Question

ma puteti ajuta cu punctul b si sa imi explicati cum se face genu asta de probleme​

Answer 1

Sper ca te-am ajutat

Answer 2

a)

$\it \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{y+x}{xy}=\dfrac{\sqrt7+\sqrt2+\sqrt7-\sqrt2}{(\sqrt7-\sqrt2)(\sqrt7+\sqrt2)}=\dfrac{2\sqrt7}{7-2}=\dfrac{2\sqrt7}{5}=\dfrac{\sqrt{28}}{5}\\ \\ \\ 25<28<36 \Rightarrow \sqrt{25}<\sqrt{28}<\sqrt{36} \Rightarrow 5<\sqrt{28}<6|_{:5} \Rightarrow \\ \\ \Rightarrow 1<\dfrac{\sqrt{28}}{5}<\dfrac{6}{5} \Rightarrow \dfrac{\sqrt{28}}{5}\in\Big(1,\ \dfrac{6}{5}\Big) \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}\in\Big(1,\ \dfrac{6}{5}\Big)$

b)

$\it (x-y+2\sqrt2)^{2022}=(\sqrt7-\sqrt2-\sqrt7-\sqrt2+2\sqrt2)^{2022}=0^{2022}=0$