Anonymous
Anonymous asked in 科學數學 · 9 years ago

中二數學identities

1) Given that x^2 + 6x 恆等 (x+p)^2 + q , find the values of the constants p and q.

Simplify the following parts:

1)

5 y-1

---- - --------

4-y y-4

2)

1 2

---------- - ----------------

x^2+x (x+1)(x+3)

3)

1 x

---------- - ------------

x-4 (4-x)^3

4)

4

---------- - 2

2 + 1

----

w

2 Answers

Rating
  • 9 years ago
    Favorite Answer

    x^2 + 6x 恆等 (x+p)^2 + q

    x^2 + 6x = x^2 + 6x + 9 - 9 = (x+3)^2 -9

    therefore p=3 & q = -9

    (1)

    5/(4-y) – (y-1)/(y-4)

    = 5/(4-y) + (y-1)/(4-y)

    = (5+y-1)/(4-y)= (4+y)/(4-y)

    (2)

    1/(x^2+x) - 2/[(x+1)(x+3)]

    = 1/[x(x+1)] - 2/[(x+1)(x+3)]

    = [(x+3)-2x]/[x(x+1)(x+3)]

    = (3-x)/[x(x+1)(x+3)]

    (3)

    1/(x-4) - x/(4-x)^3

    = -[1/(4-x) + x/(4-x)^3]

    = -[(4-x)^2+x]/(4-x)^3

    (4)

    4/(2/w +1) - 2

    = 4w/(2+w) - 2

    = [4w - 2(2+w)]/(2+w)

    = (4w - 4 - 2w)/(2+w)

    = (2w-4)/(2+w)

    = 2(w-2)/(2+w)

    2011-11-08 11:10:55 補充:

    (4) if question is 4/(2 + 1/w) - 2

    = 4w/(2w+1) - 2

    = [4w - 2(2w+1)]/(2w+1)

    = {4w - 4w -2)/(2w+1)

    = -2/(2w+1)

  • 9 years ago

    1. x^2+6x恆等(x+p)^2+q

    x^2+6x=0

    x(x+6)=0

    x=0/-6

    x=0 0=(x+p)^2+q

    0=(0+p)^2+q

    0=p^2+q

    x=-6 0=(x+p)^2+q

    0=(-6+p)^2+q

    0=p^2-12p+36+q

    p^2-12p+36+q=p^2+q

    -12p+36=0

    36=12p

    3=p

    p=3 0=p^2+q

    0=3^2+q

    0=9+q

    -9=q

    p=3,q=-9

    1. 5 y-1

    ---- - --------

    4-y y-4

    5 (y-4) y-1(4-y)

    = ------------- - --------------

    (4-y)(y-4) (y-4)(4-y)

    5(y-4)-(-1)(y-1)(y-4)

    = --------------------------

    (4-y)(y-4)

    (5-(-1)(y-1))(y-4)

    = ------------------------

    (4-y)(y-4)

    5+y-1

    = ----------

    4-y

    y+4

    = ---------

    4-y

    2. 1 2

    ---------- - ----------------

    x^2+x (x+1)(x+3)

    (x+1)(x+3)-2(x^2+x)

    = ---------------------------

    (x^2+x)(x+1)(x+3)

    x^2+4x+3-2x^2-2x

    = ---------------------------

    (x^2+x)(x+1)(x+3)

    (-1)x^2 -2x-3

    = ---------------------------

    (x^2+x)(x+1)(x+3)

    (-1)(x-3)(x+1)

    = ---------------------------

    (x^2+x)(x+1)(x+3)

    3-x

    = ---------------------------

    x^3+4x^2+3x

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