Question

*Lege de compozitie*

C-ul de la 2, va rog

Answer 1

.................................

Answer 2

Se detemină elementul neutru e.

[tex]\it x*e=x \Leftrightarrow 3(x-1)(e-1) + 1 = x \Leftrightarrow 3(x-1)(e-1) + 1 -x= 0\Leftrightarrow\\ \\ \Leftrightarrow 3(x-1)(e-1) -(x-1) = 0 \Leftrightarrow (x-1)(3e-3-1) =0 \Leftrightarrow \\ \\ \Leftrightarrow (x-1)(3e-4)=0 \Leftrightarrow 3e-4=0 \Leftrightarrow e = \dfrac{4}{3} [/tex]

Fie t' simetricul lui t față de legea dată.

$\it t*t'=e \ \ \ \ (1) \\ \\ t' = t \ \ \ \ (2) \\ \\ (1), (2) \Rightarrow t*t=e \Leftrightarrow 3(t-1)(t-1)+1=\dfrac{4}{3}\Leftrightarrow 3(t-1)^2=\dfrac{1}{3}\Leftrightarrow$

$\it \Leftrightarrow (t-1)^2 = \dfrac{1}{9} \Leftrightarrow t-1=\pm\dfrac{1}{3} \Leftrightarrow t-1\in \left \{-\dfrac{1}{3},\ \dfrac{1}{3} \right \}|_{+1}\Leftrightarrow \\ \\ \\ \Leftrightarrow t \in \left \{\dfrac{2}{3}, \ \ \dfrac{4}{3} \right \}$