Question

20.20. 1) х – х = 0; 2) х* + х = 0; 3) x4 - 8x = 0;

Answer 1

1)x-x=0

x=x

x є R

2)x+x=0

x= -x

x=0

3)x⁴-8x=0

x(x³-8)=0

x=0 или x=2

Answer 2

It seems like you have provided three mathematical equations involving the variable "x." Let's solve each of them:

1. x - x = 0: This equation simplifies to 0 = 0. This equation is an identity, which means it is true for all real values of x. In other words, x can be any real number, and the equation will still be satisfied.

2. x* + x = 0: I assume you meant to write "x^2" instead of "x*." If so, the equation becomes x^2 + x = 0. To solve this quadratic equation, you can factor it: x(x + 1) = 0

Now, you have two possible solutions: a) x = 0 b) x + 1 = 0, which means x = -1

So, the solutions to this equation are x = 0 and x = -1.

3. x^4 - 8x = 0: To solve this equation, you can factor it as follows: x(x^3 - 8) = 0

Now, you have two possible solutions: a) x = 0 b) x^3 - 8 = 0

To solve the second part of the equation (x^3 - 8 = 0), you can use the difference of cubes factorization: (x - 2)(x^2 + 2x + 4) = 0

Now, you have two additional solutions: c) x - 2 = 0, which means x = 2 d) x^2 + 2x + 4 = 0

The quadratic equation x^2 + 2x + 4 = 0 has no real solutions because its discriminant is negative. Therefore, the only real solutions to the original equation x^4 - 8x = 0 are x = 0 and x = 2.

So, the solutions to the third equation are x = 0 and x = 2.