ĐKXĐ:
\(x\ne\pm2\)
\(\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}=\dfrac{x+2}{x^2-4}+\dfrac{x-2}{x^2-4}+\dfrac{x^2+1}{x^2-4}=\dfrac{x+2+x-2+x^2+1}{x^2-4}=\dfrac{x^2+2x+1}{x^2-4}=\dfrac{\left(x+1\right)^2}{x^2-4}\)b, Để B = 2 hay \(\dfrac{\left(x+1\right)^2}{x^2-4}=2\Rightarrow x^2+2x+1=2\left(x^2-4\right)\Leftrightarrow x^2+2x+1-2x^2+8=0\Leftrightarrow2x-x^2+9=0\Leftrightarrow-\left(x^2-2x+1\right)+10=0\Rightarrow-\left(x-1\right)^2=-10\Rightarrow\left(x-1\right)^2=10\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{10}\\x-1=-\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{10}+1\\x=-\sqrt{10}+1\end{matrix}\right.\)
