Question

СРОЧНО!даю 20 балов!​

Answer 1

Первый способ :

$\displaystyle\bf\\\Big(\sqrt{9-4\sqrt{2} } +\sqrt{9+4\sqrt{2} }\Big)^{2} =\\\\\\=\Big(\sqrt{9-4\sqrt{2} }\Big)^{2} +2\cdot\sqrt{(9-4\sqrt{2} )\cdot(9+4\sqrt{2}) }+\Big(\sqrt{9+4\sqrt{2} }\Big)^{2}=\\\\\\=|9-4\sqrt{2}| +2\cdot\sqrt{9^{2}-(4\sqrt{2} )^{2} } +|9+4\sqrt{2}|=\\\\\\=9-4\sqrt{2} +2\cdot\sqrt{81-32}+9+4\sqrt{2} =18+2\sqrt{49}=18+2\cdot 7=32$

Второй способ  :

$\displaystyle\bf\\\Big(\sqrt{9-4\sqrt{2} } +\sqrt{9+4\sqrt{2} }\Big)^{2} =\Big(\sqrt{8-4\sqrt{2}+1 } +\sqrt{8+4\sqrt{2}+1 }\Big)^{2} =\\\\\\=\displaystyle\bf\\\Big(\sqrt{(2\sqrt{2})^{2} -2\cdot 2\sqrt{2}\cdot 1+1^{2} } +\sqrt{(2\sqrt{2})^{2} +2\cdot 2\sqrt{2}\cdot 1+1^{2} }\Big)^{2} =\\\\\\=\displaystyle\bf\\\Big(\sqrt{(2\sqrt{2} -1)^{2} } +\sqrt{(2\sqrt{2}+1)^{2} }\Big)^{2} =\Big(2\sqrt{2} -1+2\sqrt{2}+1\Big)^{2}=\\\\\\=\Big(4\sqrt{2} \Big)^{2} =32$