Question

Решить: x^-15=0 x^+25=0 x^+8x=0

Answer 1

1)
${ x}^{2} - 15 = 0 \\ {x}^{2} = 15 \\ x = + - \sqrt{15 } \\ x = \sqrt{15 } \\ x = - \sqrt{15}$
2)
${x}^{2} + 25 = 0 \\ { x}^{2} = - 25$
Корней нет, так как функция с четным показателем степени должна быть положительна или равна 0

3)
${x}^{2} + 8x = 0 \\ \times \times (8 + x) = 0 \\ x = 0 \\ x + 8 = 0 \\ x = 0 \\ x = - 8$

Answer 2

To solve the equations, we need to find the values of 'x' that satisfy each equation. Let's solve them one by one:

1. x^(-15) = 0

Any nonzero number raised to the power of -15 will never be equal to 0, so there are no real solutions for this equation.

2. x^(25) = 0

Again, any nonzero number raised to the power of 25 will never be equal to 0, so there are no real solutions for this equation either.

3. x^2 + 8x = 0

To find the solutions for this quadratic equation, we can factor it:

x^2 + 8x = 0 x(x + 8) = 0

Now, we have two factors: x = 0 and (x + 8) = 0

From the second factor, we get: x + 8 = 0 x = -8

So, the solutions for this equation are x = 0 and x = -8.

To summarize:

1. x^(-15) = 0 has no real solutions.
2. x^(25) = 0 has no real solutions.
3. x^2 + 8x = 0 has solutions x = 0 and x = -8.