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