Question

Rut gon :\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

Answer 1

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-1\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{x-4}{x-1}\)