Question

helpp plzzzzzzzzzzzzzzz​

Answer 1

Explicație pas cu pas:

$x○y = \frac{xy + x + y - 1}{2}$

a)

$1○2 = \frac{1 \times 2 + 1 + 2 - 1}{2} = \frac{2 + 2}{2} = \frac{4}{2} = 2$

b)

$x○x = \frac{x \times x + x + x - 1}{2} = \frac{ {x}^{2} + 2x - 1}{2}$

$x○x \leqslant 1 = > \frac{ {x}^{2} + 2x - 1}{2} \leqslant 1$

$\frac{ {x}^{2} + 2x - 1}{2} - 1\leqslant 0 \\ \frac{ {x}^{2} + 2x - 1 - 2}{2} \leqslant 0 \\ \frac{ {x}^{2} + 2x - 3}{2} \leqslant 0 \\ (x + 3)(x - 1) \leqslant 0 \\ = > - 3 \leqslant x \leqslant 1$

=> x ∈ [-3 ; 1]

c)

$( - 1)○y =\frac{(-1)y + (-1) + y - 1}{2} = \frac{ - y - 1 + y - 1}{2} = - \frac{2}{2} = - 1$

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

$y = 0○1○...○2020$

atunci:

$(-1)○0○1○...○2020 = (-1)○(0○1○...○2020) = ( - 1)○y = - 1$