Question

(x^2-x+3/2x^2-x+2)^2 -3•x^2-x+3/2x^2-x+2 +2=0

Answer 1

Как то так....... Пример 1

Answer 2

Let's solve the equation:

(1) [(x^2 - x + 3) / (2x^2 - x + 2)]^2 - 3 * [(x^2 - x + 3) / (2x^2 - x + 2)] + 2 = 0

To solve this equation, first simplify it. We'll start by setting y = [(x^2 - x + 3) / (2x^2 - x + 2)]:

(2) y^2 - 3y + 2 = 0

Now, let's solve this quadratic equation (2) for y:

(3) y^2 - 3y + 2 = (y - 2)(y - 1) = 0

The equation y^2 - 3y + 2 = 0 factors as (y - 2)(y - 1) = 0, and we can solve for y by setting each factor equal to zero:

y - 2 = 0 => y = 2 y - 1 = 0 => y = 1

Now, we need to remember that y is the expression [(x^2 - x + 3) / (2x^2 - x + 2)]. So we have two possibilities:

1. y = 2: [(x^2 - x + 3) / (2x^2 - x + 2)] = 2

2. y = 1: [(x^2 - x + 3) / (2x^2 - x + 2)] = 1

Now, let's solve each of these possibilities:

1. [(x^2 - x + 3) / (2x^2 - x + 2)] = 2

Multiply both sides by (2x^2 - x + 2):

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

x^2 - x + 3 = 4x^2 - 2x + 4

Now, move all terms to one side of the equation:

3x^2 - x - 2 = 0

We have a quadratic equation, which can be solved using the quadratic formula:

x = [1 ± sqrt(1 - 4(3)(-2))] / (2 * 3)

x = [1 ± sqrt(25)] / 6

x = [1 ± 5] / 6

This results in two possible solutions:

a) x = (1 + 5) / 6 = 6/6 = 1

b) x = (1 - 5) / 6 = -4/6 = -2/3

2. [(x^2 - x + 3) / (2x^2 - x + 2)] = 1

Multiply both sides by (2x^2 - x + 2):

x^2 - x + 3 = 2x^2 - x + 2

Now, move all terms to one side of the equation:

x^2 - 2x + 1 = 0

This is a quadratic equation that factors easily:

(x - 1)(x - 1) = 0

x - 1 = 0

x = 1

So, the solutions for the original equation (1) are:

a) x = 1 b) x = -2/3