Question

\(\dfrac{y+12-4\sqrt{3y}}{y-12}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}y>=0\\y< >12\end{matrix}\right.\)

\(\dfrac{y+12-4\sqrt{3y}}{y-12}\)

\(=\dfrac{y-2\cdot\sqrt{y}\cdot2\sqrt{3}+12}{y-12}=\dfrac{\left(\sqrt{y}-2\sqrt{3}\right)^2}{y-12}\)

\(=\dfrac{\sqrt{y}-2\sqrt{3}}{\sqrt{y}+2\sqrt{3}}\)