Question

1) IхI=12 2)IхI= -3 3)4 IхI=20

Answer 1

1) х=+-12
2) нет решений
3) /х/=5
     х=+-5

Answer 2

Второе невозможно
1)x=+-12
3)4*(+-5)=20

Answer 3

1) IхI=12: To solve this equation, we need to determine the value of the variable "I". The equation is written in a slightly unusual format, but we can assume that "IхI" means "I multiplied by I" or "I squared". So the equation can be rewritten as I^2 = 12.

To find the value of I, we need to take the square root of both sides of the equation. However, we need to consider both the positive and negative square root since the equation does not specify whether I is positive or negative.

Taking the square root of both sides, we get I = ±√12. Simplifying the square root of 12, we have I = ±2√3 or approximately ±3.464.

Therefore, the solutions to the equation IхI = 12 are I = 3.464 and I = -3.464.

2) IхI = -3: In this equation, we have I multiplied by I equal to -3. This equation has no real solutions. The reason is that when we multiply two real numbers together, the result is always positive or zero. Therefore, there are no real values of I that satisfy the equation IхI = -3.

3) 4 + IхI = 20: To solve this equation, we need to isolate the variable "I". We start by subtracting 4 from both sides of the equation:

4 + IхI - 4 = 20 - 4 IхI = 16

Now, we have the equation IхI = 16. Similar to the first equation, IхI means I multiplied by I or I squared. So the equation can be rewritten as I^2 = 16.

Taking the square root of both sides, we get I = ±√16. Simplifying the square root of 16, we have I = ±4.

Therefore, the solutions to the equation 4 + IхI = 20 are I = 4 and I = -4.