a) \(A=\left[\dfrac{x+3}{\left(x-3\right)^2}+\dfrac{6}{x^2-9}-\dfrac{x-3}{\left(x+3\right)^2}\right]\left[1:\left(\dfrac{24x^2}{x^4-81}-\dfrac{12}{x^2+9}\right)\right]\)
\(\left(ĐKXĐ:x\ne\pm3\right)\)
\(=\dfrac{\left(x+3\right)^3+6\left(x-3\right)\left(x+3\right)-\left(x-3\right)^3}{\left(x-3\right)^2\left(x+3\right)^2}\cdot\left[1:\dfrac{24x^2-12\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}\right]\)
\(=\dfrac{x^3+9x^2+27x+27+6x^2-54-x^3+9x^2-27x+27}{\left(x-3\right)^2\left(x+3\right)^2}\cdot\dfrac{\left(x^2-9\right)\left(x^2+9\right)}{24x^2-12x^2+108}\)
\(=\dfrac{24x^2\left(x^2+9\right)\left(x-3\right)\left(x+3\right)}{12\left(x^2+9\right)\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2x^2}{x^2-9}\)
b) \(B=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left[\left(x-2\right)+\dfrac{10-x^2}{x+2}\right]\)
\(=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{1}+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{x^2-4}\cdot\dfrac{x+2}{x^2-4+10-x^2}\)
\(=\dfrac{-6\left(x+2\right)}{6\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-1}{x-2}\)
phần b điều kiện xác định là \(x\ne\pm2\) nhé