a) <=> \(3x^4-9x^3+9x^2-27x=0\)
<=>\(3x\left(x^3-3x^2+3x-9\right)=0\)
<=>\(3x\left(x-3\right)\left(x^2+3\right)\)=0
<=>x=0 hoặc x=3
b) \(\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
<=>\(\left(x+3\right)\left(x^2-4x+5\right)=0\)
<=>\(\left(x+3\right)\left(\left(x-2\right)^2+1\right)=0\)
=> x=-3
a) 3x4 - 9x3 = -9x2 + 27x
3x4 - 9x3 + 9x2 - 27x = 0
3x(x3 - 3x2 + 3x - 9) = 0
3x[x2(x - 3) + 3(x - 3)] = 0
3x(x - 3)(x2 + 3) = 0
vì x2 + 3 > 0 nên:
3x = 0 hoặc x - 3 = 0
x = 0 : 3 x = 0 + 3
x = 0 x = 3
=> x = 0 hoặc x = 3
b) (x + 3)(x2 - 3x + 5) = x2 + 3x
x3 - 3x2 + 5x + 3x2 - 9x = x2 + 3x
x3 - 4x + 15 = x2 + 3x
x3 - 4x + 15 - x2 - 3x = 0
x3 - 7x + 15 - x2 = 0
(x2 - 4x + 5)(x + 3) = 0
vì x2 - 4x + 5 > 0 nên
x + 3 = 0
=> x = -3
