Giải bài 50 trang 30 – SGK Toán lớp 9 tập 1

Hướng dẫn: $\dfrac{A}{\sqrt{B}}=\dfrac{A\sqrt{B}}{B}\,\left( B>0 \right)$ 

$\begin{align} & \dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}; \\ & \dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}; \\ & \dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{20}}{3.20}=\dfrac{2\sqrt{5}}{60}=\dfrac{\sqrt{5}}{30}; \\ & \dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2\left( \sqrt{2}+1 \right)\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}; \\ \end{align}$
Với $b\ne 0, y>0$, ta có: 
$\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{\left( y+b\sqrt{y} \right)\sqrt{y}}{by}=\dfrac{y\left( \sqrt{y}+b \right)}{by}=\dfrac{\sqrt{y}+b}{b}$