Question

\(\dfrac{\sqrt{x-3}}{\sqrt{\sqrt{x}-\sqrt{3}}}\):\(\dfrac{\sqrt{\sqrt{x}-\sqrt{3}}}{\sqrt{3}}\)

Answer 1

ĐK: \(x>3\)

\(=\dfrac{\sqrt{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}}{\sqrt{\sqrt{x}-\sqrt{3}}}\cdot\dfrac{\sqrt{3}}{\sqrt{\sqrt{x}-\sqrt{3}}}=\dfrac{\sqrt{3}\sqrt{\sqrt{x}+\sqrt{3}}}{\sqrt{\sqrt{x}-\sqrt{3}}}\\ =\dfrac{\sqrt{3}\sqrt{\left(\sqrt{x}+\sqrt{3}\right)^2}}{\sqrt{x-3}}=\dfrac{\sqrt{3}\left(\sqrt{x}+\sqrt{3}\right)}{\sqrt{x-3}}\\ =\dfrac{\sqrt{3x}+3}{\sqrt{x-3}}=\dfrac{\left(\sqrt{3x}+3\right)\sqrt{x-3}}{x-3}=\dfrac{\sqrt{3x^2-9x}+3\sqrt{x-3}}{x-3}\)