Question

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Решите уравнение (х²-3)²-4(х²-3)+3=0

Answer 1

${( {x}^{2} - 3)}^{2} - 4( {x}^{2} - 3) + 3 = 0 \ сделаем : замену : переменной : : \ {x}^{2} - 3 = y \ {y}^{2} - 4y + 3 = 0 \ d = 16 - 12 = 4 \ y1 = frac{4 + sqrt{4} }{2} = frac{4 + 2}{2} = frac{6}{2} = 3 \ y2 = frac{4 - 2}{2} = frac{2}{2} = 1 \ произведем : обратную : замену \ {x}^{2} - 3 = y1 \ {x}^{2} - 3 = 3 \ {x}^{2} = 6 \ x = ± sqrt{6} \ - - - - - \ {x}^{2} - 3 = y2 \ {x}^{2} - 3 =1 \ {x}^{2} = 4 \ x = sqrt{4} \ x = ±2 \ otvet \ x = (±2 ; : : ± sqrt{6} )$