\(D=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4x}{x-1}\)ĐK:x\(\ge\)0.x khác 0
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x}{x-1}\)
\(=\dfrac{4x+4\sqrt{x}}{x-1}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)