Question

\(\left(\dfrac{18}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\)

Answer 1

\(đkx\ne3;x\ne-3\\ =\left(\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right):\left(\dfrac{x+3-1}{x+3}\right)\\ =\left(\dfrac{18+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+2}{x+3}\right)\\ =\dfrac{18+\left(x^2-4x+3x-12\right)-\left(x^2-x-3x+3\right)}{\left(x+3\right)\left(x-3\right)}.\dfrac{x+3}{\left(x+2\right)}\\ =\dfrac{18+x^2-x-12-x^2+4x-3}{x^2-9}.\dfrac{x+3}{x+2}=\dfrac{3x+6}{x-3}.\dfrac{1}{x+2}\\ =\dfrac{3\left(x+2\right)}{x-3}.\dfrac{1}{x+2}\\ =\dfrac{3}{x-3}\)