(х+1)(х+2)-(х+3)(х+4)=0

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вопрос Следующий
вопрос

To solve this equation, we will first expand the terms:

(x+1)(x+2) - (x+3)(x+4)
= x^2 + 2x + x + 2 - (x^2 + 4x + 3x + 12)
= x^2 + 3x + 2 - x^2 - 7x - 12
= -4x - 10

Now we set this expression equal to 0 and solve for x:

-4x - 10 = 0
-4x = 10
x = -10/4
x = -5/2

Therefore, the solution to the equation (x+1)(x+2) - (x+3)(x+4) = 0 is x = -5/2.