a)
\(4x\left(x-7\right)-\left(2-3x\right)^2\\ =4x^2-28x-\left(4-12x+9x^2\right)\\ =4x^2-28x-4+12x-9x^2\\ =4x^2-9x^2-28x+12x-4\)
\(=-5x^2-16x-4\)
b)
\(\dfrac{3}{x-3}+\dfrac{-6x}{x^2-9}+\dfrac{x}{x+3}\\ =\dfrac{3}{x-3}+\dfrac{-6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{-6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3\left(x+3\right)-6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x-3}{x+3}\)
