\(\dfrac{x-3}{x+3}+\dfrac{x+3}{x-3}+\dfrac{36}{9-x^2}\\ =\dfrac{x-3}{3+x}-\dfrac{x+3}{3-x}+\dfrac{36}{\left(3-x\right)\left(3+x\right)}\\ =\dfrac{\left(x-3\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{\left(x+3\right)^2}{\left(3-x\right)\left(3+x\right)}+\dfrac{36}{\left(3-x\right)\left(3+x\right)}\\ =\dfrac{3x-x^2-9+3x-x^2-6x-9+36}{\left(3-x\right)\left(3+x\right)}\\ =\dfrac{-2x^2+18}{\left(3-x\right)\left(3+x\right)}\\ =\dfrac{-2\left(x^2-9\right)}{\left(3-x\right)\left(3+x\right)}\\=\dfrac{-2\left(x^2-9\right)}{-\left(x-3\right)\left(x+3\right)}\\ =2\)
__
\(\dfrac{x}{x+3}+\dfrac{x}{x-3}+\dfrac{-18}{x^2-9}\\ =\dfrac{x}{x+3}+\dfrac{x}{x-3}+\dfrac{-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2-3x+x^2+3x-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x^2-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\\ =2\)
__
\(\dfrac{1}{x+3}+\dfrac{1}{x-3}+\dfrac{6}{x^2-9}\\ =\dfrac{1}{x+3}+\dfrac{1}{x-3}+\dfrac{6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}+\dfrac{6}{\left(x-3\right)9x+3}\\ =\dfrac{x-3+x+3+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)
__
\(\dfrac{x}{x+y}+\dfrac{y}{x-y}+\dfrac{2xy}{x^2-y^2}\\ =\dfrac{x}{x+y}+\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}+\dfrac{y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{x^2-xy+xy+y^2+2xy}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{x+y}{x-y}\)
a: \(=\dfrac{\left(x-3\right)^2+\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-6x+9+x^2+6x+9-36}{\left(x+3\right)\left(x-3\right)}=\dfrac{2x^2-18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-18}{x^2-9}=2\)
b: \(\dfrac{x}{x+3}+\dfrac{x}{x-3}+\dfrac{-18}{x^2-9}\)
\(=\dfrac{x\left(x-3\right)+x\left(x+3\right)-18}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-3x+x^2+3x-18}{x^2-9}=\dfrac{2x^2-18}{x^2-9}=2\)
c: \(\dfrac{1}{x+3}+\dfrac{1}{x-3}+\dfrac{6}{x^2-9}\)
\(=\dfrac{x-3+x+3+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)
d: \(=\dfrac{x\left(x-y\right)+y\left(x+y\right)+2xy}{x^2-y^2}\)
\(=\dfrac{x^2+y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)
