Question

Salut, as dori sa stiu daca exista o metoda mai usoara la ex. 16( niste indicatii, poate chiar o rezolvare completa).

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$f(x) = a(x+1)(x+2)(x+3)...(x+2014),\quad a\in \mathbb{R}^* \\ \\ \displaystyle \int_{x+1}^{x+2}\dfrac{f'(t)}{f(t)}\, dt =\ln\Big(f(t)\Big)\Bigg|_{x+1}^{x+2} = \ln\Big(f(x+2)\Big)-\ln\Big(f(x+1)\Big) = \\ \\ = \ln\Big(\dfrac{f(x+2)}{f(x+1)}\Big) = \ln\Big[\dfrac{a(x+3)(x+4)(x+5)...(x+2016)}{a(x+2)(x+3)(x+4)...(x+2015)}\Big] = \\ \\ = \ln\Big[\dfrac{x+2016}{x+2}\Big] = \ln(x+2016)-\ln(x+2)\\ \\ \Rightarrow \ln(x+2016)-\ln(x+2) = \ln(x+2016)-x^2 \\\\ \Rightarrow x^2 = \ln(x+2)$

Facem graficul lui x² si lui ln(x+2) in acelasi sistem de coordonate.

$g(x) = \ln(x+2) \\ g'(x) = \dfrac{1}{x+2}$

=> Ecuatia are o solutie negativa si una pozitiva.

=> n = 1, m = 1

Am atașat imaginea cu graficul.