Question

ex 1 ! coroana ms !

Answer 1

[tex]A=\{x\in Z^*-\{1\}|\frac{\sqrt{(2\sqrt2-3)^2}+\sqrt{9+4\sqrt2}}{x-1} \in Z\}\\ \sqrt{(2\sqrt2-3)^2}=|2\sqrt2-3|=3-2\sqrt2\\ \sqrt{9+4\sqrt2}=\sqrt{8+1+4\sqrt2}=\sqrt{(2\sqrt2+1)^2}=2\sqrt2+1\\ Revenind:\\ \frac{3-2\sqrt2+2\sqrt2+1}{x-1}=\frac{4}{x-1}\in Z\Rightarrow x-1\in D_4\\ x-1\in \{-4,-2,-1,1,2,4\}\\ x\in \{-3,-2,0,2,3,5\}\\ x\in Z^*\Rightarrow x\in \{-3,-2,2,3,5\}\\ A=\{-3,-2,2,3,5\}[/tex]