Question 148530
If I imagine this problem correctly, it looks like this:

{{{(((x+1)/(x+2))+(1/x))/(x/(x+2))=1/x^2}}}


First add x+1/x+2 and 1/x.  First get a common denominator: x(x+2).  So multiply x+1/x+2 by x/x and 1/x by x+2/x+2.


{{{((x(x+1)/x(x+2))+(1(x+2)/(x+2)x))/(x/(x+2))=1/x^2}}}


Simplify it


{{{((x(x+1)/x(x+2))+(1(x+2)/(x+2)x))/(x/(x+2))=1/x^2}}}
becomes (in the numerators)

{{{((x^2+x)/(x^2+2x))+(x+2)/(x^2+2x))/(x/(x+2))=1/x^2}}}

Add in the numerators and simplify

{{{((x^2+2x+2)/(x^2+2x))/(x/(x+2))=1/x^2}}}


Flip the second fraction
{{{((x^2+2x+2)/(x^2+2x))((x+2)/x)=1/x^2}}}


Multiply across

{{{(x^3+2x^2+2x+2x^2+4x+4)/(x^3+2x^2)}}}


Add the terms


{{{(x^3+4x^2+6x+4)/(x^3+2x^2)}}}


Group

{{{((x^3+4x^2)+(6x+4))/((x^3+2x^2))}}}


Factor


{{{(x^2(x+4)+2(3x+2))/(x^2(x+2))}}}


Cancel out the {{{x^2}}}'s


{{{((x+4)+2(3x+2))/(x+2)}}}