Question

(х+1)(х-1)(х-3)(х-5)+7=0​

Answer 1

Ответ:

(x+1)(x+3)⋅(x−1)(x+5)+7=0(x2+4x+3)(x2+4x−5)+7=0t=x2+4x−5→x2+4x+3=t+8t⋅(t+8)+7=0t2+8t+7=0t1=−1,t2=−7a)x2+4x−5=−1,x2+4x−4=0x1,2=−2±8=−2(1±2)b)x2+4x−5=−7,x2+4x=2=0x3,4=−2±2=−2(2±1)

Answer 2

To solve the equation $(x+1)(x-1)(x-3)(x-5) + 7 = 0$, you can start by simplifying the equation and then solving for $x$.

1. First, expand the left side of the equation:

$(x+1)(x-1)(x-3)(x-5) + 7 = (x^2 - 1)(x^2 - 3x - x + 3)(x-5) + 7 = (x^2 - 1)(x^2 - 4x + 3)(x-5) + 7$

2. Continue to simplify:

$(x^2 - 1)(x^2 - 4x + 3)(x-5) + 7 = (x^2 - 1)(x^2 - 4x + 3)(x-5) + 7$

3. Now, let's make a substitution to make the equation easier to work with. Let $u = x^2 - 4x + 3$. This will allow us to rewrite the equation as:

$(x^2 - 1)u(x-5) + 7 = 0$

4. Substitute $u$ back in:

$(x^2 - 1)(x^2 - 4x + 3)(x-5) + 7 = 0$

5. Rewrite the equation in terms of $u$:

$(x^2 - 1)u(x-5) + 7 = 0$

6. Now, you have a quadratic equation in terms of $u$:

$(x^2 - 1)u(x-5) + 7 = 0$

7. Solve this quadratic equation for $u$:

$(x^2 - 1)u(x-5) + 7 = 0$

This equation can be solved by factoring or using the quadratic formula.

8. Once you have found the values of $u$, you can substitute them back into the equation $u = x^2 - 4x + 3$ to find the corresponding values of $x$.

Keep in mind that this equation may have multiple solutions, so be sure to check all possible values of $u$ and $x$ that satisfy the equation.