Question

Aflati numarul n
a) 32 supra n la puterea 2
b)128 supra n la puterea 2
c)8 supra n-1
d)n-1 supra n+1

Answer 1

[tex] a)\frac{32}{n^2} \in N=> \frac{2^5}{n^2} \in N=>2\cdot(\frac{4}{n})^2 \in N=> \frac{4}{n} \in N=>\\ n\in D_4=\{1,2,4\}\\ b)\frac{128}{n^2} \in N=> \frac{2^7}{n^2} \in N=>2\cdot(\frac{8}{n})^2 \in N=> \frac{8}{n} \in N=>\\ n\in D_8=\{1,2,4,8\}\\ c) \frac{8}{n-1} \in N=>n-1\in D_8=\{1,2,4,8\}\\ n-1=1=>n=1+1=>n=2\\ n-1=2=>n=2+1=>n=3\\ n-1=4=>n=4+1=>n=5\\ n-1=8=>n=8+1=>n=9\\ n\in\{2,3,4,9\}\\ d) \frac{n-1}{n+1} = \frac{n+1-1-1}{n+1} =\frac{n+1}{n+1}-\frac{2}{n+1}=1-\frac{2}{n+1}=>\\ [/tex]
2/(n+1)∈N=>n+1 este divizor al numarului 2=>
n+1=1=>n=0(nu\ convine\ valoarea)
n+1=2=>n=1