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1. Chuyªn ®Ò : §a thøc
1. Chuyªn ®Ò : §a thøc1. Chuyªn ®Ò : §a thøc
1. Chuyªn ®Ò : §a thøc


  
− + − +
 
!
"
" # 
− + − +

$
   
    
− + − + + − +
#
%&
"   
− + − + − + −


(
"

" " " " " " "
− − +
)
"
"  "  "
− −


(
)
(
)
− + −
*+,
=

()*+
=
!-.(
"
− + +
,)
− +

/0-12"

(
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(
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"

+ + − − +

(
)
"
+ − +
/

(
)
(
)
+ − − −
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
(
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(
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(
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(
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(
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(
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  
− − + − − + − − = + + −
,-.2
2

(
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
+ + − = −
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
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
(
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= + +
,
(
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(
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= + +
,
(
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(
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= + +
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− − + − − + − − +
(
9000-.
= + +

$9"=1,!21$6.-0
1F?<9-1>*?-9D!-9)+F-!
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$9"21,!#21
G9=!
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2
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
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4+
2.
2. 2.
2. Chuyªn ®
Chuyªn ®Chuyªn ®
Chuyªn ®Ò
ÒÒ
Ò:
: :
: BiÓn ®æi biÓu thøc nguyªn
BiÓn ®æi biÓu thøc nguyªnBiÓn ®æi biÓu thøc nguyªn
BiÓn ®æi biÓu thøc nguyªn
I. Mét sè h»ng ®¼ng thøc c¬ b¶n
1.(a ± b)
2
= a
2
± 2ab + b
2
;
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ca ;
2
1 2 n
(a a ... a )
=
−
+ + + + + + + + + + + +
2 2 2
1 2 n 1 2 1 3 1 n 2 3 2 n n 1 n
a a ... a 2(a a a a ... a a a a ... a a ... a a )
;
2.
(a ± b)
3
= a
3
± 3a
2
b + 3ab
2
± b
3
= a
3
± b
3
± 3ab(a ± b);
(a ± b)
4
= a
4
± 4a
3
b + 6a
2
b
2
± 4ab
3
+ b
4
;
3.
a
2
– b
2
= (a – b)(a + b) ;
a
3
– b
3
= (a – b)(a
2
+ ab + b
2
) ;
a
n
– b
n
= (a – b)(a
n – 1
+ a
n – 2
b + a
n – 3
b
2
+ … + ab
n – 2
+ b
n – 1
) ;
4.
a
3
+ b
3
= (a + b)(a
2
– ab + b
2
)
a
5
+ b
5
= (a + b)(a
4
– a
3
b + a
2
b
2
– ab
3
+ b
5
) ;
a
2k + 1
+ b
2k + 1
= (a + b)(a
2k
– a
2k – 1
b + a
2k – 2
b
2
– … + a
2
b
2k – 2
– ab
2k – 1
+ b
2k
) ;
II. B¶ng c¸c hÖ sè trong khai triÓn (a + b)
n
– Tam gi¸c Pascal
§Ønh 1
Dßng 1 (n = 1) 1 1
Dßng 2 (n = 2) 1 2 1
Dßng 3 (n = 3) 1 3 3 1
Dßng 4 (n = 4) 1 4 6 4 1
Dßng 5 (n = 5) 1 5 10
10
5 1
Trong tam gi¸c n#y, hai c¹nh bªn gåm c¸c sè 1 ; dßng k + 1 ®−îc th#nh lËp tõ
dßng k (k
≥
1), ch¼ng h¹n ë dßng 2 ta cã 2 = 1 + 1, ë dßng 3 ta cã 3 = 2 + 1, 3 = 1 +
2, ë dßng 4 ta cã 4 = 1 + 3, 6 = 3 + 3, 4 = 3 + 1, …Khai triÓn (x + y)
n
th#nh tæng th×
c¸c hÖ sè cña c¸c h¹ng tö l# c¸c sè trong dßng thø n cña b¶ng trªn. Ng−êi ta gäi b¶ng
trªn l# tam gi¸c Pascal, nã th−êng ®−îc sö dông khi n kh«ng qu¸ lín. Ch¼ng h¹n, víi
n = 4 th× :
(a + b)
4
= a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
v# víi n = 5 th× :
(a + b)
5
= a
5
+ 5a
4
b + 10a
3
b
2
+ 10a
2
b
3
+ 10ab
4
+ b
5
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3
II. C¸c vÝ dô
VÝ dô 1
. §¬n gi¶n biÓu thøc sau :
A = (x + y + z)
3
– (x + y – z)
3
– (y + z – x)
3
– (z + x – y)
3
.
Lêi gi¶i
A = [(x + y) + z]
3
– [(x + y) – z]
3
– [z – (x – y)]
3
– [z + (x – y)]
3
= [(x + y)
3
+ 3(x + y)
2
z + 3(x + y)z
2
+ z
3
] – [(x + y)
3
– 3(x + y)
2
z + 3(x + y)z
2
–
z
3
] – [z
3
– 3z
2
(x – y) + 3z(x – y)
2
– (x – y)
3
] – [z
3
+ 3z
2
(x – y) + 3z(x – y)
2
+ (x – y)
3
]
= 6(x + y)
2
z – 6z(x – y)
2
= 24xyz
VÝ dô 2
. Cho x + y = a, xy = b (a
2
≥
4b). TÝnh gi¸ trÞ cña c¸c biÓu thøc sau :
a) x
2
+ y
2
; b) x
3
+ y
3
; c) x
4
+ y
4
; d) x
5
+ y
5
Lêi gi¶i
a)
x
2
+ y
2
= (x + y)
2
– 2xy = a
2
– 2b
b)
x
3
+ y
3
= (x + y)
3
– 3xy(x + y) = a
3
– 3ab
c)
x
4
+ y
4
= (x
2
+ y
2
)
2
– 2x
2
y
2
= (a
2
– 2b)
2
– 2b
2
= a
4
– 4a
2
b + 2b
2
d)
(x
2
+ y
2
)(x
3
+ y
3
) = x
5
+ x
2
y
3
+ x
3
y
2
+ y
5
= (x
5
+ y
5
) + x
2
y
2
(x + y)
Hay : (a
2
– 2b)(a
3
– 3ab) = (x
5
+ y
5
) + ab
2
⇒ x
5
+ y
5
= a
5
– 5a
3
b + 5ab
2






















VÝ dô 3
. Chøng minh c¸c h»ng ®¼ng thøc :
a)
a
3
+ b
3
+ c
3
– 3abc = (a + b + c)(a
2
+ b
2
+ c
2
– ab – bc – ca) ;
b)
(a + b + c)
3
– a
3
– b
3
– c
3
= 3(a + b)(b + c)(c + a)
Lêi gi¶i
a)
a
3
+ b
3
+ c
3
– 3abc = (a + b)
3
+ c
3
– 3abc – 3a
2
b – 3ab
2
= (a + b + c)[(a + b)
2
– (a + b)c + c
2
] – 3ab(a + b + c)
= (a + b + c) [(a + b)
2
– (a + b)c + c
2
– 3ab]
= (a + b + c)(a
2
+ b
2
+ c
2
– ab – bc – ca)
b)
(a + b + c)
3
– a
3
– b
3
– c
3
= [(a + b + c)
3
– a
3
] – (b
3
+ c
3
)
= (b + c)[(a + b + c)
2
+ (a + b + c)a + a
2
] – (b + c)(b
2
– bc + c
2
)
= (b + c)(3a
2
+ 3ab + 3bc + 3ca) = 3(b + c)[a(a + b) + c(a + b)]
= 3(a + b)(b + c)(c + a)
VÝ dô 4.
Cho x + y + z = 0.
Chøng minh r»ng : 2(x
5
+ y
5
+ z
5
) = 5xyz(x
2
+ y
2
+ z
2
)
Lêi gi¶i
V× x + y + z = 0 nªn x + y = –z ⇒ (x + y)
3
= –z
3
Hay x
3
+ y
3
+ 3xy(x + y) = –z
3
⇒ 3xyz = x
3
+ y
3
+ z
3
Do ®ã : 3xyz(x
2
+ y
2
+ z
2
) = (x
3
+ y
3
+ z
3
)(x
2
+ y
2
+ z
2
)
= x
5
+ y
5
+ z
5
+ x
3
(y
2
+ z
2
) + y
3
(z
2
+ x
2
) + z
3
(x
2
+ y
2
)
M# x
2
+ y
2
= (x + y)
2
– 2xy = z
2
– 2xy (v× x + y = –z). T−¬ng tù :
y
2
+ z
2
= x
2
– 2yz ; z
2
+ x
2
= y
2
– 2zx.
V× vËy : 3xyz(x
2
+ y
2
+ z
2
) = x
5
+ y
5
+ z
5
+ x
3
(x
2
– 2yz) + y
3
(y
2
– 2zx) + z
3
(z
3
–
2xy) = 2(x
5
+ y
5
+ z
5
) – 2xyz(x
2
+ y
2
+ z
2
)
Suy ra : 2(x
5
+ y
5
+ z
5
) = 5xyz(x
2
+ y
2
+ z
2
) (®pcm)
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4
B*i tËp:
1.
Cho a + b + c = 0 v# a
2
+ b
2
+ c
2
= 14.
TÝnh gi¸ trÞ cña biÓu thøc : A = a
4
+ b
4
+ c
4
.
2.
Cho x + y + z = 0 v# xy + yz + zx = 0. TÝnh gi¸ trÞ cña biÓu thøc :
B = (x – 1)
2007
+ y
2008
+ (z + 1)
2009
.
3.
Cho a
2
– b
2
= 4c
2
. Chøng minh r»ng : (5a – 3b + 8c)(5a – 3b – 8c) = (3a – 5b)
2
.
4.
Chøng minh r»ng nÕu:
5.
(x – y)
2
+ (y – z)
2
+ (z – x)
2
= (x + y – 2z)
2
+ (y + z – 2x)
2
+ (z + x – 2y)
2
th× x = y = z.
6.
a) Chøng minh r»ng nÕu (a
2
+ b
2
)(x
2
+ y
2
) = (ax + by)
2
v# x, y kh¸c 0 th×
a b
x y
.
b) Chøng minh r»ng nÕu (a
2 + b2 + c2)(x2 + y2 + z2) = (ax + by + cz)2
v# x, y, z kh¸c 0 th×
a b c
x y z
.
7.Cho x + y + z = 0. Chøng minh r»ng :
a)5(x3 + y3 + z3)(x2 + y2 + z2) = 6(x5 + y5 + z5) ;
b)x7 + y7 + z7 = 7xyz(x2y2 + y2z2 + z2x2) ;
c)10(x7 + y7 + z7) = 7(x2 + y2 + z2)(x5 + y5 + z5).
8.Chøng minh c¸c h»ng ®»ng thøc sau :
a)(a + b + c)2 + a2 + b2 + c2 = (a + b)2 + (b + c)2 + (c + a)2 ;
b)x4 + y4 + (x + y)4 = 2(x2 + xy + y2)2.
9.Cho c¸c sè a, b, c, d tháa m`n a2 + b2 + (a + b)2 = c2 + d2 + (c + d)2.
Chøng minh r»ng : a4 + b4 + (a + b)4 = c4 + d4 + (c + d)4
10. Cho a2 + b2 + c2 = a3 + b3 + c3 = 1.
TÝnh gi¸ trÞ cña biÓu thøc : C = a2 + b9 + c1945.
11. Hai sè a, b lÇn l−ît tháa m`n c¸c hÖ thøc sau :
a3 – 3a2 + 5a – 17 = 0 v# b3 – 3b2 + 5b + 11 = 0. H`y tÝnh : D = a + b.
12. Cho a3 – 3ab2 = 19 v# b3 – 3a2b = 98. H`y tÝnh : E = a2 + b2.
13. Cho x + y = a + b v# x2 + y2 = a2 + b2. TÝnh gi¸ trÞ cña c¸c biÓu thøc sau :
a) x3 + y3 ; b) x4 + y4 ; c) x5 + y5 ; d) x6 + y6 ;
e) x7 + y7 ; f) x8 + y8 ; g) x2008 + y2008.
3. Chuyªn ®Ò:
3. Chuyªn ®Ò:3. Chuyªn ®Ò:
3. Chuyªn ®Ò:
Ph©n tÝch ®a thøc thnh nh©n tö
Ph©n tÝch ®a thøc thnh nh©n töPh©n tÝch ®a thøc thnh nh©n tö
Ph©n tÝch ®a thøc thnh nh©n tö
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5
I. Ph−¬ng ph¸p t¸ch mét h¹ng tö th*nh nhiÒu h¹ng tö kh¸c:
Bi 1:
Ph©n tÝch c¸c ®a thøc sau th#nh nh©n tö
2 2
2 2
2 2
2 2
2 2
, 5 6 d , 1 3 3 6
, 3 8 4 e , 3 1 8
, 8 7 f, 5 2 4
, 3 1 6 5 h , 8 3 0 7
, 2 5 1 2 k , 6 7 2 0
a x x x x
b x x x x
c x x x x
g x x x x
i x x x x
− + − +
− + + −
+ + − −
− + + +
− − − −
Bi 2
: Ph©n tÝch c¸c ®a thøc sau th#nh nh©n tö:
(§a thøc ® cho cã nhiÖm nguyªn hoÆc nghiÖm h÷u tØ)
II. Ph−¬ng ph¸p thªm v* bít cïng mét h¹ng tö
1)
D¹ng 1
:
Thªm bít cïng mét h¹ng tö lm xuÊt hiÖn h»ng ®¼ng thøc hiÖu cña hai
b×nh ph−¬ng: A
2
– B
2
= (A – B)(A + B)
Bi 1
: Ph©n tÝch c¸c ®a thøc sau th#nh nh©n tö:
2
) D¹ng 2
:
Thªm bít cïng mét h¹ng tö lm xuÊt hiÖn thõa sè chung
Bi 1
: Ph©n tÝch c¸c ®a thøc sau th#nh nh©n tö:
3 2 3
3 2 3
3 2 3 2
3 2 3 2
1, 5 8 4 2, 2 3
3, 5 8 4 4, 7 6
5, 9 6 16 6, 4 13 9 18
7, 4 8 8 8, 6 6 1
x x x x x
x x x x x
x x x x x x
x x x x x x
− + − + −
+ + + − +
− + + − + −
− − + − − + +
3 2 3
3 3 2
3 2 3 2
3 3
9, 6 486 81 10, 7 6
11, 3 2 12, 5 3 9
13, 8 17 10 14, 3 6 4
15, 2 4 16, 2
x x x x x
x x x x x
x x x x x x
x x x
− − + − −
− + − + +
+ + + + + +
− −
2
3 2 3 2
3 2 3 2
3 2 4 3 2
12 17 2
17, 4 18, 3 3 2
19, 9 26 24 20, 2 3 3 1
21, 3 14 4 3 22, 2 1
x x
x x x x x
x x x x x x
x x x x x x x
− + −
+ + + + +
+ + + − + −
− + + + + + +
(
)
2
2 2 2 2
4 4
4 4
4 4 4
4 4 4 2
1, (1 ) 4 (1 ) 2, 8 36
3, 4 4, 64
5, 64 1 6, 81 4
7, 4 81 8, 64
9, 4 10,
x x x x
x x
x x
x x y
x y x x
+ − − − +
+ +
+ +
+ +
+ + +
1