Question

1) 0,1-2,2(2Х-5)=0,3(5-4Х) 2)(Х+1,2)(2,8-4Х)=0

Answer 1

Ответ:

1) 0,1 - 2,2(2х-5) = 0,3(5-4х)

0,1 - 4,4х + 11 = 1,5 - 1,2х

-4,4х + 1,2х = 1,5 - 0,1 - 11

- 3,2х = - 9,6

х = - 9,6/ (-3,2)

х = 3

2) (Х+1,2)(2,8-4Х)=0

Произведение равно нулю, если действие хотя бы в одной из скобок равно нулю. Отсюда уравнение имее два корня:

х + 1,2 = 0

Х₁ = - 1,2

2,8 - 4х = 0

4х = 2,8

х₂ = 0,7

Пошаговое объяснение:

Answer 2

Let's solve each equation one by one:

1. 0,1-2,2(2Х-5)=0,3(5-4Х)

First, let's simplify the equation by distributing the terms:

0.1 - 2.2(2X - 5) = 0.3(5 - 4X)

0.1 - 4.4X + 11 = 1.5 - 1.2X

Next, let's combine like terms on both sides:

-4.4X + 11.1 = -1.2X + 1.5

Now, let's isolate the variable terms on one side and the constant terms on the other side:

-4.4X + 1.2X = 1.5 - 11.1

-3.2X = -9.6

Dividing both sides by -3.2:

X = -9.6 / -3.2

X = 3

Therefore, the solution to the equation is X = 3.

2. (X + 1.2)(2.8 - 4X) = 0

To solve this equation, we'll set each factor equal to zero and solve for X:

X + 1.2 = 0 or 2.8 - 4X = 0

Solving the first equation:

X = -1.2

Solving the second equation:

2.8 - 4X = 0

-4X = -2.8

Dividing both sides by -4:

X = -2.8 / -4

X = 0.7

Therefore, the solutions to the equation are X = -1.2 and X = 0.7.