Question

să se rezolve ecuatia:
vă rog muult

Answer 1

Răspuns:

S=∅

Explicație pas cu pas:

$\dfrac{1}{(n-1)!} -\dfrac{2}{n!}=\dfrac{n^{2}}{(n+1)!}~|*(n+1)! ,~~~ \dfrac{(n+1)!}{(n-1)!} -\dfrac{2(n+1)!}{n!}=\dfrac{n^{2}(n+1)!}{(n+1)!},~~~\\ \dfrac{(n-1)!n(n+1)}{(n-1)!} -\dfrac{2n!(n+1)}{n!}=\dfrac{n^{2}(n+1)!}{(n+1)!}~~~$

După simplificări, obținem ecuația  

n(n+1)-2(n+1)=n², ⇒n²+n-2n-2-n²=0, ⇒-n=2, deci n=-2.

Deoarece, n este natural și n≥1, ⇒ că ecuația nu are soluție.

Answer 2

 

$\displaystyle\bf\\c)\\\\\frac{1}{(n-1)!}-\frac{2}{n!}=\frac{n^2}{(n+1)!}\\\\(n-1)!\times n=n!\\\\(n+1)!=n!\times(n+1)\\\\Prima~fractie~o~amplificam~cu~n.\\\\\frac{n}{(n-1)!\times n}-\frac{2}{n!}=\frac{n^2}{n!\times(n+1)}\\\\\\\frac{n}{n!}-\frac{2}{n!}=\frac{n^2}{n!\times(n+1)}\\\\\\\frac{(n-2)}{n!}=\frac{n^2}{n!\times(n+1)}\\\\\\\frac{(n-2)\times n!\times(n+1)}{n!}=n^2\\\\\\\frac{(n-2)\times(n+1)}{1}=n^2\\\\\\n^2-2n+n-2=n^2\\\\n^2-n-2=n^2\\\\n^2-n^2-n-2=0\\\\-n-2=0\\-n=2\\n=-2\\n\notin N$

⇒  Ecuatia nu are solutie.