Question

Como resolver a equação: (n + 6)! (n - 5)! - (n + 5)! (n - 6)! / (n + 5)! (n - 5)! = 11

Answer 1

$\dfrac{(n+6)!(n-5)!-(n+5)!(n-6)!}{(n+5)!(n-5)!}=11\\\\\\\dfrac{(n+6)(n+5)!(n-5)(n-6)!-(n+5)!(n-6)!}{(n+5)!(n-5)(n-6)!} =11\\\\\\\dfrac{(n+6)(n-5)-1}{(n-5)} =11\\\\\\(n+6)(n-5)-1 =11n-55\\\\n^2-5n+6n-30-1=11n-55\\\\n^2-10n+24=0\\\\\\n=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\n=\dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(24)}}{2(1)}\\\\\\n=\dfrac{10\pm\sqrt{4}}{2}\\\\\\n=\dfrac{10\pm 2}{2}$

$n = 5\pm1\\\\n'=5+1=6\\\\n''=5-1=4$


Como não se define fatorial de número negativo e n = 4 causaria esse efeito na expressão original, ficamos somente com a solução n = 6.