Question

Por favor resolva:

a)equação: n! / (n-2)! = 42

 

b) simplifique: n! / (n-3)!

Answer 1

n.(n-1).(n-2)! / (n-2)! = 42

n²-n = 42

n²-n-42 = 0

delta = (-1)² - 4.1.(-42)

delta = 1 + 168

delta = 169

 

n = [-(-1)+/-\/169] / 2

n = (1+/-13)/2

n' = 14/2 = 7

n" = -12/2 = -6 (não convém)

 

b) n! / (n-3)! = n.(n-1).(n-2).(n-3)! / (n-3)! = n.(n-1).(n-2)

Answer 2

a)

$<var> \frac{n! }{ (n-2)!} = 42\\ \frac{n(n-1)(n-2)!}{ (n-2)!} = 42\\n(n-1)=42\\n^2-n-42=0\\ \\D=(-1)^2-4*1*(-42)=1+168\\D=169\\ \\n_1=\frac{-(-1)+\sqrt{169}}{2*1}=\frac{1+13}{2}=\frac{14}{2}=7\\ \\n_1=\frac{-(-1)-\sqrt{169}}{2*1}=\frac{1-13}{2}=\frac{-12}{2}=-6\\ </var>$

 

b)

$<var>\frac{n!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=\\(n^2-n)(n-2)=\\n^3-2n^2-n^2+2n=\\n^3-3n^2-2n</var>$