Question

(Х^2+1)^2-6(х^2+1)+5=0​

Answer 1

Ответ:

21 будет..................

Answer 2

Ответ:x1=0 x2=2

Объяснение:

(x²+1)²-6(x²+1)+5=0

x⁴+2x²+1-6x²-6+5=0

x⁴-4x²=0(целые числа сокращаются)

x²(x²-4)=0

x²=0 x²-4=0

x1=0 x²=4

x2=2

Answer 3

To solve the quadratic equation $(x^2 + 1)^2 - 6(x^2 + 1) + 5 = 0$, you can use a substitution to simplify the equation. Let $y = x^2 + 1$, so your equation becomes:

$y^2 - 6y + 5 = 0$.

Now, you have a quadratic equation in terms of $y$. You can factor it or use the quadratic formula to solve for $y$:

$y^2 - 6y + 5 = 0$

To factor it, you look for two numbers that multiply to 5 and add up to -6. These numbers are -5 and -1. So, you can factor it as follows:

$(y - 5)(y - 1) = 0$

Now, set each factor equal to zero:

$y - 5 = 0$ or $y - 1 = 0$

Solve for $y$:

For $y - 5 = 0$, $y = 5$.

For $y - 1 = 0$, $y = 1$.

Now that you have the possible values of $y$, you need to find the corresponding values of $x$:

1. If $y = 5$, then $x^2 + 1 = 5$. Subtract 1 from both sides:

   $x^2 = 5 - 1$

   $x^2 = 4$

   Take the square root of both sides:

   $x = \pm 2$

2. If $y = 1$, then $x^2 + 1 = 1$. Subtract 1 from both sides:

   $x^2 = 1 - 1$

   $x^2 = 0$

   Take the square root of both sides:

   $x = 0$

So, the solutions to the original equation $(x^2 + 1)^2 - 6(x^2 + 1) + 5 = 0$ are $x = -2$, $x = 2$, and $x = 0$.