BỘ THÔNG TIN VÀ TRUYỀN THÔNG
HỌC VIỆN CÔNG NGHỆ BƯU CHÍNH VIỄN THÔNG
Bài giảng
GIẢI TÍCH HÀM MỘT BIẾN SỐ Biên soạn: PGS. TS. Phạm Ngọc Anh

, x
1
x
2

x
0
∈ Z x
1
, x
2
, ∈ {0, 1, 2, , 9}
k x
n
= 0 ∀n > k
x = x
0
, x
1
x
2
x
k
,
p
x = x
0
, x
1
x

k
2
x
k
p

 
p
.
PTIT
x = x
0
+
x
1
10
+ +
x
y
10
k
,
x = x
0
+
x
1
10
+ +
x

∈ Z, x
1
, x
2
, ∈ {0, 1, 2, , 9}.
x
0
x [x]
x
n
n x
m
m ≤ x < m + 1
m x ⌊x⌋
m
m − 1 < x ≤ m
m x ⌈x⌉
x, y ∈ R
x = x
0
, x
1
x
2
y = y
0
, y
1
y
2

, b = b
0
, b
1
b
2
a = b
k ∈ N
a
0
= b
0
, , a
k
= b
k
a
k+1
< b
k+1
a
k+1
> b
k+1
.
a < b a > b
a, b a < b r ∈ Q a < r < b
a = a
0
, a

1
a
k
b
k+1
b ∈ Q
1
2
(a
0
, a
1
a
k
a
k+1
9 + a
0
, a
1
a
k
b
k+1
0) b /∈ Q,
a < r < b
A ⊆ R
m ∈ R A m ≤ a ∀a ∈ A m
A m A
m = inf A

R
C i
i
2
= −1 i
•
C := {z = a + bi : a, b ∈ IR, i
2
= −1}.
PTIT
• z := a + bi a, b ∈ R z
a z z z
b z z z
√
a
2
+ b
2
z |z|
a − bi z ¯z
−a − bi z
• x = x
1
+ x
2
i, y = y
1
+ y
2
i

2
+ x
2
y
1
)i
x = y ⇔



x
1
= y
1
x
2
= y
2
(x + y) + z = x + (y + z) ∀x, y, z ∈ C.
(xy)z = x(yz) ∀x, y, z ∈ C.
x + y = y + x ∀x, y ∈ C
x(y+z) = xy+xz ∀x, y, z ∈
C.
x+ y = ¯x+ ¯y ∀x, y ∈ C
x.y = ¯x.¯y ∀x, y ∈ C
x.¯x = |x|
2
∀x ∈ C
(
x

∆
2a
.
∆ < 0 ∆ = −(−∆) = (i
√
−∆)
2
x
1,2
=
−b+i
√
−∆
2a
.
z =
a−bi
a+bi
a, b ∈ R, b = 0 Im(z) , Re(z), |z|.
a − b i
z =
(a − bi)
2
a
2
− b
2
i
2
=

a
2
+b
2
(z) =
−2ab
a
2
+b
2
|z| =

Re
2
(z) + Im
2
(z) = 1
a
1
, a
2
, , a
n
b
1
, b
2
, , b
n
|

|a
i
|
2
, b =
n

i=1
|b
i
|
2
, c =
n

i=1
a
i
.
¯
b
i
.
PTIT
b = 0 b
1
= b
2
= = b
n

− c.b
i
)(b.¯a
i
− c.
¯
b
i
)
= b
2
n

i=1
|a
i
|
2
− bc
n

i=1
a
i
.
¯
b
i
− bc
n

ϕ
z = a + bi
PTIT
z ∈ C
z
2
z+i
∈ iR z
(Oxy)
z = x + yi
z
2
z + i
=
z
2
(¯z −i)
|z + i|
2
=
1
|z + i|
2
.(x + yi)
2
(x − yi − i)
z
2
z+i
∈ iR Re

2
+ b
2
= |z|



x = |z|. cos ϕ
y = |z|. sin ϕ.
z
z = |z|.(cos ϕ + i sin ϕ),
ϕ z Arg(z) z ∈ C
z = |z|.[cos Arg(z) + i sin Arg(z)]
z
Arg(z)
ϕ z ϕ + k2π (k ∈ Z)
z Arg(z) 2π
Arg(¯z) = −Arg(z) ∀z ∈ C
z = |z|.(cos ϕ + i sin ϕ) ¯z
¯z = |z|.( cos ϕ − i sin ϕ) = |z|.[cos(−ϕ) + i sin( − ϕ)].
PTIT
Arg(¯z) = −ϕ = −Arg(z) ✷
Arg(z
1
.z
2
) = Arg(z
1
) + Arg(z
2

1
|.|z
2
|.(cos ϕ
1
+ i sin ϕ
1
).(cos ϕ
2
+ i sin ϕ
2
)
= |z
1
|.|z
2
|.[cos ϕ
1
cos ϕ
2
− si n ϕ
1
sin ϕ
2
+ i(sin ϕ
1
cos ϕ
2
+ co s ϕ
1

z
1
z
2
) = Arg(z
1
) − Arg(z
2
) ∀z
1
, z
2
∈ C
|z
1
+ z
2
| ≤ |z
1
| + |z
2
| ∀z
1
, z
2
∈ C Arg(z
1
) =
Arg(z
2

Arg(z) = 0 ↔ z ∈ R ↔ ¯z = z z = (
c−b
c−a
)
2
.
a
b
¯z = (
c − b
c − a
)
2
.
a
b
= (
c − b
c − a
)
2
.
¯a
¯
b
=

1
c
−

b
= z.
✷
PTIT
z z = |z|(cos ϕ + i sin ϕ)
z
n
= |z|
n
(cos nϕ + i sin nϕ) ∀n ∈ N
∗
, z ∈ C.
n = 1
n = k z
k
= |z|
k
(cos kϕ + i sin kϕ)
z
k+1
= z
k
.z = |z|
k
(cos kϕ + i sin kϕ).|z|(cos ϕ+i sin ϕ) = |z|
k+1
[cos kϕ cos ϕ−
sin kϕ sin ϕ+i(sin kϕ cos ϕ+cos kϕ sin ϕ)] = |z|
k+1
[cos(k+1)ϕ+i sin(k+1)ϕ]

10
cos
9
x sin x + + i
10
C
10
10
sin
10
x.
=

C
0
10
cos
10
x − C
2
10
cos
8
x sin
2
x + − C
10
10
sin
10

8
x sin
2
x + − C
10
10
sin
10
x
PTIT
cos 10x = C
1
10
cos
9
x sin x − C
3
10
cos
7
x sin
3
x + + C
9
10
cos x sin
9
x.
✷
n

0
+ i sin ϕ
0
), z
n
0
= z
|z|(cos ϕ + i sin ϕ) = [|z
0
|(cos ϕ
0
+ i sin ϕ
0
)]
n
.
|z|(cos ϕ + i sin ϕ) = |z
0
|
n
(cos nϕ
0
+ i sin nϕ
0
).






−1 C
−1 = cosπ + i sin π −1
z
k
= cos
π + k2π
3
+ i sin
π + k2π
3
k = 0, 1, 2.
−1
k = 0 ⇒ z
0
= cos
π
3
+ i sin
π
3
=
√
3
2
+
1
2
i,
k = 1 ⇒ z
1

=
n

k=1
cos(a + kb), B
n
=
n

k=1
sin(a + kb) a, b ∈ R, b /∈ 2πZ.
A
n
+ iB
n
=
n

k=1
[cos(a + kb) + i sin(a + kb)] =
n

k=1
e
i(a+kb)
= e
ai
n

k=1

i(a+b)

e
ib

n
− 1
e
ib
− 1
=
(cos(a + b) + i sin(a + b))[(cos b + i sin b)
n
− 1]
cos b + i sin b − 1
= cos(a + b +
nb
2
).
sin
nb
2
sin
b
2
+ i sin( a + b +
nb
2
).
sin

.
b /∈ 2πZ sin
b
2
= 0
A
n
B
n
sin
b
2
→
f : C → C
f(z) + zf(−z) = 1 + z z ∈ C.
PTIT
z −z
f(−z) − zf(z) = 1 − z.
f(−z)
(1 + z
2
)f(z) = 1 + z
2
.
z = i z = i
f(z) f(i) = α + iβ α, β ∈ R
z = −i z = −i
f(i) + if(−i) = 1 + i ⇔ if(−i) = 1 + i −α −βi ⇔ f(−i) = 1 −β + (α −1)i.
z = +i f(z) = 1
f(z)

2
− 1
x
2
+ 1

2
+

2x
x
2
+ 1

2
= 1,
PTIT
z = a + bi a
2
+ b
2
= 1, x ∈ R



a =
x
2
−1
x

x − i
.
✷
z ∈ C
z +
1
z
= 2 cos ϕ.
z
n
+
1
z
n
= 2 cos nϕ.
z +
1
z
= 2 cos ϕ ⇔ z
2
− 2z cos ϕ + 1 = 0 ⇔ z = cos ϕ+i sin ϕ.
z = cos ϕ + i sin ϕ ⇒ |z| = 1 ⇒ z¯z = 1 ⇒
1
z
= ¯z = cos ϕ −
i sin ϕ ⇒
1
z
n
= (¯z)

cos α + i sin α
cos α − i sin α

n
=
cos nα + i sin nα
cos nα − i sin nα
=
1 + i tan nα
1 − i tan nα
.
✷
z ∈ C
ix x ∈ R
z
3
+ (1 − 2i)z
2
+ (1 − i)z − 2i = 0.
PTIT
ix ix
−ix
3
−(1 −2i)x
2
+ ( 1 −i) i x −2i = 0 ⇔ (−x
2
+ x ) + (−x
3
+ 2 x





z
1
= i,
z
2
=
1
2

(−1 −

√
17 − 4) + (1 +

√
17 + 4)i

,
z
3
=
1
2

(−1 −


)
(
1
n
) :
1
1
,
1
2
, ,
1
n
,

(1 +
1
n
)
n

: 1 +
1
n
, 1 +
1
2
,
(x
n

) +∞ lim
n→∞
x
n
= +∞
∀M > 0 , ∃n
0
∈ N ∀n ≥ n
0
⇒ x
n
≥ M.
A (x
n
) n
(x
n
) A lim
n→∞
x
n
= A
∀ǫ > 0, ∃n
0
∈ N n ≥ n
0
⇒ |x
n
− A| < ǫ.
lim

] + 1, n > n
0
⇒ |x
n
− 0| =
|a|
n
<
|a|
n
0
<
|a|
|a|
ǫ
= ǫ.
✷
(x
n
) A A
(x
n
) A
1
A
2
A
1
= A
2

A
2
| < ǫ ∀n ≥ n
2
n
0
= max{n
1
, n
2
}



|x
n
− A
1
| < ǫ
|x
n
− A
1
| < ǫ
∀n ≥ n
0
,
PTIT
|A
1

n
= A = ∞, (x
n
)
lim
n→∞
x
n
= +∞, (x
n
)
lim
n→∞
x
n
= −∞, (x
n
)
lim
n→∞
x
n
= A = ∞,
∀ǫ > 0, ∃n
0
∈ N, ∀n ≥ n
0
⇒ |x
n
− A| < ǫ.

n→∞
x
n
= +∞,
∀M > 0, ∃n
0
∈ N ∀n ≥ n
0
⇒ x
n
≥ M.
m = min{x
0
, x
1
, , x
n
0
−1
, M}, x
n
≥ m ∀n ∈ N ✷
lim
n→∞
x
n
= A lim
n→∞
y
n

A ≤ B
a ∈ R, A = lim
n→∞
< a ǫ = a−A > 0
∀ǫ > 0, ∃n
0
∈ N, ∀n ≥ n
0
⇒ |x
n
− A| < ǫ,
x
n
< A + ǫ = a.
A = lim
n→∞
x
n
< a A = lim
n→∞
x
n
∈ (a, b)
PTIT
A > B ǫ =
A−B
2
n
2
, n





a
n
≤ b
n
,
|x
n
− A| < ǫ,
|y
n
− B| < ǫ,
∀n ≥ n
0
⇒



y
n
< B + ǫ =
A+B
2
A − ǫ =
A+B
2
< x

α
α
√
a = 1 + h a > 1
h > 0
α
√
a
n
= (1 + h)
n
=
n

k=0
C
k
n
h
k
≥ 1 + nh +
n(n + 1)
2
h
2
≥
n(n + 1)
2
h
2

n→∞
x
n
= lim
n→∞
z
n
= A,
lim
n→∞
y
n
= A.
ǫ > 0 lim
n→∞
x
n
= lim
n→∞
z
n
= A
n
1
, n
2



|x