Question

Решите уравнения (c-1) x (3-c) =0 9c-21=9c+18 (35-x)x=0 256x=1024

Answer 1

$1)(c-1)(3-c)=0 \\ c-1=0\\c=1 \\ 3-c=0\\c=3 \\ \\ 2)9c-21=9c+18 \\ 9c-9c=18+21 \\ 0c=39$
нет корней

$3)(35-x)x=0\\35-x=0\\x=35\\x=0 \\ \\ 4)256x=1024\\x=4$

Answer 2

1. (c-1)(3-c) = 0

To solve this equation, we need to find the values of c that make either factor equal to zero. So, we have:

c-1 = 0 or 3-c = 0

c = 1 or c = 3

Therefore, the solutions to the equation are c = 1 and c = 3.

2. 9c-21 = 9c+18

Subtracting 9c from both sides of the equation, we get:

-21 = 18

This is a contradiction, since -21 is not equal to 18. Therefore, there are no solutions to this equation.

3. (35-x)x = 0

To solve this equation, we need to find the values of x that make either factor equal to zero. So, we have:

35-x = 0 or x = 0

x = 35 or x = 0

Therefore, the solutions to the equation are x = 35 and x = 0.

4. 256x = 1024

Dividing both sides of the equation by 256, we get:

x = 4

Therefore, the solution to the equation is x = 4.