Question

(х^2-3х+1)(х^2-3х+3)=3​

Answer 1

Ответ:

x^4-3x^3+3x^2-3x^3+9x^2-9x+x^2-3x+3=3

x^4-3x^3+3x^2-3x^3+9x^2-9x+x^2-3x=0

x^4-6x^3+13x^2-12x=0

x(x^3-6x^2+13x-12)=0

x(x^3-3x^2-3x^2+9x+4x-12)=0

x(x^2(x-3)-3x(x-3)+4(x-3))=0

x(x-3)(x^2-3x+4)=0

x=0       x-3=0         x^2-3x+4=0

              x=3              Д<0

ответ x1=0   x2=3

Пошаговое объяснение:

Answer 2

Ответ:

Чисто реальный ответ x=1,5+-sqet(17), комплексны x=(3+-i*sqrt(15))/2

Пошаговое объяснение:

(х^2-3х+1)(х^2-3х+3)=3​

(х^2-3х+1)=t

(х^2-3х+1)(х^2-3х+3)=t(t+2)=3

t^2+2t-3=0

t=-1+-4=3;-5

х^2-3х+6=0 D<0 => решение комплексное, не думаю что оно нужно

x=(3+-i*sqrt(15))/2

х^2-3х-2=0

x=1,5+-sqet(17)

Answer 3

To solve the equation $(x^2 - 3x + 1)(x^2 - 3x + 3) = 3$, you can follow these steps:

1. Expand the left side of the equation:

$(x^2 - 3x + 1)(x^2 - 3x + 3) = x^4 - 3x^3 + x^2 - 3x^3 + 9x^2 - 3x + x^2 - 3x + 3$

Simplify this expression:

$x^4 - 6x^3 + 11x^2 - 6x + 3 = 3$

2. Subtract 3 from both sides of the equation to move all terms to the left side:

$x^4 - 6x^3 + 11x^2 - 6x = 0$

3. Now, let's factor the equation:

$x(x^3 - 6x^2 + 11x - 6) = 0$

4. The expression $x^3 - 6x^2 + 11x - 6$ can be factored further. We can use synthetic division or simply observe that $x = 1$ is a root of this cubic polynomial:

$(x - 1)(x^2 - 5x + 6) = 0$

5. Factor the quadratic equation:

$(x - 1)(x^2 - 5x + 6) = 0$

$(x - 1)(x - 2)(x - 3) = 0$

6. Now, set each factor equal to zero and solve for x:

a) $x - 1 = 0$ => $x = 1$ b) $x - 2 = 0$ => $x = 2$ c) $x - 3 = 0$ => $x = 3$

So, the solutions to the equation $(x^2 - 3x + 1)(x^2 - 3x + 3) = 3$ are $x = 1$, $x = 2$, and $x = 3$.