Let `Delta(x)=|[3,3x,3x^2+2a^2] , [3x, 3x^2+2a^2, 3x^3+6a^2x] , [3x^2+2a^2, 3x^3+6a^2x, 3x^4+12a^2x^2+2a^4]|` then

Delta(x)=det[[3,3x,3x^(2)+2a^(2)3x,3x^(2)+2a^(2),3x^(3)+6a^(2)x3x^(2)+2a^(2),3x^(3)+6a^(2)x,3x^(4)+12a^(2)x^(2)+2a^(4)]]

" If " f(x) = |{:(3,,3x,,3x^(2)+2a^(2)),(3x,,3x^(2)+2a^(2),,3x^(3)+6a^(2)x),(3x^(2)+2a^(2),,3x^(3)++6a^(2)x,,3x^(4)+12a^(2)x^(2)+2a^(4)):}| then

2x^2(x^3-x)-3x(x^4+2x)-2(x^2-3x^4)

(3x^2+2)(4x-3x^2)

(3x + 2x^2 + 4x^3) / (x-4)

12x - [3x^(2) + 5x^(2) - {7x^(2) - (4 - 3x - x^(3)) + 6x^(3)}-3x]

|[2,4,6], [2+3x, 4+3y,6+3z] , [2x,2y,2x]|=

If (3x + 4)^2 + (3x - 2)^2 = (6x + 5)(3x - 2) + 12 , then the value of x is:

Find the intervals in which the following function are increasing or decreasing. f(x)=10-6x-2x^2 f(x)=x^2+2x-5 f(x)=6-9x-x^2 f(x)=2x^3-12 x^2+18 x+15 f(x)=5+36 x+3x^2-2x^3 f(x)=8+36 x+3x^2-2x^3 f(x)=5x^3-15 x^2-120 x+3 f(x)=x^3-6x^2-36 x+2 f(x)=2x^3-15 x^2+36 x+1 f(x)=2x^3+9x^2+20 f(x)=2x^3-9x^2+12 x-5 f(x)=6+12 x+3x^2-2x^3 f(x)=2x^3-24 x+107 f(x)=-2x^3-9x^2-12 x+1 f(x)=(x-1)(x-2)^2 f(x)=x^3-12 x^2+36 x+17 f(x)=2x^3-24+7 f(x)=3/(10)x^4-4/5x^3-3x^2+(36)/5x+11 f(x)=x^4-4x f(x)=(x^4)/4+2/3x^3-5/2x^2-6x+7 f(x)=x^4-4x^3+4x^2+15 f(x)=5x^(3/2)-3x^(5/2),x >0 f(x)==x^8+6x^2 f(x)==x^3-6x^2+9x+15 f(x)={x(x-2)}^2 f(x)=3x^4-4x^3-12 x^2+5 f(x)=3/2x^4-4x^3-45 x^2+51 f(x)=log(2+x)-(2x)/(2+x),xR