Wypisz wszystkie liczby całkowite spelniajace nierownosc
x(x+2) < 8
prosze o konkretne wypisanie tych liczb!!!
x = -4 bo x(x + 2) = (-4)* (-4 + 2) = (-4) * (-2) = 8 = 8 czyli (-4) odpada
x = -3 bo x(x + 2) = (-3)* (-3 + 2) = (-3) * (-1) = 3 < 8
x = -2 bo x(x + 2) = (-2)* (-2 + 2) = (-2) * 0 = 0 < 8
x = -1 bo x(x + 2) = (-1) * (-1 +2) = (-1) * 1 = - 1 < 8
x = 0 bo x(x + 2) = 0 * (0 + 2) = 0 * 2 = 0 < 8
x = 1 bo x(x + 2) = 1 * (1 + 2) = 1 * 3 = 3 < 8
x = 2 bo x(x + 2) = 2 * (2 + 2) = 2 * 4 = 8 a to równe jest 8 zatem x = 2 odpada
odp. Są to liczby: -3, -2, -1, 0, 1
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x = -4 bo x(x + 2) = (-4)* (-4 + 2) = (-4) * (-2) = 8 = 8 czyli (-4) odpada
x = -3 bo x(x + 2) = (-3)* (-3 + 2) = (-3) * (-1) = 3 < 8
x = -2 bo x(x + 2) = (-2)* (-2 + 2) = (-2) * 0 = 0 < 8
x = -1 bo x(x + 2) = (-1) * (-1 +2) = (-1) * 1 = - 1 < 8
x = 0 bo x(x + 2) = 0 * (0 + 2) = 0 * 2 = 0 < 8
x = 1 bo x(x + 2) = 1 * (1 + 2) = 1 * 3 = 3 < 8
x = 2 bo x(x + 2) = 2 * (2 + 2) = 2 * 4 = 8 a to równe jest 8 zatem x = 2 odpada
odp. Są to liczby: -3, -2, -1, 0, 1