Question

Уравнение x^2+x^-2=2

Answer 1

$x^2+x^{-2}=2 \\ x^2+ \frac{1}{x^2} =2 (\cdot x^2) \\ x^4+1=2x^2 \\ x^4-2x^2+1=0$
Заменяем переменную
$x^2 = t$
$t^2-2t+1=0 \\ D = 4 -4 = 0 \\ t_1=t_2= \frac{2}{2} = 1$
Не забываем про замену
$x^2 = t \\ t_1 = t_2 = 1 \\ x = \pm 1$

Answer 2

Х^2+1/х^2=2
(х-1/2)^2=0
х^2-1/х=0
х^2-1=0
х^2=1
х=1
х=-1

Answer 3

To solve the equation x^2 + x^(-2) = 2, we can follow these steps:

Step 1: Get rid of the negative exponent by multiplying both sides of the equation by x^2:

x^2 * (x^2 + x^(-2)) = x^2 * 2

Step 2: Simplify the left side:

x^4 + 1 = 2x^2

Step 3: Move all the terms to one side of the equation:

x^4 - 2x^2 + 1 = 0

Now, we have a quadratic equation in terms of x^2. Let's make a substitution to solve it more easily:

Let y = x^2.

Now, the equation becomes:

y^2 - 2y + 1 = 0

This is a simple quadratic equation that can be factored:

(y - 1)^2 = 0

Now, we solve for y:

y - 1 = 0

y = 1

However, remember that y is actually x^2. So, substitute back:

x^2 = 1

Now, take the square root of both sides:

x = ±√1

x = ±1

So, the solutions to the original equation are x = 1 and x = -1.