Question

Exercitiile 5,6 va rog ! Dau 13p​ + coroana

Answer 1

$\it 5)\ a=x^2-6x+10=x^2-6x+9+1=(x-3)^2+1\geq1\Rightarrow a\in\mathbb{R}_{+}\\ \\ \\ 6)\ 7+2\sqrt{10}=5+2+2\sqrt{10} =(\sqrt5)^2+2\sqrt{10}+(\sqrt2)^2=(\sqrt5+\sqrt2)^2\ \ \ (1)\\ \\ Analog,\ \ 7-2\sqrt{10} =(\sqrt5-\sqrt2)^2\ \ \ \ (2)\\ \\ \\ (2\sqrt5-5)^2=(5-2\sqrt5)^2\ \ \ \ \ (3)\\ \\ \\ (1),\ (2),\ (3)\Rightarrow b=\sqrt{(\sqrt5+\sqrt2)^2}+\sqrt{(\sqrt5-\sqrt2)^2}+\sqrt{(5-2\sqrt5)^2}=\\ \\ =|\sqrt5+\sqrt2| +|\sqrt5-\sqrt2|+|5-2\sqrt5| =\sqrt5+\sqrt2+\sqrt5-\sqrt2+5-2\sqrt5=\\ \\ = 5\in\mathbb{Z}$