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8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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Chuyn Ph ng Trnh & H Ph ng Trnh
A. Cc lo i ph ng trnh v h ph ng trnh c bn
I.Ph ng trnh b c nht1.1 Dng : a x+b=0 1.2 Cch gi i:
a 0 : ph ng trnh c m t nghim bxa
=
a=0 :+ : ph ng trnh v nghi mb 0+ : ph ng trnh c nghi m x ty b 0=
1.3 Bi t pBi 1:Gii ph ng trnh
x a x b x c 1 1 12( )
bc ac ab a b c
+ + = + + (*)
Ly VT-VP (*) c phn tch thnh1 1 1
(x a b c)( ) 0bc ac ab
+ + =
Nu1 1 1
0bc ac ab
+ + th ph ng trnh c nghi m l : x a b c= + +
Nu1 1 1
0bc ac ab
+ + = th ph ng trnh trn ng v i mi x
Bi 2:Gii ph ng trnh
a b x a c x c b x 4x1
c b a a b c
+ + + + + ++ +
=
Cng 3 vo 2 v ca ph ng trnh ta c :1 1 1 a b c x
(a b c x)( ) 4a b c a b c
+ + + + + + =+ +
1 1 1 4(a b c x)( ) 0
a b c a b c+ + + + =
+ +
Vy x a b c= + +II. Ph ng trnh b c hai2.1 Dng : 2ax bx c 0(a 0)+ + = 2.2 Cch gi i:
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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a=0 : ph ng trnh suy bi n thnh b c 1a 0 : l p 2b 4ac =
0 < : ph ng trnh v nghi m
0 = : ph ng trnh c nghi m kp 1 2b
x x2a
= =
0 > : ph ng trnh c hai nghi m phn bi t : 1,2bx
2a =
Ch :
Nu 1b
x2a
= v 1b
x2a
+ = th :
khi1x x< 22
a 0>khi1x x> a 0 ph ng trnh b c hai c 2nghim phn bi t
2.3 Hthc Vi-eti) Nu ph ng trnh c 2 nghi m th2ax bx c 0+ + = 1x & x2 1 2 1 2
b cx x & x x
a a+ = =
ii) o li cho 2 s bt k , ,khi chng l nghi m ca ph ng trnh2x Sx P 0 + = v i S= + v P =
nh l 1 : Cho tam th c bc 2 2f(x)=ax bx c+ + i) Nu tm c s af( ) 0 th tam th c c nghi m ,cn n u af( ) 0 <
th tam th c c 2 nghi m phn bi tii) Nu tm c s , sao cho f( )f( ) 0 th tam th c c nghi m ,nu
f( )f( ) 0 < th tam th c c 2 nghi m phn bi t
nh l 2 : ph ng trnh 2ax bx c 0+ + = c nghi m hu t iu kin cn v l bit s l 1 schnh ph ng
nh l 3 : Nu 0p
xq
= l nghi m hu tca ph ng trnh 2ax bx c 0+ + = trong
th q l c ca a v p l c ca(p,q) 1= c2.3 Bi t pBi 1:Gii ph ng trnh
2 2
2x x 33x x 1 3x 4x 1 2
+ = + + (*)
T p xc nh 1R \ 1,3
x 0= : khng l nghi m
x 0 : (*) 2 11 1 23x 1 3x 4x x
+ + +
3=
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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t 1y 3xx
= + ta quy v ph ng trnh b c hai 23y 21y 30 0 + = gii ra ta c
nghim v t tm c nghimy 2= y 5= 5 13x6
=
Bi 2:Gii ph ng trnh
2 2 2
1 1 1x 9x 40 x 11x 30 x 13x 42 18
+ ++ + + + + +
1=
1 1 1 1 1 1x 4 x 5 x 5 x 6 x 6 x 7 18
+ + =+ + + + + +
1
3
2x 11x 26 0 + = Gii ra ta c nghim vx 1= x 2= Bi 3:Gii ph ng trnh
(x a)(x b) (x b)(x c) (x a)(x c ) 1c(c a)(c b) a(a b)(a c ) b(b a)(b c ) x
+ + =
Trong a, l 3 skhc nhau v khc 0b,c
(x a)(x b) (x b)(x c) (x a)(x c) 1
0c(c a)(c b) a(a b)(a c) b(b a)(b c ) x
+ +
=
R rng ta th y l 3 nghi m phn bi t ca ph ng trnh trn . Khi quy ng mu s (k ) ,vtri ph ng trnh s l mt a thc khc 0 do ph ng trnh c khngqu 3 nghi m .Vy ph ng trnh cho c 3 nghi m
a,b,cx 0
1 2 3x a,x b,x c= = =
Bi 4:a. Gis l hai nghi m ca ph ng trnh .Hy tnh
theo a1 2x , x
2x ax+1=07
7 1 2S x x= +7
b. Tm a thc bc 7 c h snguyn nh n s 7 735 3
= + 5
2
2
a
l nghi m
a. Ta c 2 2 2 22 1 2 1 2 1 2S x x (x x ) 2x x a= + = + =
2 2 2 2 2 2 24 1 2 1 2S (x x ) 2x x (a 2)= + =
2 2 33 1 2 1 1 2 2 2S (x x )(x x x x ) a(S 1) a 3= + + = =
3 3 4 4 3 3 7 5 37 1 2 1 2 1 2 1 2 3 4 1S (x x )(x x ) x x (x x ) S S S a 7a 14a 7a= + + + = = +
b. t 7 71 23 5
x ,x5 3
= =
Theo Vi-et l nghi m ca ph ng trnh1 2x ,x2x x 1 0 + = m ta c
7 77 1 2S x x= +
7 5 3a 7a 14a 7a= + = 3 55 3
+
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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7 5 315 105 210 105 34 0 + =
Vy a thc cn tm l 7 5 315x 105x 210x 105x 34 0 + =III. Ph ng trnh b c ba3.1 Dng 3 2ax bx cx d 0+ + + =3.2 Cch gi i ph ng trnh b c ba tng qut c a CardanoCi ny i v i hc sinh THCS khng quan tr ng lm nn tc gi ch xin a link bnno mu n tm hi u th tham kh o http://www.york.ac.uk/depts/maths/histstat/cubic.pdf hoc http://en.wikipedia.org/wiki/Cubic_equation .3.3 Hthc Vi-eti) Nu ph ng trnh b c ba 3 2ax bx cx d 0+ + + = (a 0) c ba nghi m th :1 2 3x ,x ,x
1 2 3
1 2 2 3 3 1
1 2 3
bx x x
ac
x x x x x xa
dx x x
a
+ + =
+ + =
=
ii) o li nu 3 s u, tha mnv,w u v w=m,uv+vw+wu=n,uvw=p+ + th lnghim ca ph ng trnh
u,v,w3 2t mt nt p 0 + =
nh l Bezout :Cho 1 a thc P( ,nu P( c mt nghim lx) x) th P( chia ht cho (xx) ) cngh a l P(x) (x ).Q(x) = (bc Q(x) P(x)< l 1 )3.4 Cc ph ng php chung gi i ph ng trnh b c ba3.4.1 N u bit tr c mt nghim 0x x= th phn tch ph ng trnh
20(x x )(ax bx c) 0 + + =
11
c bit nu :a b c d 0+ + + = th (*)0x =a b c d 0 + = th (**)0x =
3d c( )a b
= th 0c
xb
= (***)
Sau tm a, ta c th sdng php chia a thc hoc sdng s Horner hayng nht ha hai v tm
b,c
3.4.1 Bit mt hthc gia cc nghi m th ta dng Vi-et3.4.2 Dng h ng ng thc bin i vph ng trnh tch3.5 Bi t pBi 1:Gii ph ng trnh
3 2x 2x 4x 8 0+ + = (1)
G i sdng (***) gi i ra ta c x 2= Bi 2:Gii ph ng trnh
http://www.york.ac.uk/depts/maths/histstat/cubic.pdfhttp://en.wikipedia.org/wiki/Cubic_equationhttp://en.wikipedia.org/wiki/Cubic_equationhttp://www.york.ac.uk/depts/maths/histstat/cubic.pdf8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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3 2x 3x x 1 0 + + = (2)G i sdng (*) ta th y l nghi m a thc vtri suy ra (2)
.ng nht ha 2 v ta cx 1=
2(x 1)(ax bx c) 0 + + =2(x 1)(x 2x 1) 0 =
Sau gii ra nghi mBi 3:Gii ph ng trnh
3 212x 14x 17x 6 0+ + = Bit ph ng trnh c 2 nghi m m tch b ng 1
G i sdng hthc Vi-et tm c mt nghim sau lm t ng tbi trnIV. Ph ng trnh b c bn4.1 Dng : 4 3 2ax bx cx dx e 0(a 0)+ + + + = 4.2 Cch gi i ph ng trnh b c bn tng qut c a FerrariCi ny i v i hc sinh THCS khng quan tr ng lm nn tc gi ch xin a link bnno mu n tm hi u th tham kh o http://www.york.ac.uk/depts/maths/histstat/cubic.pdf hoc http://en.wikipedia.org/wiki/Quartic_equation 4.3 Cc ph ng php chung gi i ph ng trnh b c bn4.3.1 D ng 4 2ax bx c 0(a 0)+ + = Cch gi i: t ,ta c2t x 0= 2at bt c 0+ + = +Gii tm t 0+V i mi nghim , ph ng trnh c hai nghi m phn bi t :0t > 0 1 0 2x t ,x= = 0t 4.3.2 D ng v i(x a)(x b)(x c )( x d) k+ + + + = a b c d+ = + Cch gi i : t ta c ph ng trnh b c 2 theo .Gii ra tm ,sau ti p tc gii tm
t (x a)(x b= + + ) t tx
4.3.3 D ng 4 4(x a) (x b) k+ + + =
Cch gi i : t a bt x2
+= + ta c ph ng trnh trng ph ng theo t
4.3.4 D ng 4 3 2ax bx cx bx a 0+ + + =Cch gi i :+Xt c l nghi m ca ph ng trnh hay khngx 0=+Xt chia 2 v cho ph ng trnh tr thnhx 0 2x
2 21 1
a(x ) b(x ) c 0x x
+ + + =
t1
t x x= ta c ph ng trnh b c 2 theo t 4.4 Bi t pBi 1:Gii ph ng trnh
(x 1)(x 6)(x 5)(x 2) 252+ + + + =Bi 2:Gii ph ng trnh
http://www.york.ac.uk/depts/maths/histstat/cubic.pdfhttp://en.wikipedia.org/wiki/Quartic_equationhttp://en.wikipedia.org/wiki/Quartic_equationhttp://www.york.ac.uk/depts/maths/histstat/cubic.pdf8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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4 3 29x 9x 52x 9x 9 0 + =
Bi 3:Gii ph ng trnh
2 2
11 251
x (x 5) =
+
iu kin x - 5&x 0 t .Ta cx 5 y 0+ = 2 2x y 10y 2= + 5Thay vo ph ng trnh ta c
4 3 2y 10y 39y 250y 625 0 + + = .Do y 0 nn ta c2
2
25 25(y ) 10(y ) 39 0
y y+ + + =
t25
z y y= + ph ng trnh tr thnh2
z 10z 11 0 =z 11z 1
= =
Nghim loi v dthy | Z z = 1 | 10T tnh c y v xBi 4:Gii ph ng trnh
22
2
4xx 5
(x 2)+ =
iu kin x 2
2 2 2
2 22x 4x x 4x(x ) 5 ( ) 5x 2 x 2 x 2 x 2 + = =
Ta t2x
yx 2
=
12
2
y 1y 4y 5 0
y 5
= = =
T ta tm c nghim v x 2x 1= = V. H i xng5.1.1 H i xng loi 1:L loi hph ng trnh ch a n x,y m khi ta hon v x v y th m i ph ng trnh c a h khng thay i5.1.2 Cch gi i :
t ,bin i h cho v hv i hai n S, S x yP xy
= + =
P
Gii htm .S,PV i mi c p (S th x v y l hai nghi m ca ph ng trnh:,P)
2 2x Sx P 0( S 4P + = = )
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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5.2.1 H i xng loi 2:L loi hph ng trnh ch a n x,y m khi ta hon v x v y th ph ng trnh ny bi nthnh ph ng kia c a h 5.2.2 Cch gi i:Tr vv i vca 2 ph ng trnh c a hta c ph ng trnh c d ng
(x y)g(x,y) 0 =
T ta c hai h .Gii hny ,trong c m t h i xng loi 15.3 Bi t pBi 1:Gii hph ng trnh
2 22 2
1 1x y 49
x y1 1
x y 5x y
+ + + =
+ + + =
Bi 2:
Gii hph ng trnh 3 34 4
x y 1(1)
x y 1(2
+ =
+ = )
4
T(2) ta c | x v | y | 1 | 1 Nu | x th t(1) suy ra ,t ng t ta c| 1 y 0 x 0 Nu th .V l!0 x 1< < 3 4 3x x ,y y> >Vy ch c th hocx 0,y 1= = x 1, y 0= = Bi 3:Gii hph ng trnh
1 12 2yx
1 12 2
xy
+ =
+ =
VI. H ng c p bc hai6.1 Dng :
2 2
2 2
ax bxy cy d
a'x b ' xy c 'y d'
+ + =
+ + =
6.2 Cch gi i :Xt c l nghi m ca hkhngx 0,y 0= =V i khng l nghi m ca hta tx 0,y 0= = x ty= t ta c mt ph ng trnh b chai theo t .Gi i tm t sau ta suy ra x,y6.3 Bi t pBi 1:
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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Gii hph ng trnh2 2
2 2
x 3xy y 1
3x xy 3y 13
+ =
+ =
Bi 2:
Gii hph ng trnh 22 2
3x 2xy 160
x 3xy 2y 8
=
=
Bi 3:Gii hph ng trnh
2 2x 2xy 3y 0x | x | y | y | 2
+ =
+ =
B. Mt sph ng trnh v h ph ng trnh khng m u mcBi 1:
1x 2
x1
y 2y1
z 2z
=
=
=
y
z
x
Hny tho t nhn ta thy y l h hon v vng quay nh ng v i nhng cch nh githng th ng th ta khng th tm nghi m bi ton ny c .Bi ny c gii nhsau :
t x cot g , (0, ),2
=
Ta thu c 2y cot g tg 2cot g2 = = Suy ra z cot g4 x cot g8 = =
Ta nh n c kcot g cot g87
= =
K t h p v i iu kin ta c cc nghi m sau2 4
x cot g , y cot g ,z cot g7 7 7 7
= = = =
2 2 4x cot g ,y cot g ,z cot g7 7 7
87
= = = =
3 3 6x cot g ,y cot g ,z cot g
7 7 712
7
= = = =
4 4 8x cot g ,y cot g ,z cot g
7 7 716
7
= = = =
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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5 5 10 20x cot g ,y cot g ,z cot g
7 7 7 7
= = = =
6 6 12 24x cot g ,y cot g ,z cot g
7 7 7 7
= = = =
Bi 2:Gii ph ng trnh
2 4 4 4 44 x 2 x
x 2 x3 3
++ = +
iu kin 0 x 2 Ph ng trnh vi t li4 4 4 4 4 4
x 2x x 2(2 x) (2 x) 2xx x 2 x
3 3 3
+ + ++ + = + +
Sdng B T
4 4 4 4 4 4a 2b b 2c c 2a
a b c 3 3 3
+ + ++ + = + +
Ta suy ra
4 4 4 4 4 4x 2x x 2(2 x) (2 x) 2x
x x 2 x3 3 3
+ + ++ + + +
Du ng thc xy ra khi v ch khi x 2 x x 1= = Bi 3:Gii ph ng trnh
34x 3x 2+ =
Xt 3f(x) 4x 3x 2= +
suy ra ng bin ,suy ra ph ng trnh c duy nh t mtnghim
2f '(x) 12x 3 0= + > f(x)
t 1 1x (t )2 t
= >, t 0
Thay vo ph ng trnh ta nh n c3
3
1t 4
t =
t ta c3X t 0= >
2 X 2 5X 4X 1 0X 2 5(L) = + =
=
T ta tnh c sau tnh ct
33
1 1x ( 2 5
2 2 5= +
+)
Bi 4:Gii ph ng trnh
8/14/2019 Chuyen De Phuong Trinh va He Phuong Trinh
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3
24
4x 3x1 x
16x 12x 1
= +
t x cos , [0, ], = ta c sin5 cos3 =
C. K thut t n ph quy v h i xng loi hai trong bi ton c ch acn
Bi 1:Gii ph ng trnh
2 2x 3(2 3x ) 2+ =
t ta thu c h i xng loi 22y 2 3x=
2
2
y 2 3x
x 2 3y
=
= n y th cng vi c hon ton n ginBi 2:Gii ph ng trnh
3 3x 1 2 2x 1+ =
t 3y 2x= 11
1
ta thu c h i xng loi 23
3
y 2x
x 2y
=
=
Nhn xt: trong nh ng bi ton trn vi c t n phhon ton d dng c thquy v h i xng loi hai ,nh ng trong nh ng bi ton kh h n vic t n phdo ph n xhon ton khng p d ng c.Vy chng ta ph i lm th no? Chng ta c th hiu rvk thut t n phvh i xng trong v d sau yBi 3:Gii ph ng trnh
24x 2x 1 6 12x+ + + =
t 2x 1 ay b+ = + v i a, l hng s.Ta c :b
2
2 2 2
4x 12x ay b 5 0
a y 2aby 2x b 1 0
+ + + =
+ + =
Xc nh a,b sao cho h trn l h i xng loi 2 tc l :2 2
a 2ab 2 b 15
a 2b 3
= =4 12 a b = = =
+
Vy ta t 2x 1 2y 3 0+ = + ta thu c h i xng
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2
2
2
2x 6x y 4 0
2y 6y x 4 0
(x y)(2x 2y 5) 0
2x 6x y 4 0
+ =
+ = + =
+ =
n y th vi c gii ph ng trnh hon ton mang tnh th tcBi 4:Gii ph ng trnh
2 4x 97x 7x28
++ =
Bi 5:Gii ph ng trnh
25 x 1 2x x 3 1 + + + + =
Bi 6:Gii ph ng trnh3 225 x 3 3+ + =
Bi 7:Gii ph ng trnh
2(x 2) x x 20 6x 20+ = +