Question

Va rog dau coroana exercitiul 4 , urgent​
@augustindevian

Answer 1

Explicație pas cu pas:

a)

15(x²-2) + 3(x²-2) + 15 + 2 = -1

18(x² - 2) = -18

x² - 2 = -1 <=> x² = 1 => x = ±1

b)

$x \circ y = 3xy + 3x + 3y + 3 - 1 = 3x(y+1) + 3(y+1) - 1 = \\ = (3x+3)(y+1) - 1 = \bf 3(x+1)(y+1) - 1$

c)

$x \circ (-1) = (-1) \circ y = -1$

$\underbrace{(-2009) \circ (-2008) \circ...\circ (-2)}_{x} \circ (-1) \circ \underbrace{0 \circ 1 \circ...\circ 2008 \circ 2009}_{y} = x \circ (-1) \circ y = (-1) \circ y = -1$

d)

element neutru:

$x \circ e = e \circ x = x$

$x \circ e = x \iff 3ex + 3x + 3e + 2 = x \\ 3e(x + 1) + 2(x + 1) = 0 \\ (x + 1)(3e + 2) = 0 \\ 3e + 2 = 0 \implies e = - \frac{2}{3}$

$e \circ x = x \iff 3ex + 3e + 3x + 2 = x \\ 3e(x + 1) + 2(x + 1) = 0 \\ (x + 1)(3e + 2) = 0 \\ 3e + 2 = 0 \implies e = - \frac{2}{3}$

simetricul lui x = -3 în raport cu legea "°"

$x \circ x^{\prime} = x^{\prime} \circ x = e$

$(- 3) \circ x^{\prime} = - \dfrac{2}{3}$

$3( - 3)x^{\prime} + 3( - 3) + 3x^{\prime} + 2 = - \dfrac{2}{3} \\ - 6x^{\prime} = \dfrac{19}{3} \implies x^{\prime} = - \dfrac{19}{18}$

$x^{\prime} \circ ( - 3) = - \dfrac{2}{3}$

$3( - 3)x^{\prime} + 3x^{\prime} + 3( - 3) + 2 = - \dfrac{2}{3} \\ - 6x^{\prime} = \dfrac{19}{3} \implies x^{\prime} = - \dfrac{19}{18}$

Answer 2

Răspuns:

Explicație pas cu pas: