Question

determinati numerele natural n, astfel incat c) 8/15<4/2n+1 mai mic sau egal cu 2/3​

Answer 1

$\frac{8}{5} < \frac{4}{2n+1}\leq \frac{2}{3} \\\frac{8}{5} < \frac{4}{2n+1} = > 20 < 16n+8 = > n < \frac{3}{4} (1) \\\frac{4}{2n+1}\leq \frac{2}{3} = > 4n+2\leq 12 = > n\leq \frac{5}{2}(2)\\ (1)\ n \ apartine (-inf,3/4)\\(2) n\ apartine (-inf,5/2]\\= > n\ apartine\ (3/4,5/2] \\cum\ este\ natural$

=> n∈(3/4,5/2] ∩Ν

=> n={0,1,2}