Vũ Chí Công
Giới thiệu về bản thân
Ta có
f
(
x
)
=
100
x
100
x
+
10
⇒
⎧
⎨
⎩
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
{
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
Thế
a
+
b
=
1
a
+
b
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇔
f
(
a
)
+
f
(
b
)
=
1
Ta có
f
(
x
)
=
100
x
100
x
+
10
⇒
⎧
⎨
⎩
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
{
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
Thế
a
+
b
=
1
a
+
b
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇔
f
(
a
)
+
f
(
b
)
=
1
Ta có
f
(
x
)
=
100
x
100
x
+
10
⇒
⎧
⎨
⎩
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
{
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
Thế
a
+
b
=
1
a
+
b
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇔
f
(
a
)
+
f
(
b
)
=
1
Ta có
f
(
x
)
=
100
x
100
x
+
10
⇒
⎧
⎨
⎩
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
{
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
Thế
a
+
b
=
1
a
+
b
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇔
f
(
a
)
+
f
(
b
)
=
1
Ta có
f
(
x
)
=
100
x
100
x
+
10
⇒
⎧
⎨
⎩
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
{
f
(
a
)
=
100
a
100
a
+
10
f
(
b
)
=
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
⇒
f
(
a
)
+
f
(
b
)
=
100
a
100
a
+
10
+
100
b
100
b
+
10
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
(
100
b
+
10
)
+
100
b
(
100
a
+
10
)
100
b
(
100
a
+
10
)
+
10
(
100
a
+
10
)
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
.100
b
+
100
a
.10
+
100
b
,
100
a
+
100
b
.10
100
b
.100
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
=
100
a
+
b
+
100
a
.10
+
100
b
+
a
+
100
b
.10
100
b
+
a
+
100
b
.10
+
100
a
.10
+
100
Thế
a
+
b
=
1
a
+
b
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇒
100
+
100
a
.10
+
100
+
100
b
.10
100
+
100
b
.10
+
100
a
.10
+
100
=
1
⇔
f
(
a
)
+
f
(
b
)
=
1