^huorng phdp gidi. -*
Dya vao djnh li ve da'u ciia tam thuc bac hai de xet da'u ciia bieu thtfc chiia no.
* Doi voi da thuc bac cao P(x) ta lam nhu sau
• Phan tich da thiic P(x) thanh tich cac tam thuc bac hai (hoac c6 ca nhj thu^
bac nha't)
• Lap bang xet da'u ciia P(x) . Tu do suy ra dau ciia no . P(x)
* D61 voi phan thuc (^rorig do P(x), Q(x) la cac da thuc) ta lam nhu sau
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^ phan tich da thuc P(x), Q(x) thanh tich cac tam thuc b|ic hai (ho|c c6 ca
nhi thuc bac nha't) , S ...yj |
p(x)
Lap bang xet da'u ciia ———. Tu do suy ra da'u ciia no.
) U c ••- f i t -1 u, t i v
P 1. C A C V f D U M I N H H O A
Vi dv 1: Xet da'u ciia cac tam thuc sau 3, xV Q ( x H c 0
a) - X +4x + 5 b) 3 x 2- 2 x - 8 c) -2x^ + 6x - 5
a) Ta C O -x^ + 4x + 5 = 0 •ằ x = - l
Laigiai , ^x|l ij-.. ^
X = 5 V(|^T^rTV=;'5,X :.iy • %.r*-:ml - ' fl I
X - 0 0 5 +00
-x^ +4x + 5 - 0 + 1 -
Suyra -x'^+4x + 5>0<=> X G( - 1 ; 5 ) va - x ^ + 4X + 5 < 0 < ằX 6( - ^ ; - 1 )U( 5 ; -H ằ )
X = 2 : . - - - - j~ " - - T r i- r - - " 7 -
b) Taco 3x^ - 2 x - 8 = 0c:> __4
^ ~ 3
X - 0 0
3 2 . +00
3x2 _ 2 x - 8 + 0 1 +
Suy ra
3x2 - 2 x - 8 > 0 < ^ x e 4^
- 0 0 ; — u (2; +ô) va 3x2 - 2x - 8 < 0 x e c) Ta C O A' = -1 <0, a <0 suy ra -2x2 + 6 x - 5 < 0 V x e R
Nhan xet: < r , , ,
Cho tam thiic bac hai ax^ + bx + c. Xet nghiem ciia tam thuc, ne'u:
V6 nghiem khi do tam thiic b|c hai f (x) = ax2 + bx + c cung da'u voi a voi
moi X r ' j
Nghiem kep khi do tam thiic bac hai f (x) = ax2 + bx + c ciing da'u voi a voi moi X
Co hai nghiem f(x) cung da'u voi a khi va chi khi x e ( - c o;x 2a , ) u(x2;+'ằ) (ngoai hai nghiem) va f(x) trai da'u voi a khi va chi khi x e(xi;x2) (trong hai nghi§m)(ta C O the nho cau la trong trai ngoai cung)
Phdn loai va phumig pitdp gidi Dai so 10
V i d\ 2: T i i y theo gia t r i ciia tham so'm, hay xet dau ciia cac bieu thuc f(x) = x^ + 2 m x + 3 m - 2
;<f Lcngidi T a m thuc f(x) c6 a = 1 > 0 va A' = - 3m + 2 .
* N e u l < m < 2 = > A ' < 0 = > f(x) > 0 Vx e R . m = l
m = 2 m > 2
* N e u
* N e u
U f S h 1 ' ' . - ; : ; v - .
A' = 0 => f(x) > 0 Vx e R va f(x) = 0 x = - m S;'^
m < 1 • A' > 0 => f(x) CO hai nghiem
xj = - m - V m ^ - 3 m + 2 va X2 = - m + Vm^ - 3 m + 2 . K h i do:
+) f(x)>0<:> X € ( - o o ; X i ) u ( X 2 ; + < ằ ) +) f ( x ) < 0 < = > x e ( x i; x 2 ) .
V i dvi 3: Xet dau ciia cac bieu thuc sau
a) (-x2 + x - l) ( 6 x 2 - 5 x + l ) b) x ^ - x - 2 - x ^ + 3 x + 4 Lcri gidi
a) Ta CO - x ^ + x - 1 = 0 v 6 nghiem, 6x^ - 5 x + l = 0<=>x = — hoac x = —
' ^ • ' 2 3 Jang xet dau
-00 1
3
2
3 +00
- X ^ + X - 1 0 - 1 -
6x^-5\ l + 1 - 0 +
(-x2 + x - l) { 6 x ^ - 5 x + l ) - 0 + 0 -
Suy ra (-x^ + x - l)(6x^ - 5x + 1 j d u o n g k h i va chi k h i x € V | - x ^ + x - l j ^ 6 x ^ - 5 x + l j a m k h i va chi k h i x e
3'2)
b) Ta CO x^ - X - 2 = 0 o x = - l
x = 2 , - x ^ + 3 x + 4 = 0 o x = - l x = 4 Bang xet dau
X -00 - 1 2 4 +00
+ 0 - 0 + 1 +
- x^ + 3x + 4 0 + 1 + 0 -
x^-x-2 1 1 - 0 + II -
-x^ + 3 x + 4
X - X - 2
ra — ^ d u o n g k h i va chi k h i x e (2; 4 ) , - x ^ + 3 x + 4 x^ - x - 2
- x ^ + 3 x + 4
p 2 . B A I T A P L U Y E N T A P g^i 4.79: Xet dau cac tam thuc sau a) ii^)
am k h i va chi khi x e (-oo; -1) u ( - 1 ; 2) u (4; +oo) .
= -2x^ + 3 x - l
Huang dan gidi
b) g(x) = - x 2 - x + 1 c) h(x) = - 2 x 2 + x - l .
• uk}> l',>
j) Tam thuc f ( x ) c6 a = - 2 < 0, c6 hai nghiem x j = —; X2 = 1 i i
* f(x) > 0 (trai dau v o i a) <=> x € (—;1)
* f(x) < 0 (cung dau v o i a) <=> x e (-oo; ^) u (1; +oo).
b) Tam thuc g(x) c 6 a = ^ > 0 , c 6 A = 0 => g(x) > 0 (cung d a u v o i a) Vx ;t i
v a g ( | ) - 0 .
c) Tam thuc g(x) c6 a = - 2 > 0 , c6 A = - 7 < 0 = > g ( x ) < 0 (ciing d a u v o i a) V x e R .
Bai 4.80: Xet dau cac bieu thuc sau < *' a)f(x) = ( x 2 - 5 x + 4)(2-5x + 2x2) b) f(x) = x ^- 3 x - 2 - 8
a) Ta c6:
Huang dan gidi x ^ - 5 x + 4 = 0 o x = l ; x = 4 2 - 5 x + 2x^ =0<=>x = 2;x = - Bangxet dau:
x^ - 3 x
X -00 1
2 1 2 4 +00
x^ - 5 x + 4 + 1 + 0 - 1 - 0 +
__2x2-5x + 2 + 0 1 0 + 1 +
f(x) + 0 0 + 0 - 0 +
m til
b) Ta c6: f(x) = (x^ - 3x)^ -2(x^ - 3 x ) - 8 _ (x^ - 3 x + 2)(x^ - 3 x- 4 )
x^ - 3 x - 3 x
Bang xet dau
X - 0 0 - 1 0 1 2 3 4 +00
x^ - 3 x + 1 + 0 - 1 - 1 - 0 + 1 +
x^ - 3 x - 4 + 0 - 1 - 1 - 1 - 1 - 0 +
x 2 - 3 x + 2 + 1 + 1 + 0 - 0 + 1 + 1 +
f(x) + M - 0 + I I - I I + 0 -- I I +
Bai 4.81: Xet dau cac bieu thuc sau 1 1 1
a) x + 9 X 2
X 3x + 7 ^
- 2 r + 5 x - x - 2
b) x"^ - 4 x + l . d) X - 3x + 2
a) T a c o : f(x) =
Huang dan gidi 2 x - 2 ( x + 9 ) - x ( x + 9) _ - x ^ - 9 x - 1 8
2x(x + 2) 2x(x + 9)
f(x) > 0 o x e (-6; -3) u (2; 0) f (x) < 0 <=> -6) u (-3; 2) u (0; +oo)
b) Ta c6: f(x) = x'' + 2x2 + 1 - 2( ^ 2 + 2x +1) = (x^ +1)^ -[ ^ / 2( x +1) f(x) = (x^ - V2x + 1 - 7 2 ) ( x 2 + ^/2x + 1 +
V 2-N/W 2^1 172 + V 4 V 2 - 2
00 :>i
-\2 X I
=> f(x) > 0 <=> X G - 0 0 ; -;+oo
£(x) < 0 o
, 5 x 2- 2 x- 3
> 0 < ằ x e( - o o; - l) u u( 2; + o o )
Va 5x^ - 2 x - 3
x 2- x - 2 < 0< i = > X € -1;-^ u( l; 2 ) d) f (x) = (x - l)2(x + 2) => f (x) > 0 o X G (-2;+oo) \}
f ( x ) < 0 < = > x e ( - o o; - 2 )
Bai 4.82: T u y theo gia t r i cua tham so m , hay xet dau cua bieu thuc , g(x) = ( m - l) x 2+ 2 ( m - l ) + m - 3
Cty TNHH M I V l ) \ Khang Vift
Huang dan gidi [sjgu m = 1 =>g(x) = - 2 < 0 V X G R • X -
fsleu m ?t 1 , l<hi d o g ( x ) l a tam thiic bac hai c6 a = m - 1 va A ' = 2(m - 1 ) , d o Jo ta CO cac t r u a n g hop
^ > 1 CO hai nghiem phan bi^t
m - l- V 2( m - l ) , m - l + 7 2( m - l )
^ = va X T = . ( A
"1 m - 1 m - 1
g ( x ) > 0 o x€ ( - o o; x i ) u( x 2 ; + c o ) ; g(x) < 0 ci> x e (x,; X j ) . a < 0
* m < l = > i . . „ = > g ( x ) < 0 V x e R .
D A N G T O A N 2: B A / T O A N CHUA T H A M S O L I E N QUANDEN T A M THUC B A C H A I LUON MANG MOT D A U
ai.CAC Vf DU MINH HOA
Vi du 1: C h u n g m i n h rang voi moi gia trj ciia m t h i
>',fH lr^^ Is. •> m m . a) Phuang trinh m x - ( 3 m + 2) x + 1 = 0 luon c6 nghiem
b ) Phuong trinh [ m ^+ 5 ) x 2- ( N / 3 m - 2 ) x + l = 0 luon v6 nghiem Lai gidi
a) Voi m = 0 p h u o n g trinh tro thanh -2x + 1 = 0 o x = - suy ra p h u a n g trinh C O nghiem
Voi m ^ 0 , ta CO A = (3m + i f - 4 m = 9m2 + 8 m + 4
V i t a m thuc 9 n r + 8 m + 4 CO a^^ = 9> 0 , A',^ = : - 2 0 < 0 _^ ^
"en + 8 m + 4 > 0 v o i m p i m , ^ „ [ [ . , • .. • Do do p h u o n g trinh da cho luon c6 nghiem v o i m p i m .
Taco A - ( x/ 3 m- 2) ^ - 4 ( m 2+ 5 ) = - m 2 - 4 7 3 m- 1 6
^1 tam thuc - m ^ - 4>y3m - 1 6 c6 a ^ = - 1 < 0, A = - 8 < 0 .) d .;
l e n - m ^ - 4\/3m - 1 6 < 0 v o i moi m
'^o do p h u o n g trinh da cho luon v 6 nghiem voi m o i m . I 2: T i m cac gia trj cua m de bieu thuc sau luon a m
l[ ( x ) = mx2 - x - 1 b) g ( x ) = ( m - 4) x 2 + ( 2 m - 8 ) x + m - 5
Ph&n loai va phuang phdp giai D(ii so 10'
Lai giai
a) V o i m = 0 thi f ( x ) = - x - l lay ca gia trj d u o n g ( c h i n g han f ( - 2 ) = l ) ^.
m = 0 khong thoa man yeu cau bai toan
V 6 i m ;^ 0 t h i f (x) = mx^ - x - 1 la tarn thuc bac hai do do > <
m < 0
A2
f ( x ) < 0 , Vx a = m < 0 A = l + 4 m < 0 <=>
I o m < —1 ft m < - - 4
4
Vay v o i m < - - thi bieu thuc f (x) luon am. ^.
b) V o i m = 4 thi g ( x ) = - 1 < 0 thoa man yeu cau bai toan *
V o i m 9 ^ 4 t h i g ( x ) = ( m - 4 ) x ^ + ( 2 m - 8 ) x + m - 5 la tam thuc bac hai do a = m - 4 < 0
A' = ( m - 4 ) ^ - ( m - 4 ) ( m - 5 ) < 0 ' I m i
do g ( x ) < a Vx<=>
f m < 4 ^ ^".OH Hô/!fM n o SV
<=> <^ <=> m < 4
>- [ m - 4 < 0
' Vay v a i m < 4 thi bieu thuc g ( x ) luon anr
V i d u 3: T i m cac gia trj ciia m de bieu thuc sau luon d u o n g -x^ + 4( m + l ) x + l- 4 m ^
-4x2 + 5 x- 2
a) h ( x ) = b) k ( x ) = N/x2 - x + m - 1
^, Lcrigidi snvn/rta u
a) Tam thuc -4x2 + 5x - 2 c6 a = - 4 < 0, A = - 7 < 0 suy ra -4x^ + 5x - 2 < 0 Vx D o do h ( x ) luon d u o n g k h i va chi khi h ' ( x ) = - x 2 + 4( m + l ) x + l- 4 n r lu6n am , ,
O
a = -i<o r : ' 5
, - X <=> 8 m+ 5 < 0 <=> m < —
A' = 4( m + l) 2+ ( l- 4 m 2 ) < 0 8
Vay v o i m < — thi bieu thuc h ( x ) luon duong.
8
b) Bieu thuc k ( x ) luon d u o n g <=> Vx^ - x + m - 1 > 0, Vx
x + m > 1, Vx <=> • x 2- x + m> 0 !-~^y~-rnF\A-'^-- / Vx
x - x + m - l > 0
= 1 > 0
^ < A i = 1 - 4m < 0 <=> • A2 = l - 4 ( m - l ) < 0
1
m > - _ 4 5 <=> m > —
5 4 7 - i •• ' :
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.- v:"-^'^
y^y vdi iTi > ~ thi bieu thuc k ( x ) luon duong.
^ 2 . B A l T A P L U Y e N T A P r - - u , ^ i t „ i : 0 fl^i4.83: C h u n g m i n h rang v o i moi gia trj ciia m thi ' - r - r j j^j,/ r-
,) phuong trinh x^ - 2 ( m + 2)x - ( m + 3) = 0 luon c6 nghiem s^,^^ . ^ h)phuong trinh (m^ + 1 jx^ + (VSm - 2 ) x + 2 = 0 luon v6 nghi|m....'-^p,,;
Huang dan giai
. 7 7 HI f-/.mi' '/m =•(/.)}(::
,)TacoA = ( m + 2) + m + 3 = m'^+5m + 7
Vi tam thuc m ^ + S m + y c6 a^, = 1 > 0, A ' ^ = - 2 < 0 nen x = - 4 , x = 0 v o i mpi m
Do do p h u o n g trinh da cho luon c6 nghiem v a i moi m . '
:^ Ta CO A = ( V 3 m - 2)^ - 8{m2 +1) = -Sm^ - 4^/3m - 4 '
Vi tam thuc - 5 m 2 - 4 V 3 m- 4 c6 a „ =-5<0, A'„ <0 nen -Sm^-4V3m-4<0 voi moi m . D o do p h u o n g trinh da cho luon v6 nghiem voi m p i m .
Bai 4.84: T i m cac gia trj cua m de bieu thuc sau luon am
^'fW = - x 2- 2 x - m b) g ( x ) = 4 m x 2- 4 ( m - l ) x + m - 3
Huang dan giai % mmilMJ, ' f { x ) < 0 , Vx<^< a = - l < 0 1
A =1 - 4 m < 0 4
% vol m > thi bieu thuc f (x) luon am. 1
^oi m = 0 khong thoa man yeu cau bai toan
'^oj m thi g ( x ) = 4mx2 - 4( m - l ) x + m- 3 la tam thuc bac hai do do . . . f a = 4 m < 0
Đ ( x ) < 0 , V x ằ - ^
A' = 4( m - 1 ) - 4 m( m- 3) < 0 m < 0
' ( I + rf:i)
4m + 4 < 0 l m < - l , o m < - 1
voi m < - 1 thi bieu thuc g ( x ) luon am. / i s m Li.'C'i y
Plidit loiti I'll pillion^ pluip giat uat so lu
Bai 4.85: C h u n g m i n h r^ng ham so sau c6 tap xac djnh la IR v o i mpi gj^
cua m .
a) y = V m ^ - 4mx + m ^ - 2m + 5 b) y = 2x + 3m
, ^ Jx^ + 2 ( l - m ) x + 2m^"T3 Huong dan giai
a) D K X D : m^x^ - 4 m x + - 2m + 5 > 0 (*) Vol m = 0 thi dieu kien (*) d u n g voi moi x
2 2 2
Vol m ?i 0 xet tam thuc bac hai f ( x ) = m^x - 4 m x + m - 2 m + 5 Ta CO a = m > 0
A' = 4 m ^ - m ^ | m ^ - 2m + sj = - m ^ |m^ - 2m + ij = - m ^ ( m -1 ) ^ < 0 Suyra f (x) = m^x^ -4 m x + - 2 m + 5 > 0 Vx e iR
Do do voi m o i m ta c6 m^x^ - 4 m x + m^ - 2m + 5 > 0, Vx G IR Vay tap xac djnh cua ham SO la D = R .
b) D K X D : x^ + 2(1 - m ) x + 2m^ + 3 > 0
Xet tam thuc bac hai f ( x ) = x^ + 2 ( l - m ) x + 2m^ + 3 , T a c 6 a = l > 0 , A ' = ( l - m ) ^ - ( 2 m 2+ 3 ) = - m^ - 2 m - 2 < 0
,. (Vi tam thuc bac hai t ( m ) = - m ^ - 2m - 2 c6 a^, = - l < 0 , A'^, = - l < 0 Suy ra voi moi m ta c6 x^ + 2 ( 1 - m ) x + 2m^ + 3 > 0, Vx € IR
Vay tap xac djnh cua ham so la D = R.
Bai 4.86: T i m m de
a) 3x2 - 2(m + l)x - 2m2 + 3 m - 2 > 0 Vx 6 R '
b) H a m so y = sj{m + \)x^ - 2 ( m - l)x + 3m - 3 c6 nghla voi moi x.
if)
y i
c) x + m
X + X + 1 <1 Vx G R
Huang dan giai a) 3 x 2 - 2( m + l ) x- 2 m 2 + 3 m - 2 > 0 G R
<=> A' = ( m +1)2 + 3(2m2 - 3 m + 2) < 0 7 m 2 - 7 m + 7 < 0 bpt v6 nghi?m Vay khong c6 m thoa man yeu cau bai toan
b) H a m so c6 nghia voi moi x
<r>(m + l) x 2- 2 ( m - l ) x + 3 m - 3 > 0 V X G R (1)
* m = - 1 khong thoa man
>• m 1 G > I ' + ;
f m + 1 > 0
• ^ ^ - ' ^ ^ ' ) ^ l A ' = ( m - l ) ( - 2 m - 4 ) . 0 ^ ^ ^ f a CO X + X + 1 > 0 Vx e
X + m
<2 + X + 1
X + m
< l < = > - l< - 2 < 1 o X + X + 1
(1) ' x^ + 2x + m + 1 > 0 (2) x^ + l - m > 0
2 dung V x 6 K < = > l - m > 0 o m < l
/2) d u n g Vx G R <=> A ' = - m <0<=> m > 0 , . , Vay 0 < m < 1 la nhOng gia trj can tim.