helpp plzzzzzzzzzzzzzzz​

Helpp Plzzzzzzzzzzzzzzz class=

Răspuns :

Explicație pas cu pas:

[tex]x○y = \frac{xy + x + y - 1}{2} [/tex]

a)

[tex]1○2 = \frac{1 \times 2 + 1 + 2 - 1}{2} = \frac{2 + 2}{2} = \frac{4}{2} = 2 [/tex]

b)

[tex]x○x = \frac{x \times x + x + x - 1}{2} = \frac{ {x}^{2} + 2x - 1}{2}[/tex]

[tex]x○x \leqslant 1 = > \frac{ {x}^{2} + 2x - 1}{2} \leqslant 1 [/tex]

[tex]\frac{ {x}^{2} + 2x - 1}{2} - 1\leqslant 0 \\ \frac{ {x}^{2} + 2x - 1 - 2}{2} \leqslant 0 \\ \frac{ {x}^{2} + 2x - 3}{2} \leqslant 0 \\ (x + 3)(x - 1) \leqslant 0 \\ = > - 3 \leqslant x \leqslant 1[/tex]

=> x ∈ [-3 ; 1]

c)

[tex]( - 1)○y =\frac{(-1)y + (-1) + y - 1}{2} = \frac{ - y - 1 + y - 1}{2} = - \frac{2}{2} = - 1[/tex]

legea "○" este asociativă, deci putem să notăm:

[tex]y = 0○1○...○2020[/tex]

atunci:

[tex](-1)○0○1○...○2020 = (-1)○(0○1○...○2020) = ( - 1)○y = - 1[/tex]