Question

1)V5-2x=1-x2)v2-x^2=x3)v3x+4=x4)v4+x+v4=4

Answer 1

Ответ:

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Answer 2

It looks like you've provided a set of four equations. Let's solve each of them one by one:

1. $5 - 2x = 1 - x^2$:

First, let's move all the terms to one side to set the equation to zero:

$5 - 2x - 1 + x^2 = 0$

Combine like terms:

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

Now, we can factor the quadratic equation:

$(x - 2)(x + 2) = 0$

Now, we have two possible solutions:

$x - 2 = 0$ or $x + 2 = 0$

Solving for $x$ in each case:

1. $x - 2 = 0$ gives $x = 2$.
2. $x + 2 = 0$ gives $x = -2$.

So, the solutions to the first equation are $x = 2$ and $x = -2$.

2. $v^2 - x^2 = x^3$:

Let's rearrange the terms:

$v^2 = x^3 + x^2$

Now, we can take the square root of both sides:

$v = \sqrt{x^3 + x^2}$

3. $v^3x + 4 = x^4$:

Rearrange the terms:

$v^3x - x^4 + 4 = 0$

4. $v^4 + x + v^4 = 4$:

Combine like terms:

$2v^4 + x = 4$

Now, if you have specific values for $x$ or $v$, you can substitute them into these equations to find the corresponding solutions. Otherwise, these equations are in terms of variables, and you can't find specific numerical solutions without more information or additional equations.