Lời giải:
a.
\(A=\frac{x-4}{x^3-3x^2+x-3}: \frac{x^2-5x+4}{(x-2)(x-3)}=\frac{x-4}{(x-3)(x^2+1)}:\frac{(x-1)(x-4)}{(x-2)(x-3)}\)
\(=\frac{x-4}{(x-3)(x^2+1)}.\frac{(x-2)(x-3)}{(x-1)(x-4)}=\frac{x-2}{(x-1)(x^2+1)}\)
b/
\(D=\frac{x^2-16}{3x^3-3x}:\frac{(x-1)^2-6(x-1)+9}{3x^2-3x}\\ =\frac{(x-4)(x+4)}{3x(x^2-1)}:\frac{(x-1-3)^2}{3x(x-1)}\\ =\frac{(x-4)(x+4)}{3x(x-1)(x+1)}.\frac{3x(x-1)}{(x-4)^2}=\frac{x+4}{(x+1)(x-4)}\)
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