Question

(2x+1)*(x+1)*(7x-0,21)=0 б) | х-4 | = -2​

Answer 1

Ответ:

Х1=-1. Х2=-0.5. Х3=0.03. Б) утверждение ложно для любого значениях, поскольку функция модуля всегда положительная или 0.

Пошаговое объяснение:

(2х+1)*(х+1)*(7х-21/100)=0. 2х+1=0. Х+1=0. 7х-21/100=0. Х=-0.5. Х=-1. Х=3/100

Answer 2

Let's solve the given equations step by step:

a) (2x+1)(x+1)(7x-0.21) = 0

To find the solutions, we need to set each factor equal to zero and solve for x.

Setting 2x+1 = 0: 2x = -1 x = -1/2

Setting x+1 = 0: x = -1

Setting 7x-0.21 = 0: 7x = 0.21 x = 0.21/7 x = 0.03

Therefore, the solutions to the equation (2x+1)(x+1)(7x-0.21) = 0 are x = -1/2, -1, and 0.03.

b) |x-4| = -2

The absolute value of any real number is always non-negative. Therefore, the absolute value of x-4 cannot be equal to a negative value (-2 in this case). Hence, there are no solutions to the equation |x-4| = -2.

In summary: a) The solutions to the equation (2x+1)(x+1)(7x-0.21) = 0 are x = -1/2, -1, and 0.03. b) The equation |x-4| = -2 has no solutions.