Question

(x-5)^4-3(x-5)^2-4=0

Answer 1

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

1) Это уравнение легко привести к квадратному уравнению.

Для этого выполним замену:

   $(x-5)^2=y$        =>   $(x-5)^4=y^2$

и получим квадратное уравнение с ОДЗ:  $y\geq 0$

   $y^2-3y-4=0$

2) Решим по теореме Виета:

  $y_1=-1$  не удовлетворяет ОДЗ.  

  $y_2=4$

3) Выполним обратную замену:

  $(x-5)^2=4$

 4) Получили два уравнения:

   $x-5=-2$      и  

   $x-5=2$

5) Решим их и получим:

 $x_1=3$

 $x_2=7$

Ответ:  {3;  7}

Answer 2

Ответ:

Объяснение:

вот и все

Answer 3

To solve the equation $(x-5)^4-3(x-5)^2-4=0$, we can use a substitution to simplify the equation. Let $u = (x-5)^2$. Then, our equation becomes:

$u^2 - 3u - 4 = 0$.

Now, we can solve this quadratic equation for $u$:

$u^2 - 3u - 4 = 0$.

To solve this, we can use factoring:

$(u-4)(u+1) = 0$.

Now, set each factor to zero and solve for $u$:

$u - 4 = 0$ or $u + 1 = 0$.

So, we get two possible solutions for $u$:

$u = 4$ or $u = -1$.

Now, recall that $u = (x-5)^2$, so we have:

$(x-5)^2 = 4$ or $(x-5)^2 = -1$.

For the second equation, $(x-5)^2 = -1$, there are no real solutions because the square of any real number cannot be negative.

Now, let's solve the first equation, $(x-5)^2 = 4$:

Taking the square root of both sides:

$x - 5 = \pm 2$.

Now, we get two cases:

1. $x - 5 = 2$: Adding 5 to both sides: $x = 2 + 5$, $x = 7$.

2. $x - 5 = -2$: Adding 5 to both sides: $x = -2 + 5$, $x = 3$.

So, the solutions to the original equation are $x = 7$ and $x = 3$.