Question

У(2у-1)/12-у^2+1/6=у ​

Answer 1

Відповідь:

Пояснення:

Answer 2

To solve the equation, let's simplify and rearrange the terms:

(2y - 1)/12 - y^2 + 1/6 = y

First, let's eliminate the fractions by multiplying the entire equation by the common denominator, which is 12:

12 * [(2y - 1)/12] - 12 * (y^2) + 12 * [1/6] = 12 * y

Simplifying further:

2y - 1 - 12y^2 + 2 = 12y

Combine like terms:

-12y^2 + 2y - 1 + 2 = 12y

-12y^2 + 2y + 1 = 12y

Rearrange the terms to set the equation equal to zero:

-12y^2 - 10y + 1 = 0

Now we have a quadratic equation. To solve it, we can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -12, b = -10, and c = 1. Substituting these values into the quadratic formula:

y = (-(-10) ± √((-10)^2 - 4(-12)(1))) / (2(-12))

Simplifying further:

y = (10 ± √(100 + 48)) / (-24)

y = (10 ± √148) / (-24)

y = (10 ± √(4 * 37)) / (-24)

y = (10 ± 2√37) / (-24)

Simplifying the expression:

y = (5 ± √37) / (-12)

So the solutions to the equation are:

y ≈ (5 + √37) / (-12)

y ≈ (5 - √37) / (-12)

These are the approximate solutions since we used the quadratic formula, which involves taking a square root.