Đáp án+Giải thích các bước giải:
$a)ĐKXĐ:x\neq$$±1;x\neq3$
$A=$$(\dfrac{1}{x+1}+$ $\dfrac{1}{1-x}+$ $\dfrac{2}{1-x^2}):$ $\dfrac{x-3}{x-2}$
$=$$\dfrac{1-x+1+x+2}{(1-x)(1+x)}:$ $\dfrac{x-3}{x-2}$
$=$$\dfrac{4}{(1-x)(1+x)}.$ $\dfrac{x-2}{x-3}$
$=$$\dfrac{4(x-2)}{(1-x^2)(x-3)}$
$b)A=$$\dfrac{1}{3}⇒\dfrac{4(x-2)}{(1-x^2)(x-3)}=$ $\dfrac{1}{3}$
$⇒12(x-2)=(1-x^2)(x-3)$
$⇒ 12x-24=-x^3+3x^2+x-3$
$⇒x^3-3x^2+11x-21=0$
$⇒x=6,321$
$c)A<1⇒\dfrac{4(x-2)}{(1-x^2)(x-3)}<1$
$⇒$$\dfrac{4(x-2)-(1-x^2)(x-3)}{(1-x^2)(x-3)}<0$
$⇒4x-8+x^3-3x^2-x+3<0$
$⇒ x^3-3x^2+3x-5<0$