F(x) = x³( 2x-1) в точці x0=1

Question

Алгебра: F(x) = x³( 2x-1) в точці x0=1

hamkochan · Accepted Answer

1) 3x² + 9 - 12x + x² = 0 4x² - 12x + 9 = 0 D = b² - 4ac = 144 - 16×9 = 0 x = -b/2a x = 12/8 x = 1,5 2) 5x² + 1 - 6x + 4x² = 0 9x² - 6x + 1 = 0 D = b² - 4ac = 36 - 36×1 = 0 x = -b/2a x = 6/18 x = 1/3 3) x² + 2x - 3 = 0 D = b² -4ac = 4 - 4×(-3) = 26 = 4² x1 = ( - 2 + 4) / 2 = 1 x2 = ( - 2 - 4) / 2 = - 3 4) x² + 3x -4 = 0 D = b²- 4ac = 9 - 4×(-4) = 25 = 5² x1 = ( - 3 + 5) / 2 = 1 x2 = ( - 3 - 5) / 2 = - 4 5) x² - 5x + 4 = 0 D = b² - 4ac = 25 - 4×4 = 9 = 3² x1 =( 5 + 3) / 2 = 4 x2 = ( 5 - 3) / 2 =...

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