\(\left(\dfrac{\sqrt{x}}{7-7\sqrt{x}}+\dfrac{\sqrt{7}}{\sqrt{7x}-x}\right):\dfrac{1}{7\sqrt{x}-x\sqrt{7}}\)
\(=\left(\dfrac{\sqrt{x}}{7\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{7}}{\sqrt{x}\left(\sqrt{7}-\sqrt{x}\right)}\right)\cdot x\sqrt{7}\left(\sqrt{7}-\sqrt{x}\right)\)
\(=\dfrac{x\left(\sqrt{7}-\sqrt{x}\right)+7\sqrt{7}\left(1-\sqrt{x}\right)}{7\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{7}-\sqrt{x}\right)}\cdot x\sqrt{7}\left(\sqrt{7}-\sqrt{x}\right)\)
\(=\dfrac{x\sqrt{7}-x\sqrt{x}+7\sqrt{7}-7\sqrt{7x}}{\sqrt{7x}\cdot\left(1-\sqrt{x}\right)}\)
\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=2+\sqrt{x}-1=\sqrt{x}+1\)
