Question

REPEDE, VA ROG:
n(n+1)(n+2)=210 (ecuatia sa fie rezolvata in multimea numerelor naturale)

Answer 1

   
[tex]n(n+1)(n+2)=210 \\ n(n^2+3n+2)=210 \\ n^3+3n^2+2n=210 \\ n^3+3n^2+2n-210=0 \\ Artificiul1: ~~~ 3n^2 = -5n^2 + 8n^2\\ Artificiul2: ~~~ 2n = -40n + 42n \\ \\ n^3-5n^2+8n^2-40n+42n-210=0 \\ n^2(n-5)+8n(n-5)+42(n-5)=0 \\ (n-5)(n^2+8n+42)=0 \\ n-5=0~~~~=\ \textgreater \ ~~n_1 = \boxed{5} \\ n^2+8n+42=0 \\ \Delta = 8^2-4\cdot42 = 64-168=-104 \ \textless \ 0~~~\Longrightarrow n_{23} \notin N \\ \Longrightarrow ~~~n=\boxed{5}~~solutie ~unica. [/tex]

Answer 2

Deoarece n, n+1 si n+2 sunt trei numere naturale consecutive, incercam sa-l scriem si pe 210 ca produsul a trei numere naturale consecutive.
n(n+1)(n+2)=5*6*7
n=5