Question

Aratati ca 2x+6 supra x²+4x+3=2 supra x+1, pentru orice x∈R{-1;-3},va rog sa explicati

Answer 1

   
[tex]\displaystyle \bf\\ \frac{2x+6}{x^2+4x+3}= \frac{2(x+3)}{x^2+3x+x+3}=\\\\\\ = \frac{2(x+3)}{x(x+3)+(x+3)}= \frac{2(x+3)}{(x+3)(x+1)}=\boxed{\bf \frac{2}{x+1}}\\\\cctd [/tex]