Х(х+1)(х+2)(х+3)+(х+1)(х+2)(х+3)(х+4)=24

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вопрос

Let's first expand both terms:

(х+1)(х+2)(х+3) = х^3 + 6х^2 + 11х + 6

(х+1)(х+2)(х+3)(х+4) = (х^3 + 6х^2 + 11х + 6)(х+4)
= х^4 + 6х^3 + 11х^2 + 6х + 4х^3 + 24х^2 + 44х + 24
= х^4 + 10х^3 + 35х^2 + 50х + 24

Now, we add the two terms:

х^3 + 6х^2 + 11х + 6 + х^4 + 10х^3 + 35х^2 + 50х + 24 = 24

Rearranging the terms and simplifying, we get:

х^4 + 11х^3 + 41х^2 + 61х + 30 = 24

Subtracting 24 from both sides, we get:

х^4 + 11х^3 + 41х^2 + 61х + 6 = 0

Therefore, the solution to the equation is х = -2.