Question

ΙxΙ-2Ιx+1Ι+3Ι2x-4Ι=1

Answer 1

х=0; х+1=0; х=-1; 2х-4=0;х=2;

разобьем точками х=0, х=-1, х=2 числовую прямую и на каждом промежутке раскроем подули, получим.

_____-1________0_______2______

(-∞;-1]; (-1;0]; (0;2]; (2;+∞)

1)х∈(-∞;-1]; -х+2х+2-6х+12=1; -5х=-13; х=2.6∉(-∞;-1];

2)х∈(-1;0]; -х-2х-2-6х+12=1; -9х=-9;х=1∉(-1;0];

3)х∈ (0;2];х-2х-2-6х+12=1; -7х=-9; х=1 2/7∈ (0;2];

4)х∈ (2;+∞); х-2х-2+6х-12=1; 5х=15; х=3∈ (2;+∞);

Ответ х=1 2/7; х=3.

Answer 2

Ответ:

Объяснение:

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Answer 3

The equation you provided is:

|x| - 2|x + 1| + 3|2x - 4| = 1

To solve this equation, we'll break it down into different cases based on the absolute value expressions.

Case 1: x ≥ 0 In this case, the absolute value expressions simplify to their positive form.

For |x|, we have x. For |x + 1|, we have x + 1. For |2x - 4|, we have 2x - 4.

Substituting these values back into the equation, we get:

x - 2(x + 1) + 3(2x - 4) = 1

Simplifying the equation:

x - 2x - 2 + 6x - 12 = 1 5x - 14 = 1 5x = 15 x = 3

So, x = 3 is a solution for the case when x ≥ 0.

Case 2: x < 0 In this case, the absolute value expressions simplify to their negative form.

For |x|, we have -x. For |x + 1|, we have -(x + 1). For |2x - 4|, we have -(2x - 4).

Substituting these values back into the equation, we get:

-x - 2(-(x + 1)) + 3(-(2x - 4)) = 1

Simplifying the equation:

-x + 2(x + 1) - 3(2x - 4) = 1 -x + 2x + 2 - 6x + 12 = 1 -5x + 14 = 1 -5x = -13 x = 13/5

So, x = 13/5 is a solution for the case when x < 0.

Therefore, the equation |x| - 2|x + 1| + 3|2x - 4| = 1 has two solutions: x = 3 and x = 13/5.