Đáp án:
$A=4\ln(\sqrt{3}+2)\\ B=2$
Giải thích các bước giải:
$A=\ln(\sqrt{3}+2)^{2021}+\ln(2-\sqrt{3})^{2017}\\ =2021\ln(\sqrt{3}+2)+2017\ln(2-\sqrt{3})\\ =4\ln(\sqrt{3}+2)+2017\ln(\sqrt{3}+2)+2017\ln(2-\sqrt{3})\\ =4\ln(\sqrt{3}+2)+2017\ln\left((\sqrt{3}+2)(2-\sqrt{3})\right)\\ =4\ln(\sqrt{3}+2)+2017\ln\left(1\right)\\ =4\ln(\sqrt{3}+2)\\ B=\log_5\sqrt{3}-\dfrac{1}{2}\log_512+\log_550\\ =\log_5\sqrt{3}-\log_5\sqrt{12}+\log_550\\ =\log_5\left(\dfrac{\sqrt{3}}{2\sqrt{3}}.50\right)\\ =\log_5\left(25\right)\\ =2$
