Question

mă ajutați, am foarte multe teme de vacanță? determinați elementele ​

Answer 1

Explicație pas cu pas:

g)

$\frac{2x - 3}{x + 5} = \frac{2x + 10 - 13}{x + 5} = \frac{2(x + 5) - 13}{x + 5} = 2 - \frac{13}{x + 5}\in \mathbb{Z}$

$\implies \frac{13}{x + 5}\in \mathbb{Z} \iff (x + 5) ∈ D_{13} \\ \iff (x + 5) ∈ \{ - 13; - 1;1;13 \}$

$x + 5 = - 13 \implies x = - 18$

$x + 5 = - 1 \implies x = - 6$

$x + 5 = 1 \implies x = - 4$

$x + 5 = 13 \implies x = 8$

h)

${x}^{2} + x = 2 \cdot (1 + 2 + 3 + ... + 2021) \\ x(x + 1) = 2 \cdot \frac{2021 \cdot 2022}{2} \\ x(x + 1) = 2021 \cdot 2022 \\ x(x + 1) = 2021 \cdot (2021 + 1) \\ x = 2021 \implies \bf H = \{2021\}$

Answer 2

g)

$\it x\in\mathbb{Z};\ \ \dfrac{2x-3}{x+5}\in\mathbb{Z},\ \ x=?\\ \\ \\ \dfrac{2x-3}{x+5}=\dfrac{2x+10-13}{x+5}=\dfrac{2(x+5)}{x+5}-\dfrac{13}{x+5}=2-\dfrac{13}{x+5}\in\mathbb{Z} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{13}{x+5}\in\mathbb{Z} \Rightarrow x+5\Big|13 \Rightarrow x+5\in\{-13,\ -1,\ 1,\ 13\}\Big|_{-5} \Rightarrow\\ \\ \\ \Rightarrow x\in\{-18,\ \ -6,\ \ -4,\ \ 8\}=G$

h)

$\it x\in\mathbb{N}\\ \\ x^2+x=2(1+2+3+\ ...+ 2021) \Rightarrow x^2+x=2\cdot\dfrac{2021\cdot2022}{2} \Rightarrow\\ \\ \\ \Rightarrow x(x+1)=2021\cdot2022 \Rightarrow x=2021 \Rightarrow H=\{2021\}$