Question

(x-2)^4-5(x^2-4)^2+4(x+2)^4=0

Answer 1

(x-2)^4-5(x^2-4)^2+4(x+2)^4 = 24x^3+160x^2 + 96x  =  8x(3x^2+20x+12)=0

x=0

3x^2+20x+12=0

 D=400-4*3*12 =V256=16

  x=-20+16/ 6=-4/6=-2/3

  x=-20-16 / 6=-6

Ответ   0  -2/3   -6

Answer 2

(x-2)^4-5(x^2-4)^2+4(x+2)^4=0

x^4-8x^3+24x^2-32x+16-5x^4+40x^2-80+4*(x+2)^4=0

-4x^4-8x^3+64x^2-32x-64+4x^4+32x^3+96x^2+128x+64=0

-8x^3+64x^2-32x-64+32x^3+96x^2+128x+64=0

24x^3+160x^2-32x-64+128x+64=0

24x^3+160x^2+96x=0

Разложим на множители 24x^3+160x^2+96x:

8x*(x+6)*(3x+2)=0

 

8x=0

x1=0

 

x+6

x2=-6

 

3x+2=0

x3=-2/3