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  • 8/2/2019 alg aa

    1/4

    (....)

    2012 _3.1()

    : 1 4

    : :

    : 1 2012

    A.1. 34.

    A.2. 125.

    .3. . . .

    . .

    .1. f

    4(x 1) 0+ , x IR .

    4(x 2) 0 x IR

    x 1 0+ x 2 0 , x 1 x 2 f { }A IR 1, 2=

    .2. { }x A IR 1, 2 = f

    2 22 2

    4 4 (x 1) (x 2)(x 1) (x 2)f (x)

    x 1 x 2 x 1 x 2

    + +

    = = =

    + +

  • 8/2/2019 alg aa

    2/4

    (....)

    2012 _3.1()

    : 2 4

    2 2 2 2

    (x 1) (x 2) (x 1) (x 2) x 1 (x 2) x 1 x 2 3x 1 x 2 x 1 x 2

    + +

    = = = + = + + =+ +

    { }x A IR 1, 2 = : f(x) = 3. f(2012) = 3.

    .3. 18 3x f ( 2012) :

    18 3x f (2012) 3(6 x) 3 3 (6 x) 3

    1 1 x 6 1 x 1 66 x x 6 1 1 6 + +

    x [5, 7] .

    .1.1. :

    x + 1 = 2

    || . x || . x + x = 2

    1

    (|| + 1).

    x = ||2

    1

    x = || + 1 = ||

    2 1 = (|| 1) (|| + 1).

    IR 0 1 1 0 + > .

    IR = || + 1 0,

    IR , x,

    ( )( )( )

    ( ) ( )( )

    2 1 x 1 1

    1 x 1 x 1 1 1

    + +

    + = = =

    + +

    : x 1= , IR .

    .1.2. 3, 2, :

    d(x,3) 2 x 3 2 1 3 2 4 2

    4 4 2 2 62 2 2 4 + +

    6 6 6

    [ 6, 2] [2, 6]

    2 2 2

  • 8/2/2019 alg aa

    3/4

    (....)

    2012 _3.1()

    : 3 4

    2.

    2

    1: y ( 4)x 1 = + + , IR

    2

    2: y ( 4 3)x 2 = + + , IR

    1 1 = 1 = 2

    4,

    1 1 xx, 2 2 = 2 =

    2+ 4 3 2

    2 xx.

    1 xx :90 < 1 < 180 1 ,

    1 = 1 < 0, 1 < 0 2

    4 < 0 ( 2, 2)

    - 4

    - +

    0 - +2

    +

    -2

    0

    2

    2 xx :0 < 2 < 90, 2

    2 = 2 > 0,

    2 > 0 2

    + 4 3 > 0 (1, 3)

    - + - 3 4

    - +

    0 -2

    +

    -1

    0

    3

    -

    1 xx 2

    xx :

    1

    2

    0 ( 2, 2)

    (1, 2)

    0 (1,3)

    <

    >

  • 8/2/2019 alg aa

    4/4

    (....)

    2012 _3.1()

    : 4 4

    .1. A ( ) , *

    :

    = 1 + ( 1) , * , = 7 :7 = 1 + (7 1) = 1 + 6 7 = 11,

    1 + 6 = 11 1 + 6.

    (2) = 11 1 = 12 11 = 1, 4 = 1 + 3 = 1 + 3(2) = 5, f(x) = 1x

    2+ 4x + 1, ,

    f(x) = x2

    5x + 1.

    .2. K f(x) = 0 x2

    5x + 1 = 0 x1, x2 x

    2 5x + 1 = 0 Vieta:

    1 2

    5S x x 5

    1

    = + = = =

    1 2P x x 1

    = = =

    .

    :

    2 2

    1 2 2 1 1 2 1 2x x x x x x (x x ) 5 = + = + =

    2 2 2 2 2 2

    1 2 1 2 1 2 1 2 1 2 1 2

    2 1 1 2 1 2 1 2 1 2

    x x x x x x x x 2x x 2x xB

    x x x x x x x x x x

    + + + = + = + = =

    2 2

    1 2 1 2

    1 2

    (x x ) 2x x 5 1B 23

    x x 1

    + = = =

    331 2 1 2

    400(x x ) 2012x x 12 400 4 2012 1 12 = + + = +

    32000 2012 12 0 = + =

    .3. : 2x B 2 x A + = :

    2 2x 23 2 x 5 0 x 25 x 5 0 + = + =

    || + || = 0 = 0 = 0.

    2 2x 25 x 5 0 x 25 0 + = = x 5 = 0

    x = 5.