Question

Va rog, URGENT, EX 6, DAU CORONITA

Answer 1

$\it 5)\ x^2-6x+10=x^2-6x+9+1=(x-3)^2+1\geq1 \Rightarrow x^2-6x+10\in\mathbb{R}_{+}$

$\it 6)\ x-\dfrac{1}{x}=2\sqrt2\Rightarrow \Big(x-\dfrac{1}{x}\Big)^2=(2\sqrt2)^2\Rightarrow x^2-2\cdot x\cdot\dfrac{1}{x}+\Big(\dfrac{1}{x}\Big)^2=4\cdot2 \Rightarrow\\ \\ \\ \Rightarrow x^2+\dfrac{1}{x^2}-2=8|_{+2} \Rightarrow x^2+\dfrac{1}{x^2}=10$