(x+3)^2-(x-2)(x+2)=13+6x

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вопрос

To solve this equation, we first need to expand the left side of the equation:

(x+3)^2 = (x+3)(x+3) = x^2 + 6x + 9

(x-2)(x+2) = x^2 + 2x - 2x - 4 = x^2 - 4

Putting it all together:

x^2 + 6x + 9 - (x^2 - 4) = 13 + 6x

x^2 + 6x + 9 - x^2 + 4 = 13 + 6x

6x + 13 = 13 + 6x

Subtracting 6x from both sides, we get:

13 = 13

This equation is always true, meaning that the original equation is an identity and holds true for all values of x.