ĐK: $x \neq \pm 1$.
Ta có
$(\dfrac{x+2}{x+1})^2 + (\dfrac{x-2}{x-1})^2 - \dfrac{5}{2}(\dfrac{x^2-4}{x^2-1}) = 0$
$<-> (\dfrac{x+2}{x+1})^2 + (\dfrac{x-2}{x-1})^2 - \dfrac{5}{2} (\dfrac{(x-2)(x+2)}{(x-1)(x+1)}) = 0$
Ta đặt $a = \dfrac{x+2}{x+1}, b = \dfrac{x-2}{x-1}$. Khi đó ptrinh trở thành
$a^2 + b^2 - \dfrac{5}{2} ab = 0$
$<-> 2a^2 - 5ab + 2b^2 = 0$
$<-> 2a^2 - 4ab - ab + 2b^2 = 0$
$<-> 2a(a-2b) - b(a-2b) = 0$
$<->(2a-b)(a-2b) = 0$
Vậy $2a = b$ hoặc $a = 2b$
TH1: $2a = b$
Khi đó
$2 \dfrac{x+2}{x+1} = \dfrac{x-2}{x-1}$
$<-> 2(x+2)(x-1) = (x-2)(x+1)$
$<-> 2(x^2 +x - 2) = x^2 -x -2$
$<-> x^2 +3x -2 = 0$
Vậy $x = \dfrac{-3\pm \sqrt{17}}{2}$
TH2: $a = 2b$
Khi đó
$\dfrac{x+2}{x+1} = 2\dfrac{x-2}{x-1}$
$<-> (x+2)(x-1) = 2(x-2)(x+1)$
$<-> 2(x^2 -x - 2) = x^2 +x -2$
$<-> x^2 -3x -2 = 0$
Vạy $x = \dfrac{3 \pm \sqrt{17}}{2}$
