(3x-x^2):2+(2x^2-x):6=x

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вопрос

To solve the given equation, we need to simplify both sides and then solve for x.

Starting with the left side of the equation:

(3x - x^2)/2 + (2x^2 - x)/6

First, we simplify each fraction:

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

(2x^2 - x)/6 = x(2x - 1)/6

Now, we add the two simplified fractions together:

x(3 - x)/2 + x(2x - 1)/6

To combine the fractions, we need a common denominator of 6:

(3x(3 - x) + 2x(2x - 1))/6
(9x - 3x^2 + 4x^2 - 2x))/6
(4x^2 + 7x)/6

Now, our equation becomes:

(4x^2 + 7x)/6 = x

Multiplying both sides by 6 to clear the fraction:

4x^2 + 7x = 6x

Rearranging and simplifying the equation:

4x^2 + 7x = 6x
4x^2 + 7x - 6x = 0
4x^2 + x = 0

Factoring out an x:

x(4x + 1) = 0

Setting each factor to zero:

x = 0 or 4x + 1 = 0

Solving for x:

x = 0 or 4x = -1, x = -1/4

Therefore, the solutions to the equation are x = 0 or x = -1/4.