ĐKXĐ: x\(\ne\pm2\)
a, A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
= \(\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
= \(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x^2-4+10-x^2}\)
= \(\dfrac{-6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)
= \(\dfrac{-1}{x-2}\)
Vậy A=\(\dfrac{-1}{x-2}\) với x\(\ne\pm2\)
b, Để A<0 thì \(\dfrac{-1}{x-2}\)<0
<=> x-2>0 <=> x>2 (thỏa mãn ĐKXĐ)
Vậy để A<0 thì x>2
