Решите уравнение х(х^2+4х+4)=3(х+2)
X(x+2)²-3(x+2)=0 (x+2)(x²+2x-3)=0 x+2=0 x=-2 x²+2x-3=0 x1+x2=-2 U x1*x2=-3 x1=-3 U x2=1 x={-3;-2;1}
X( x^2 + 4x + 4) = 3( x + 2) x( x + 2)(x + 2) - 3( x + 2) = 0 ( x + 2)( x^2 + 2x - 3) = 0 x + 2 = 0 x = - 2 x^2 + 2x - 3 = 0 D = b^2 - 4ac = 4 + 12 = 16 = 4^2 x1 = ( - 2 + 4) / 2 = 1 x2 =( - 2 - 4) / 2 = - 3 Ответ x1 = - 2, x2 = 1, x3 = - 3