Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 552: 141

True

The statement is true because the math holds: \begin{equation} \begin{aligned} \frac{4 \sqrt{x}}{\sqrt{x}-y}&=\frac{4 \sqrt{x}}{\sqrt{x}-y}\cdot \frac{\sqrt{x}+y}{\sqrt{x}+y} \\ &=\frac{4 \sqrt{x}\left( \sqrt{x}+y \right)}{x-y^2}\\ &=\frac{4x+ 4y\sqrt{x}}{x-y^2}\\ \end{aligned} \end{equation}