a)Thay x = 3, ta có
\(A=\dfrac{2.3}{3-1}=\dfrac{6}{2}=3\)
b)ĐKXĐ:\(x\ne2,-2\)\(B=\dfrac{x}{x+2}-\dfrac{x^2+8}{x^2-4}+\dfrac{3}{x-2}=\dfrac{x^2-2x+3x+6}{x^2-4}-\dfrac{x^2+8}{x^2-4}=\dfrac{x^2+x+6-x^2-8}{x^2-4}=\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}\)
c)\(A.B=1\Leftrightarrow3.\dfrac{1}{x+2}=1\Leftrightarrow3=x+2\Leftrightarrow x=1\left(tm\right)\)