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# 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

- williamLv 59 years agoFavorite 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