Question

решите уравнения:
1) (3х-1)(12х+1)-10=(6х+2)^2
2) х^4-17х^2+16=0
3) 2х^4-х^3=0
4) (х^2-3)^2+х^2-3=2
5) 9х^3+18х^2-х-2=0
6) (х^2+х-5)(х^2+х+1)=-9

Answer 1

$(3x - 1)(12x+1)-10=(6x+2)^2 36x^2 - 9x - 11 = 36x^2 + 24x + 4 - 15 - 33x = 0 -3(11x + 5) = 0 11x = -5 x = -11/5 x = - 2,2$

$x^4 - 17x^2 + 16 = 0 x = y^2 y^2 - 17y + 16 = 0 (y - 16)(y - 1) = 0 y = 16 = x^2 ; y = 1 = x^2 x1 = 4 ; x2 = - 4 ; x3 = 1 ; x4 = -1$

$2x^4 - x^3 = 0 x^3(2x - 1) = 0 x^3 = 0 ; 2x - 1 = 0 x1 = 0 ; x2 = 0,5$

$(x^2 - 3)^2 + x^2 - 3 = 2 -3 + x^2 +(x^2 - 3)^2 = 2 x^4 - 5x^2 + 6 = 2 x^4 - 5x^2 + 4 = 0 x^2 = y y^2 - 5y + 4 = 0 (y - 4)(y - 1) = 0 y = 4 = x^2 ; y = 1 = x^2 x1 = 2 ; x2 = - 2 ; x3 = 1 ; x4 = - 1$

$9x^3 + 18x^2 - x - 2 = 0 (x + 2)(3x - 1)(3x + 1) = 0 x1 = - 2 ; x2 = 1/3 ; x3 = - 1/3$

$(x^2 + x - 5)(x^2 + x + 1) = -9 x^4 + 2x^3 - 3x^2 - 4x - 5 = -9 x^4 + 2x^3 - 3x^2 - 4x + 4 = 0 (x - 1)^2(x + 2)^2 = 0 x - 1 = 0 ; (x + 2)^2 = 0 x - 1 = 0 ; x + 2 = 0 x1 = 1 ; x2 = -2$