X(x-1)(x+5)=(x+3)^2(x-2)

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вопрос

To solve this equation, we need to expand both sides and simplify:

Expanding the left side:
X(x-1)(x+5)
= X(x^2 + 5x - x - 5)
= X(x^2 + 4x - 5)

Expanding the right side:
(x+3)^2(x-2)
= (x+3)(x+3)(x-2)
= (x^2 + 6x + 9)(x-2)
= x^3 + 6x^2 + 9x - 2x^2 - 12x - 18
= x^3 + 4x^2 - 3x - 18

Setting the two expanded expressions equal to each other:
X(x^2 + 4x - 5) = x^3 + 4x^2 - 3x - 18

Multiplying out the left side:
X*x^2 + 4Xx - 5X = x^3 + 4x^2 - 3x - 18

Rearranging the equation:
X*x^2 + 4Xx - 5X - x^3 - 4x^2 + 3x + 18 = 0
x^3 + (4X-1)x^2 + (3-4X)x + 18 - 5X = 0

Comparing this with the general cubic equation ax^3 + bx^2 + cx + d = 0, we can see that:
a = 1
b = 4X - 1
c = 3 - 4X
d = 18 - 5X

For this equation to hold true, the coefficients of the cubic equation must be equal. Hence,
1 = 1
4X - 1 = 4
3 - 4X = -3
18 - 5X = 0

Solving these equations, we find that X = 3.

Therefore, the solution to the equation X(x-1)(x+5) = (x+3)^2(x-2) is X = 3.