Đáp án:
$\dfrac{1}{x(x+2)}-\dfrac{1}{(x+1)^2}=\dfrac{1}{12}$
$\text{ĐKXĐ : $x \neq 0 ; x \neq -2 ; x \neq -1$ }$
$⇔\dfrac{12(x+1)^2}{12x(x+2)(x+1)^2} - \dfrac{12x(x+2)}{12x(x+2)(x+1)^2} =\dfrac{x(x+2)(x+1)^2}{12x(x+2)(x+1)^2}$
$⇔12(x+1)^2 - 12x(x+2) = x(x+2)(x+1)^2$
$⇔12(x^2 +2x+1) -12x^2 -24x = x(x+2)(x^2+2x+1)$
$⇔12x^2 +24x+12 -12x^2-24x =x^4+4x^3 +5x^2+2x$
$⇔x^4 +4x^3 +5x^2 +2x -12=0$
$⇔x^4 +3x^3 +x^3 +3x^2 +2x^2 +6x-4x -12=0$
$⇔x^3(x+3)+x^2(x+3)+2x(x+3)-4(x+3)=0$
$⇔(x+3)(x^3 +x^2 +2x -4)=0$
$⇔(x+3)(x^3 -x^2 +2x^2-2x+4x -4)=0$
$⇔(x+3)[ x^2(x-1) -2x(x-1)-4(x-1)] =0$
$⇔(x+3)(x-1)(x^2-2x-4)=0$
Ta có : $x^2-2x-4=0$
$⇔x^2 -2x +1 +3 =0$
$⇔(x-1)^2 +3 =0$
Vì $(x-1)^2 ≥ 0$
Nên $(x-1)^2 +3 > 0$ (loại)
$⇔(x+3)(x-1)=0$
$⇔$\(\left[ \begin{array}{l}x+3=0\\x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-3(nhận)\\x=1(nhận)\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S = {-3 ; 1 }}$
