Question

Answer 1

a: \(\dfrac{x^2+3x}{x+3}=\dfrac{x\left(x+3\right)}{x+3}=x\)

b: \(\dfrac{x-2}{x^2-5x+6}=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\)

c: \(\dfrac{1}{x+y}+\dfrac{y}{x^2-y^2}\)

\(=\dfrac{x-y+y}{x^2-y^2}=\dfrac{x}{x^2-y^2}\)

Answer 2

\(\left(a\right)\dfrac{x^2+3x}{x+3}=\dfrac{x\left(x+3\right)}{x+3}=x\\ \left(b\right)\dfrac{x-2}{x^2-5x+6}=\dfrac{x-2}{x^2-2x-3x+6}=\dfrac{x-2}{x\left(x-2\right)-3\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\\ \left(c\right)\dfrac{1}{x+y}+\dfrac{y}{x^2-y^2}=\dfrac{x-y+y}{x^2-y^2}=\dfrac{x}{x^2-y^2}\)