa
´
y liˆen ho
.
.
pph´u
.
cmˆo
.
tlˆa
`
nn˜u
.
a ta c´o
Q(a)=Q(a)=0 ⇒ Q(a)=0.
D
iˆe
`
u n`ay vˆo l´y. B˘a
`
ng c´ach d˘a
.
t t = z,t`u
.
(2.8) thu du
.
o
.
.

V´ı d u
.
3. Ch´u
.
ng minh r˘a
`
ng nˆe
´
u a l`a nghiˆe
.
mbˆo
.
i m cu

adath´u
.
cv´o
.
i
hˆe
.
sˆo
´
thu
.
.
c P(z)=a
0
z
n

.
v´ıdu
.
1, 2
+
ta c´o
P (z)=P (z) (2.9)
v`a do a l`a nghiˆe
.
mbˆo
.
i m cu

a n´o nˆen
P (z)=(z − a)
m
Q(z) (2.10)
trong d
´o Q(z)l`adath´u
.
cbˆa
.
c n − m v`a Q(a) =0.
Ta cˆa
`
nch´u
.
ng minh r˘a
`
ng

.
ng minh Q( a) = 0. Thˆa
.
tvˆa
.
yv`ı Q(a) =0nˆen
Q(a) =0v`adod´o Q(a) =0v`ıdˆo
´
iv´o
.
id
ath´u
.
cv´o
.
ihˆe
.
sˆo
´
thu
.
.
cth`ı
Q(t)=Q(t). 
V´ı d u
.
4. Gia

iphu
.

´
nguyˆen dˆe
`
ul`au
.
´o
.
ccu

asˆo
´
ha
.
ng tu
.
.
do a = −3. Sˆo
´
ha
.
ng tu
.
.
do
50 Chu
.
o
.
ng 2. D
-

3
−4z
2
+4z − 3=(z − 3)(z
2
− z +1)
=(z −3)(z −
1
2
+ i

3
2

z −
1
2
− i

3
2

hay l`a phu
.
o
.
ng tr`ınh d˜a cho c´o ba nghiˆe
.
ml`a
z

(z)=z
6
− 3z
4
+4z
2
− 12 du
.
´o
.
ida
.
ng:
1
+
t´ıch c´ac th`u
.
asˆo
´
tuyˆe
´
n t´ınh;
2
+
t´ıch c´ac th`u
.
asˆo
´
tuyˆe
´

− 3z
4
+4z
2
− 12 = (z
2
− 3)(z
4
+4)
nˆen r˜o r`ang l`a
z
1
= −

3,z
2
=

3,z
3
=1+i,
z
4
=1− i, z
5
= −1+i, z
6
= −1 −i.
T`u
.

i c´ac
nghiˆe
.
mph´u
.
c liˆen ho
.
.
pv´o
.
i nhau ta thu du
.
o
.
.
c
P
6
(z)=(z −

3)(z +

3)(z
2
− 2z + 2)(z
2
+2z +2). 
V´ı d u
.
6. T`ım d

.
c 51
Gia

i. V`ıdath´u
.
cchı

c´o hˆe
.
sˆo
´
thu
.
.
cnˆen c´ac nghiˆe
.
mph´u
.
c xuˆa
´
thiˆe
.
n
t`u
.
ng c˘a
.
p liˆen ho
.

(x +1)
n
− (x − 1)
n
th`anh c´ac th`u
.
asˆo
´
tuyˆe
´
n t´ınh.
Gia

i. Ta c´o
P (x)=(x +1)
n
−(x − 1)
n
=[x
n
+ nx
n−1
+ ] −[x
n
− nx
n−1
+ ]=2nx
n−1
+
Nhu

a n´o:
x
k
= icotg

n
,k=1, 2, ,n− 1.
Do d´o
(x +1)
n
− (x − 1)
n
=2n

x −icotg
π
n

x −icotg

n

···

x −icotg
(n −1)π
n

.
Khi phˆan t´ıch dath´u

.
ntrˆenc`ung tru
.
`o
.
ng P d´o. Nh˜u
.
ng dath´u
.
cn`aydu
.
o
.
.
cgo
.
il`ada
th´u
.
cbˆa
´
t kha

quy.
Ch˘a

ng ha
.
n: d
ath´u


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