Монгол Бодлогын Сан

Заавар:$$a\sqrt{b}=\sqrt{a^2b}$$

Бодолт:\begin{align*} \text{Илэрх.}&=\big((5\sqrt2+2\sqrt5)\cdot(\sqrt{50}-\sqrt{20})\big)^2\\ &=\big((\sqrt{50}+\sqrt{20})\cdot(\sqrt{50}-\sqrt{20})\big)^2 & & \color{red}{\leftarrow a\sqrt{b}=\sqrt{a^2\cdot b}}\\ &=\big((\sqrt{50})^2-(\sqrt{20})^2\big)^2 & & \color{red}{\leftarrow (a-b)(a+b)=a^2-b^2}\\ &=(50-20)^2=900 & & \color{red}{\leftarrow (\sqrt{a})^2=a} \end{align*}