Question

Cine ma ajuta si pe mine la 1 si 3 va rog mult?

Answer 1

   
[tex]\displaystyle 1) \\ \texttt{Aplicam formulele:} \\ \log_ab + \log_ac = \log_a bc \\ \log_ab - \log_ac = \log_a b/c \\ \log_63 + \log_610 - \log_65 = \log_6 \left(\frac{3\times 10}{5} \right) = \log_6 \left(\frac{30}{5} \right) =\log_66 =1 [/tex]



[tex]\displaystyle 3) \\ (2x+1)^2 = 4 \\ (2x+1)^2 - 4 =0\\ (2x+1)^2 - 2^2 =0\\ \\ (2x+1 + 2)(2x+1-2)=0 \\ (2x+3)(2x-1)=0 \\ \\ 2x+3 = 0~~sau~~2x-1=0 \\ \\ x_1 = \boxed{- \frac{3}{2} } \\ \\ x_2 = \boxed{\frac{1}{2} } [/tex]