ĐK $x\ne \pm 1$
PT $\Leftrightarrow \dfrac{x^2(x+1)^2 +x^2 (x-1)^2}{(x^2-1)^2}=\dfrac{10}{9}$
$\Leftrightarrow \dfrac{2x^2 (x^2+1)}{(x^2-1)^2}=\dfrac{10}{9}$ đặt $x^2-1=a$
Ta có $\dfrac{2(a+1)(a+2)}{a^2}=\dfrac{10}{9}$
$\Leftrightarrow 24a^2 +27a+18=0$
+ $a= -6=x^2-1$ vô nghiệm
+ $a=-\dfrac{3}{4}=x^2-1 \Rightarrow x=\pm \dfrac{1}{2}$