Монгол Бодлогын Сан

Заавар: Нэмэгдэхүүн тус бүрийг хялбарчил.

Бодолт: \begin{gather} \sqrt{146.5^2-109.5^2+27\cdot 16^2}=\sqrt{(146.5-109.5)(146.5+109.5)+27\cdot 16^2}=\\ =\sqrt{37\cdot 256+27\cdot 256}=\sqrt{64\cdot 256}=8\cdot 16=128,\\ \Bigl(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\Bigr)^2=12+2\sqrt{11}-2\sqrt{12+2\sqrt{11}}\cdot \sqrt{12-2\sqrt{11}}+{}\\ {}+12-2\sqrt{11}=24-2\sqrt{12^2-(2\sqrt{11})^2}=24-2\sqrt{144-44}=24-2\cdot 10=4,\\ \sqrt 3\Bigl(\sqrt{5+2\sqrt 6}+\sqrt{5-2\sqrt 6}\Bigr)=\sqrt3\left(\sqrt{(\sqrt3+\sqrt2)^2}+\sqrt{(\sqrt3-\sqrt2)^2}\right)=\\ =\sqrt3(\sqrt3+\sqrt2+\sqrt3-\sqrt2)=\sqrt3\cdot2\sqrt3=6 \end{gather} тул $$\dfrac{\sqrt{146.5^2-109.5^2+27\cdot 16^2}}{\Bigl(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\Bigr)^2}+\sqrt 3\Bigl(\sqrt{5+2\sqrt 6}+\sqrt{5-2\sqrt 6}\Bigr)=\dfrac{128}{4}+6=38$$