Question

(3х+2)^2-2х=(3х-1)(3х+1)2х-7+3х=4+х-2х(х+5)=0​

Answer 1

$1) (3x+2)^2-2x=(3x-1)(3x+1)\\9x^2+12x+4-2x=9x^2-1\\10x=-5\\x=-0,5\\\\2) 2x-7+3x=4+x-2\\4x=9\\x=2,25\\\\3) x(x+5)=0\\x=0;x-5=0\\x_1=0;x_2=5$

Answer 2

Let's solve each equation step by step:

1. (3x+2)^2 - 2x = (3x-1)(3x+1)

Expanding the left side of the equation:

9x^2 + 12x + 4 - 2x = (3x-1)(3x+1)

Combine like terms:

9x^2 + 10x + 4 = 9x^2 - 1

Subtract 9x^2 from both sides:

10x + 4 = -1

Subtract 4 from both sides:

10x = -5

Divide both sides by 10:

x = -1/2

So the solution to the first equation is x = -1/2.

2. 2x - 7 + 3x = 4 + x - 2x

Combine like terms:

5x - 7 = 4 + x

Subtract x from both sides:

4x - 7 = 4

Add 7 to both sides:

4x = 11

Divide both sides by 4:

x = 11/4 or 2.75

So the solution to the second equation is x = 11/4 or 2.75.

3. 2x(x + 5) = 0

We have a product of two factors equal to zero, so either 2x = 0 or (x + 5) = 0.

1. 2x = 0 Divide both sides by 2: x = 0

2. x + 5 = 0 Subtract 5 from both sides: x = -5

So the solutions to the third equation are x = 0 and x = -5.

In summary, the solutions to the given equations are: First equation: x = -1/2 Second equation: x = 11/4 or 2.75 Third equation: x = 0 or x = -5