Question

Pe multimea numerelor reale se considera legea de compozitie x*y=-xy-x-y-2
A)arătați ca x*y=-(x+1)(y+1)-1
B)determinati elementul neutru al legii de compozitie
C) este in poza​

Answer 1

Explicație pas cu pas:

a)

x*y = -(x+1)(y+1)-1 = -(xy+x+y+1)-1 = -xy-x-y-1-1 = -xy-x-y-2

sau:

x*y = -xy-x-y-2 = -xy-x-y-1-1 = -x(y+1) -(y+1) - 1 = -(x+1)(y+1) - 1

b)

x*e = x

-x•e-x-e-2 = x

-x•e-2x-e-2 = 0 <=> x•e+2x + e+2 = 0

x•(e+2) + e+2 = 0

(x+1)(e+2) = 0

▪︎dacă x = -1 => e∈R

▪︎dacă x ≠ -1 => e = -2

c)

(dacă sunt logaritmi...)

x*(-1) = (-1)*x = -1, ∀ x∈ℝ

$log_{2} \dfrac{1}{10} = log_{2} {10}^{ - 1} = - 1$

$\underbrace{log_{2} \dfrac{1}{100} \ast log_{2} \dfrac{1}{99} \ast ... \ast log_{2} \dfrac{1}{11}}_{x} \ast log_{2} \dfrac{1}{10} \ast \underbrace{log_{2} \dfrac{1}{9} \ast ... \ast log_{2} \dfrac{1}{2} \ast log_{2} \dfrac{1}{1}}_{y} =$

$= \underbrace{log_{2} \dfrac{1}{100} \ast log_{2} \dfrac{1}{99} \ast ... \ast log_{2} \dfrac{1}{11}}_{x} \ast { \red {\bf (- 1)}} \ast \underbrace{log_{2} \dfrac{1}{9} \ast ... \ast log_{2} \dfrac{1}{2} \ast log_{2} \dfrac{1}{1}}_{y}$

$= \underbrace{x \ast (-1)}_{-1} \ast y = (-1) \ast y = \bf -1$