Question

(n+6)!/(n+8)!=1/360

Answer 1

Ответ:

Объяснение:

нету решения у этого уравнения если надо найти n

Answer 2

To solve the equation:

$\frac{(n+6)!}{(n+8)!} = \frac{1}{360}$

We can simplify the expression on the left-hand side by canceling out the common terms in the factorials:

$\frac{(n+6)!}{(n+8)!} = \frac{(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n!}{(n+8)(n+7)(n+6)(n+5)(n+4)(n+3)(n+2)(n+1)n!} = \frac{1}{(n+8)(n+7)}$

Now, we have the equation:

$\frac{1}{(n+8)(n+7)} = \frac{1}{360}$

To solve for $n$, we can cross-multiply:

$360 = (n+8)(n+7)$

Expand the right side:

$360 = n^2 + 15n + 56$

Rearrange the equation:

$n^2 + 15n + 56 - 360 = 0$

Simplify:

$n^2 + 15n - 304 = 0$

Now we have a quadratic equation in standard form. To solve for $n$, we can factor the quadratic or use the quadratic formula:

$n^2 + 19n - 16n - 304 = 0$

$n(n + 19) - 16(n + 19) = 0$

$(n - 16)(n + 19) = 0$

This gives us two possible solutions:

$n - 16 = 0$ or $n + 19 = 0$

$n = 16$ or $n = -19$

However, we need to check if these solutions are valid. Plugging in $n = -19$ back into the original equation:

$\frac{(n+6)!}{(n+8)!} = \frac{(-19+6)!}{(-19+8)!} = \frac{(-13)!}{(-11)!}$

Factorials of negative numbers are not defined, so $n = -19$ is not a valid solution.

Thus, the only valid solution is $n = 16$.