Answer:
[tex]\large \boxed{\sf \ \ \dfrac{1}{x^2}+\dfrac{1}{x^2+x}=\dfrac{2x+1}{x^2(x+1)} \ \ }[/tex]
Step-by-step explanation:
Hello,
This is the same method as computing for instance:
[tex]\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{3+2}{2*3}=\dfrac{5}{6}[/tex]
We need to find the same denominator.
Let's do it !
For any x real different from 0, we can write:
[tex]\dfrac{1}{x^2}+\dfrac{1}{x^2+x}=\dfrac{1}{x^2}+\dfrac{1}{x(x+1)}\\\\=\dfrac{x+1+x}{x^2(x+1)}=\dfrac{2x+1}{x^2(x+1)}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
