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Решите задачу:

$1)\; \sqrt{2+\sqrt3}\cdot \sqrt{2+\sqrt{2+\sqrt3}}\cdot \\\\\cdot \left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt3}}}\, \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt3}}\right )=\\\\=\sqrt{2+\sqrt3}\cdot \sqrt{2+\sqrt{2+\sqrt3}}\cdot \sqrt{4-(2+\sqrt{2+\sqrt3})}=\\\\=\sqrt{2+\sqrt3}\cdot \left (\sqrt{2+\sqrt{2+\sqrt3}}\cdot \sqrt{2-\sqrt{2+\sqrt3}}\right )=\\\\=\sqrt{2+\sqrt3}\cdot \sqrt{4-(2+\sqrt3)}=\sqrt{2+\sqrt3}\cdot \sqrt{2-\sqrt3}=\sqrt{4-3}=\sqrt1=1$

$2)\; \; (x^2-9)(\sqrt{6-5x}-x)=0\\\\x^2-9=0\; \; ili\; \; \sqrt{6-5x}-x=0\\\\x=\pm3\; \; ili \\\\ \sqrt{6-5x}=x,\; 6-5x=x^2,\; x^2+5x-6=0,\; x_1=-1,x_2=6\\\\+3-3-1+6=5$

$3)\; \; y=\frac{1}{2x-x^2-3}\\\\y'=-\frac{2-2x}{(2x-x^2-3)^2}=0\; \; \to \; \; 2-2x=0,\; x=1\\\\2x-x^2-3=-x^2+2x-3\ \textgreater \ 0\; ,\; \; t.k.\; \; D=4-12\ \textless \ 0\\\\Znaki\; \; y':\; \; ---(1)+++\\\\x_{min}=1\\\\y_{min}=y(1)=-\frac{1}{2}$