\(D=\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(=\frac{\left[\left(x+2\right)+\sqrt{x^2-4}\right]^2}{\left(x+2-\sqrt{x^2-4}\right)\left(x+2+\sqrt{x^2-4}\right)}+\frac{\left[\left(x+2\right)-\sqrt{x^2-4}\right]^2}{\left(x+2-\sqrt{x^2-4}\right)\left(x+2+\sqrt{x^2-4}\right)}\)
\(=\frac{\left[\left(x+2\right)+\sqrt{x^2-4}\right]^2+\left[\left(x+2\right)-\sqrt{x^2-4}\right]^2}{\left(x+2-\sqrt{x^2-4}\right)\left(x+2+\sqrt{x^2-4}\right)}\)
\(=\frac{\left[\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+\left(\sqrt{x^2-4}\right)^2\right]+\left[\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+\left(\sqrt{x^2-4}\right)^2\right]}{\left(x+2\right)^2-\left(\sqrt{x^2-4}\right)^2}\)
\(=\frac{\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+x^2-4+\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+x^2-4}{x^2+4x+4-\left(x^2-4\right)}\)
\(=\frac{2\left(x^2+4x+4+x^2-4\right)}{x^2+4x+4-x^2+4}\)\(=\frac{2\left(2x^2+4x\right)}{4x+8}\)\(=\frac{4x\left(x+2\right)}{4\left(x+2\right)}=x\)
Vậy \(D=x\).
