How do you simplify #(3 + 1/(x + (1/(x + 2/x))))/(3/(x + 2))#?

There is a neat method which might be of use in this fraction.

# 1/(x/y) = y/x" " and " "(a/b)/(c/d) = (ad)/(bc)#

Let's do this bit-by-bit....

#(3 + 1/(x + (color(red)(1/(x + 2/x)))))/(3/(x + 2))#

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Simplify the red part:
#color(red)(x/1 + 2/x) = color(red)((x^2+2)/x)" " rArr color(red) (1/((x^2+2)/x) = x/(x^2+2))#

Now we have:
#(3 + 1/(x + (color(red)(x/(x^2+2)))))/(3/(x + 2))#
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#(3 + 1/color(blue)(x + (x/(x^2+2))))/(3/(x + 2))#

Simplify the blue part:

#color(blue)[x/1 + x/(x^2+2) = (x^3 + 2x+x)/(x^2+2) = (x^3 + 3x)/(x^2+2)]#

#color(blue)(1/(x + (x/(x^2+2)))= (x^2+2)/(x^3+3x) #
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Now we have:

#[3 + color(blue)((x^2+2)/(x^3+3x))]/(3/(x + 2))#

#color(green)((3 + (x^2+2)/(x^3+3x))]/(3/(x + 2))#

Simplify the green part:
#color(green)((3 /1+ (x^2+2)/(x^3+3x))]#

=#color(green)((3x^3+9x+x^2+2)/(x^3+3x) = (3x^3+9x+x^2+2)/(x(x^2+3))#
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Now we have
#((3x^3+9x+x^2+2)/(x(x^2+3)))/(3/(x + 2))#

=#((3x^3+9x+x^2+2)(x+2))/(3x(x^2+3)#

Would this simplify further?