What is [(x/2)-(2/x) ]over [(1/2)+(1/x)] simplified?

[(x/2)-(2/x) ] / [(1/2)+(1/x)] => common denominator:

= [(x^2 - 4)/2x) ] / [(x + 2)/2x)] => factor, simplify

= [ 2x (x - 2)(x + 2) ] / [2x(x + 2)] => 2x's and (x+2) cancel , you are left with:

= x - 2

Edit: added explanation.

Step 1: You should first simplify the numerator and denominator..

Therefore: x/2 - 2/x = (x^2 - 4)/2x ---> numerator; where 2x is common denominator.

1/2 + 1/x = (2 + x)/2x ----> denominator: where 2x is common denominator also.

Step 2: Rewrite the equation.. it will be....

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

then.. [ (x^2 - 4)/2x ] * 2x/(2 + x) ---- cancel the 2x because 2x/2x = 1

therefore.. (x^2 - 4) / (2 + x)

Step 3: Simplify (x^2 - 4)

...as you can see.. x^2 - 4 can be rewriten into (x + 2)(x - 2) right? therefore....

(x + 2)(x - 2) / (2 + x)

now you cancel the (x + 2).. and the equation will left with (x - 2)..

Cheeers!!!!! ^_~

(x/2) - (2/x) = (x^2 - 4)/(2*x) = (x + 2)*(x - 2)/(2*x)

1/2 + 1/x = (x + 2)/(2*x)

(x + 2)*(x - 2)/(2*x)/((x + 2)/(2*x)) = x - 2

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

as long as x ≠ 0 or x≠ -2 ; multiply top by x and bottom by x:

= (x²/2 - 2)/ (x/2 +1)

multiply top and bottom by 2:

= (x² - 4)/ (x +2)

use difference of squares on top:

= (x - 2)(x +2)/ (x +2)

cancel out common terms

= x -2

remember x ≠ 0, -2 here.

I am 14 years old...

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