Question

Poate cineva sa ma ajute cu cele 3 ex?​

Answer 1

Explicație pas cu pas:

pe 2 nu mai stiu cum sa il fac sunt la liceu si nu am mai fc destea si le am uitat

Answer 2

$\bf 3.\\ \\ \it \dfrac{2a+3b}{5a+2b}=\dfrac{\dfrac{2a+3b}{b}}{\dfrac{5a+2b}{b}}=\dfrac{2\dfrac{a}{b}+3}{5\dfrac{a}{b}+2}=\dfrac{2\cdot\dfrac{3}{7}+3}{5\cdot\dfrac{3}{7}+2}=\dfrac{\dfrac{6}{7}+3}{\dfrac{15}{7}+2}=\dfrac{\dfrac{27}{7}}{\dfrac{29}{7}}=\dfrac{27}{29}$

$\bf4\\ \\ \it A=\{x\in\mathbb{Z}|\ -2 < x < 2\}=\{-1,\ 0,\ \ 1\}\\ \\ A\cup B=\{-1,\ 0,\ 1\} \cup \{0,\ 1,\ 3\}=\{-1,\ 0,\ 1,\ 3\}$

$\bf 5\\ \\ \it x=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2\\ \\ y=\sqrt6-\dfrac{^{\sqrt2)}2}{\ \ \sqrt2}=\sqrt6-\dfrac{\not2\sqrt2}{\not2}=\sqrt6-\sqrt2\\ \\ \\ m_g=\sqrt{x\cdot y}=\sqrt{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\sqrt{6-2}=\sqrt4=2$