Answer:
[tex]\boxed{\boxed{\dfrac{2}{x^2+x}-\dfrac{1}{x}=\dfrac{1-x}{x(x+1)}}}[/tex]
Step-by-step explanation:
The given expression is,
[tex]\dfrac{2}{x^2+x}-\dfrac{1}{x}[/tex]
Factoring [tex]x^2+x[/tex],
[tex]x^2+x=x(x+1)[/tex]
Hence,
[tex]=\dfrac{2}{x(x+1)}-\dfrac{1}{x}[/tex]
[tex]\mathrm{Least\:Common\:Multiplier\:of\:}x\left(x+1\right),x = x\left(x+1\right)[/tex]
So,
[tex]=\dfrac{2-(x+1)}{x(x+1)}[/tex]
[tex]=\dfrac{2-x-1}{x(x+1)}[/tex]
[tex]=\dfrac{1-x}{x(x+1)}[/tex]
