ĐKXĐ: $ x\ne$ $\pm 2$; $x\ne1$
a)$B$=$(\dfrac{16x-x^2}{x^2-4}$+$\dfrac{3+2x}{2-x}$+$\dfrac{3x-2}{x+2})$.$\dfrac{x^2+4x+4}{x-1}$
$\to$ $B=\dfrac{16x-x^2-(3+2x)(x+2)+ (3x-2)(x-2)}{(x+2)(x-2)}$.$\dfrac{(x+2)^2}{x-1}$
$\to$ $B=\dfrac{16x-x^2-2x^2-7x-6+3x^2-8x+4}{(x+2)(x-2)}$.$\dfrac{(x+2)^2}{x-1}$
$\to$ $B=\dfrac{x-2}{(x+2)(x-2)}$.$\dfrac{(x+2)^2}{x-1}$=$\frac{x+2}{x-1}$
b) Để B=0
$\to$ $\dfrac{x+2}{x-1}=0$
$\to$ $x+2=0$ $\to$ $x=-2$
