решите: 1) n!/(n-1)! 2) (2л+1)!.(2л-1)! 3) n!=7(n-1)! 4) (m+17)!=420(m+15)!. Created by Mizukichi. algebra-ru

(n-1)! 2) (2л+1)!.(2л-1)! 3) n!=7(n-1)! 4) (m+17)!=420(m+15)!...

Mizukichi @Mizukichi

July 2022 1 9 Report

решите:

1) n!/(n-1)!

2) (2л+1)!.(2л-1)!

3) n!=7(n-1)!

4) (m+17)!=420(m+15)!

Answers & Comments

Voxman

$1) \frac{n!}{(n-1)!} = \frac{1*2*...*(n-1)*n}{1*2*...*(n-1)} = n\\\\ 2) \frac{(2n+1)!}{(2n-1)!} = \frac{1*2*...*(2n-1)*2n*(2n+1)}{1*2*...*(2n-1)} = 2n(2n+1) = 4n^2+2n\\\\ 3) \ n!=7(n-1)!\\ \frac{n!}{(n-1)!} = 7\\ \frac{1*2*...*(n-1)*n}{1*2*...*(n-1)} = 7\\ n = 7$

$4) (m+17)!=420(m+15)!\\ \frac{(m+17)!}{(m+15)!} = 420\\ \frac{1*2*...*(m+15)*(m+16)*(m+17)}{1*2*...*(m+15)} = 420\\ (m+16)*(m+17) = 420\\ m^2+16m+17m+16*17 = 420\\ m^2+33m + 272 - 420 = 0\\ m^2+33m-148 = 0\\ m_1*m_2 = -148\\ m_1+m_2 = -33\\ m_1 = -37, m_2 = 4\\ m = 4$

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Answer

Mizukichi August 2022 | 0 Ответы

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