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Proszę o pomoc w jednym zadaniu, zadanie 105 daje naj



Proszę O Pomoc W Jednym Zadaniu Zadanie 105 Daje Naj class=

Odpowiedź :

a)  3 - p

[tex]dla \ \ p = 0\\3-p = 3-0 = 3\\\\dla \ \ p = 17\\3-p=3-17 = -14\\\\dla \ \ p = -4\\3-p = 3-(-4) = 3+4 = 7[/tex]

b)  -2p + 1

[tex]dla \ \ p = 4\\-2p+1 = -2\cdot4 =+1 = -8+1 = -7\\\\dla \ \ p = 0\\-2p+1 = -2\cdot0+1 = 1\\\\dla \ \ p = -3\\-2p+1 = -2(-3)+1 = 6+1 = 7[/tex]

c)  [tex]\frac{p-1}{2}[/tex]

[tex]dla \ \ p = 13\\\frac{p-1}{2} = \frac{13-1}{2} = \frac{12}{2} = 6\\\\dla \ \ p = 6\\\frac{p-1}{2} = \frac{6-1}{2} = \frac{5}{2} = 2,5\\\\dla \ \ p = -6\\\frac{p-1}{2} = \frac{-6-1}{2} = \frac{-7}{2} = -3,5[/tex]

 [tex]d) \ \frac{2(1-3p)}{3}\\\\dla \ \ p = 0\\\frac{2(1-3p)}{3} = \frac{2(1-3\cdot0)}{3} = \frac{2\cdot1}{3} = \frac{2}{3}\\\\dla \ \ p = 1\\\frac{2(1-3p)}{3} = \frac{2(1-3\cdot1)}{3} = \frac{2(1-3)}{3} = \frac{2\cdot(-2)}{3} = \frac{-4}{3} = -\frac{4}{3} = -1\frac{1}{3}\\\\dla \ \ p = -1\\\frac{2(1-3p)}{3} = \frac{2(1-3\cdot(-1))}{3} = \frac{2(1+3)}{3} = \frac{2\cdot4}{3} = \frac{8}{3} = 2\frac{2}{3}[/tex]