Question

Bună! Mă puteți ajuta, va rog? Mulțumesc!
Măcar un exercițiu.​

Answer 1

Răspuns:

x = -2

Explicație pas cu pas:

-------PROBLEMA 8----------------------$(x+3x+5x+7x+.....+2019x)-(2x+4x+6x+8x+.......+2020x)=2020\\x(1+3+5+7+........+2019)-x(2+4+6+8+.......+2020) = 2020\\x [(1+3+5+7+........+2019)-(2+4+6+8+.......+2020)]=2020$aplicam suma lui gauss in fiecare paranteza rotunda

$x{[(2019+1)*1010:2] - [(2020+2)*1010:2]} =2020\\x(2020 *505 - 2022 * 505)=2020\\x * 505(2020-2022) = 2020\\\\x *(-2) = 4\\-2x=4\\x=-2$

Observatie pentru a afla cati termeni sunt intr-un sir (suma) este o formula: (cel mai mare termen - cel mai mic termen) : pas +1

Answer 2

$\textbf{8.}\,\,\,(x+3x+5x+...+2019x) - (2x+4x+6x+...+2020x) = 2020$

$(1+3+5+...+2019)x - (2+4+6+...+2020)x = 2020$

$(1-2+3-4+5-6+...+2019-2020)x = 2020$

$(-1-1-1-\underset{\text{de 1010 ori}}{\underbrace{...}}-1)x = 2020$

$-1010x = 2020\Rightarrow x = \dfrac{2020}{-1010}\Rightarrow \boxed{x = -2}$

$\textbf{9.}\,\,\,(2x-3)(y+2) = -14 \Rightarrow (3-2x)(y+2) = 14 \Rightarrow$

$\Rightarrow y+2 = \dfrac{14}{3-2x}$

$\Rightarrow 3-2x\in D_{14}$

$\Rightarrow 3-2x\in \{-14,-7,-1,1,7,14\} \Big|-3$

$\Rightarrow -2x\in \{-17,-10,-4,-2,5,11\}\Big|:(-2)$

$\Rightarrow x \in \left\{\frac{17}{2}, \frac{10}{2}, \frac{4}{2},\frac{2}{2}, \frac{-5}{2},\frac{-11}{2}\right\}$

$\Rightarrow x\in \{1,2,5\} \Rightarrow y+2\in \{-14,-2, 14\}\Big|-2 \Rightarrow y\in \{-16, -4, 12\}$

$\Rightarrow \boxed{(x,y) = \left\{(1,12); (2,-16);(7,-4)\right\}}$