Question

Answer 1

\(a,=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}=\dfrac{3}{3-x}\)

\(b,=\dfrac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x+2\right)^2}{8}=\dfrac{8\left(x+2\right)}{8\left(x-2\right)}=\dfrac{x+2}{x-2}\\ c,=\dfrac{9x^2+3x+2x-6x^2}{\left(3x+1\right)\left(1-3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}=\dfrac{x\left(3x+5\right)\left(1-3x\right)}{2x\left(3x+1\right)\left(3x+5\right)}=\dfrac{1-3x}{2\left(3x+1\right)}\)