Pre Calculus Help?

Question

Simplify expression.

(x+h)^-3 -x^-3
__________
h


x^-1+y-1
_______
(x+y)^ -1



This one is hard
1+___1_____
1+ 1/ 1+ x


and this one

http://i51.photobucket.com/albums/f399/Raymondxv12/math.jpg

Answer 1

1ª)

(x+h)^-3 -x^-3
__________ =
h

[x^3-(x+h)^3] / [h · x^3 · (x+h)^3] =
[-(3hx^2+3h^2x+h^3)]/ [h · x^3 · (x+h)^3] =

-(3x^2+3hx+h^2)/ [ x^3 · (x+h)^3]

2º)
x^-1+y-1
_______ =
(x+y)^ -1

(y-x)(x+y)/(x·y) = (y^2-x^2) / (x·y)

3º)
1+___1_____ =
1+ 1/ 1+ x

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

Saludos.

Answer 2

1. [1/(x + h)^3 - 1/(x^3)]/h = [x^3 - (x + h)^3]/(x^3)*((x + h)^3)*h = [x^3 - (x^3 + 3(x^2)*h + 3x*h^2 + h^3)]/(x^3)*((x + h)^3)*h =
(-3(x^2)*h - 3x*h^2 - h^3)/(x^3)*((x + h)^3)*h =
(-3x^2 - 3xH + h^2)/(x^3)*(x + h)^3.

2. Given expression = (1/x + 1/y)/[1/(x + y)] =
[(y + x)/xy]/[1/(x + y)] = [(x + y)^2]/xy = x/y + 2 + y/x.

3. I think you mean 1 +{1/[1 + 1/(1 + x)]}. If that is your question, that is = 1 + {1/[(1 + x + 1)/(1 + x)]} =
1 +(1 + x)/(2 + x) = (2 + x + 1 + x)/(2 + x) = (3 + 2x)/(2 + x).

4. Your photobucket link is illegible.