2018高考数学一轮复习第6章不等式及其证明第1节不等式的性质与一元二次不等式课件
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3
b24ac
>0
0
<0
yax2bxc(a>0)
ax2bxc0 (a>0)
(a>0) ax2bxc<0 (a>0)
x1x2(x1<x2)
_{x_|_x_<_x_1 ___x_>_x_2_} _{_x_|x_1_<_x_<_x_2}_
a>1 1a<1x1a(x1)<0 1a<x<1
0<a<1 1a>1
x1a(x1)<0
1 1<x<a.12
a<0 xx<1ax>1
a0 {x|x>1} 0<a<1 x1<x<1a
a1
a>1 x1a<x<(3) 2 (1) 0 0 0 (2) 0 (3)
x1x22ba _{_x_|x___x_1}_
ax2bxc>0
R ___
1()("""�") (1)a>bac2>bc2.( )
(2)a>b>0c>d>0ad>bc.( ) (3) ax2bxc<0 (x1x2) a>0.( ) (4) ax2bxc0(a0) ax2bxc>0 R.( ) [] (1)� (2) (3) (4)�
f(x)mx2mx1. x[1,3]f(x)<m5 m
[] f(x)<m5 x[1,3] mx12234m6<0 x [1,3].4
g(x)mx12234m6x[1,3] m>0 g(x)[1,3]
g(x)maxg(3)7m6<0
m<67 0<m<679
m0 6<0
m<0 g(x)[1,3]
{x|x<1 x>3} [x2(k4)x42k>0 g(k)(x2)k(x24x4)>0 k[1,1] g(1)>0 g(1)>0xx22 53xx 62>>00 x<1 x>3.]
[ 1] (1)(2017� 2 )1a<1b<0
( )
Aa2<b2
Bab<b2
Cab<0
D|a||b|>|ab|
(2) 2<<< ( )
A.3232
B.320
C.032
D.20
(1)D (2)B [(1) b<a<0 ABC |a||b|ab |ab| D D.
(2)2<<<<2 32<<32. <<0 32<<0.]
x2bxa0 x25x60 x2 x3.]
1 f(x)0(xR)
(2017�)(a2)x22(a2)x4<0 xR a __________
(2,2] [ a20 a2 4<0 xR
a2 a24<a02216a2<0
a<22<a<2 2<a<2. a (2,2]]
2 f(x)0x[ab]
g(x)maxg(1)m6<0 m<6 m<0.
m mm<67
.14
x2x1x12234>0
m(x2x1)6<0
6 m<x2x1.6
yx26x1x126234[1,3]67 m<67
m mm<67
.14
3 f(x)0( m[ab]) x
k[1,1] f(x)x2(k4)x42k x __________
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1 (1)a>bab>0 (2)abab0 (3)a<bab<0. 2 (1)a>b_b_<_a__() (2)a>bb>ca>c()
(3)a>bac_>_bc() a>bc>d_a___c_>_b___d___() (4)a>bc>0ac_>_bc a>bc<0ac_<_bc a>b>0c>d>0ac_>_bd() (5)a>b>0an_>_bn(n2nN)()
[ 2] (2016�) ax2bx1>0
x|
12<x<13
x2bxa0
(
)
A{x|2<x<3}
B{x|x2 x3}
1 1
C.x|
3<x<2
D.x|
x<13x>12
B
[
ax2bx1>0
x|
12<x<13
ax2bx10 x112 x213 a<0
1212�13ba131a
ab 5. 6
0m<1.]
(1) xyR x>y>0( )
A.1x1y>0
Bsin xsin y>0
C.12x12y<0
Dln xln y>0
(2) f(x)ax2bx 1f(1)2,2f(1)4 f(2)
(1)C [ y12x (0) x>y>0 12x<12y 12x12y<0 C y1x(0) x>y>01x<1y 1x1y<0 A ysin x (0) x>y>0 sin x sin y B x>y>0xy>0/ ln(xy)> 0/ ln xln y0 D ]
C [ a1b2 ABD. C.]
4x23x4>0 ________________()
(4,1) [x23x4>0 x23x4<04<x<1x2
3x4>0 (4,1)]
5 mx22mx1>0 R m __________.
[0,1) [ m0 1>0
51062180
m0 m>04m24m<0 0<m<1
2()( ) a>bc<dac>bd a>b>0c<d<0ac>bd
a>b>03 a>3 b a>b>0a12>b12. A C
B D
D [ ac<bd yx
11 a>b>0 a2>b2>0a2<b2]
3(2017�) abR a>b( )
Aa2>b2
a B.b>1
C2a>2b
Dlg(ab)>0
(1)32xx20 (2)x2(a1)xa<0. 51062181 [] (1) x22x30 (x3)(x1)0 {x|1x3}.7 (2)(xa)(x1)<0 a>1 (1a) a1 a<1 (a,1).14
[] (2) ax2(a1)x1<0 [] a0x1<0 x>1. a<0x1a(x1)>0 x<1a x>1.4 a>0x1a(x1)<0. a1 1a1x1a(x1)<0
(2) f(1)abf(1)ab f(2)4a2b.3 m(ab)n(ab)4a2b mm nn 42 nm31 10 f(2)(ab)3(ab)f(1)3f(1).12 1f(1)2,2f(1)4 5f(2)10 f(2)[5,10].14
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2 3 a<f(xy)<bc<g(xy)<d F(xy) F(xy)mf(xy)ng(xy) mn F(xy)
3
b24ac
>0
0
<0
yax2bxc(a>0)
ax2bxc0 (a>0)
(a>0) ax2bxc<0 (a>0)
x1x2(x1<x2)
_{x_|_x_<_x_1 ___x_>_x_2_} _{_x_|x_1_<_x_<_x_2}_
a>1 1a<1x1a(x1)<0 1a<x<1
0<a<1 1a>1
x1a(x1)<0
1 1<x<a.12
a<0 xx<1ax>1
a0 {x|x>1} 0<a<1 x1<x<1a
a1
a>1 x1a<x<(3) 2 (1) 0 0 0 (2) 0 (3)
x1x22ba _{_x_|x___x_1}_
ax2bxc>0
R ___
1()("""�") (1)a>bac2>bc2.( )
(2)a>b>0c>d>0ad>bc.( ) (3) ax2bxc<0 (x1x2) a>0.( ) (4) ax2bxc0(a0) ax2bxc>0 R.( ) [] (1)� (2) (3) (4)�
f(x)mx2mx1. x[1,3]f(x)<m5 m
[] f(x)<m5 x[1,3] mx12234m6<0 x [1,3].4
g(x)mx12234m6x[1,3] m>0 g(x)[1,3]
g(x)maxg(3)7m6<0
m<67 0<m<679
m0 6<0
m<0 g(x)[1,3]
{x|x<1 x>3} [x2(k4)x42k>0 g(k)(x2)k(x24x4)>0 k[1,1] g(1)>0 g(1)>0xx22 53xx 62>>00 x<1 x>3.]
[ 1] (1)(2017� 2 )1a<1b<0
( )
Aa2<b2
Bab<b2
Cab<0
D|a||b|>|ab|
(2) 2<<< ( )
A.3232
B.320
C.032
D.20
(1)D (2)B [(1) b<a<0 ABC |a||b|ab |ab| D D.
(2)2<<<<2 32<<32. <<0 32<<0.]
x2bxa0 x25x60 x2 x3.]
1 f(x)0(xR)
(2017�)(a2)x22(a2)x4<0 xR a __________
(2,2] [ a20 a2 4<0 xR
a2 a24<a02216a2<0
a<22<a<2 2<a<2. a (2,2]]
2 f(x)0x[ab]
g(x)maxg(1)m6<0 m<6 m<0.
m mm<67
.14
x2x1x12234>0
m(x2x1)6<0
6 m<x2x1.6
yx26x1x126234[1,3]67 m<67
m mm<67
.14
3 f(x)0( m[ab]) x
k[1,1] f(x)x2(k4)x42k x __________
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1 (1)a>bab>0 (2)abab0 (3)a<bab<0. 2 (1)a>b_b_<_a__() (2)a>bb>ca>c()
(3)a>bac_>_bc() a>bc>d_a___c_>_b___d___() (4)a>bc>0ac_>_bc a>bc<0ac_<_bc a>b>0c>d>0ac_>_bd() (5)a>b>0an_>_bn(n2nN)()
[ 2] (2016�) ax2bx1>0
x|
12<x<13
x2bxa0
(
)
A{x|2<x<3}
B{x|x2 x3}
1 1
C.x|
3<x<2
D.x|
x<13x>12
B
[
ax2bx1>0
x|
12<x<13
ax2bx10 x112 x213 a<0
1212�13ba131a
ab 5. 6
0m<1.]
(1) xyR x>y>0( )
A.1x1y>0
Bsin xsin y>0
C.12x12y<0
Dln xln y>0
(2) f(x)ax2bx 1f(1)2,2f(1)4 f(2)
(1)C [ y12x (0) x>y>0 12x<12y 12x12y<0 C y1x(0) x>y>01x<1y 1x1y<0 A ysin x (0) x>y>0 sin x sin y B x>y>0xy>0/ ln(xy)> 0/ ln xln y0 D ]
C [ a1b2 ABD. C.]
4x23x4>0 ________________()
(4,1) [x23x4>0 x23x4<04<x<1x2
3x4>0 (4,1)]
5 mx22mx1>0 R m __________.
[0,1) [ m0 1>0
51062180
m0 m>04m24m<0 0<m<1
2()( ) a>bc<dac>bd a>b>0c<d<0ac>bd
a>b>03 a>3 b a>b>0a12>b12. A C
B D
D [ ac<bd yx
11 a>b>0 a2>b2>0a2<b2]
3(2017�) abR a>b( )
Aa2>b2
a B.b>1
C2a>2b
Dlg(ab)>0
(1)32xx20 (2)x2(a1)xa<0. 51062181 [] (1) x22x30 (x3)(x1)0 {x|1x3}.7 (2)(xa)(x1)<0 a>1 (1a) a1 a<1 (a,1).14
[] (2) ax2(a1)x1<0 [] a0x1<0 x>1. a<0x1a(x1)>0 x<1a x>1.4 a>0x1a(x1)<0. a1 1a1x1a(x1)<0
(2) f(1)abf(1)ab f(2)4a2b.3 m(ab)n(ab)4a2b mm nn 42 nm31 10 f(2)(ab)3(ab)f(1)3f(1).12 1f(1)2,2f(1)4 5f(2)10 f(2)[5,10].14
[] 1. """"
2 3 a<f(xy)<bc<g(xy)<d F(xy) F(xy)mf(xy)ng(xy) mn F(xy)