Question

ПОЖАЛУЙСТА Срочно!!!!!! ​

Answer 1

Ответ:   D) .

$log_{2+\sqrt3}(2-\sqrt3)+log_327=log_{2+\sqrt3}\Big(\dfrac{1}{2+\sqrt3}\Big)+log_3\, 3^3=\\\\\\=log_{2+\sqrt3}(2+\sqrt3)^{-1}+3log_33=-1\cdot log_{2+\sqrt3}(2+\sqrt3)+3=-1+3=2\\\\\\\star \ \ (2-\sqrt3)=\dfrac{(2-\sqrt3)(2+\sqrt3)}{2+\sqrt3}=\dfrac{2^2-(\sqrt3)^2}{2+\sqrt3}=\dfrac{4-3}{2+\sqrt3}=\dfrac{1}{2+\sqrt3}=\\\\=(2+\sqrt3)^{-1}\ \ \star \\\\\star\ \ log_{a}\, a^{k}=k\ \ ,\ \ a>0\ ,\ a\ne 1\ \ \star$