Đáp án:

$5)\\ A=4\\ B=9\\ C=3\\ D=\dfrac{10085}{16}$

Giải thích các bước giải:

$5)\\ A=\left(\lg\sqrt{10}+\ln \sqrt{e}-\ln \dfrac{1}{e}\right).\dfrac{2 \log_2 3}{\log_49}\\ =\left(\log_{10} 10^\tfrac{1}{2}+\ln e^\tfrac{1}{2}-\ln e^{-1}\right).\dfrac{2 \log_2 3}{\log_{2^2}3^2}\\ =\left(\dfrac{1}{2}+\dfrac{1}{2}+1\right).\dfrac{2 \log_2 3}{\log_{2}3}\\ =2.2\\ =4\\ B=e^{\ln 3 - \ln 2}+e^{2\ln \sqrt{5} -3 \ln \sqrt[3]{2} }-5\ln e^{-1}\\ =e^{\ln \tfrac{3}{2}}+e^{\ln \sqrt{5}^2 - \ln \sqrt[3]{2}^3 }-5.(-1)\ln e\\ = \dfrac{3}{2}+e^{\ln 5 - \ln 2 }+5\\ = \dfrac{3}{2}+e^{\ln \tfrac{5}{2} }+5\\ = \dfrac{3}{2}+\dfrac{5}{2} +5\\ =9\\ C=\left(81^{\tfrac{1}{4}-\sin \tfrac{\pi}{6} \log_94}\right). 49^{\log_72}\\ =\left(81^{\tfrac{1}{4}-\tfrac{1}{2} \log_{3^2}2^2}\right). 7^{2\log_72}\\ =\left(3^{4\left(\tfrac{1}{4}-\tfrac{1}{2}\log_32\right)}\right). 7^{\log_72^2}\\ =\left(3^{1-2\log_32}\right). 4\\ =\left(3^{\log_33-\log_32^2}\right). 4\\ =\left(3^{\log_3 \tfrac{3}{4}}\right). 4\\ =\tfrac{3}{4}. 4\\ =3\\ D=2017 \left[49^{\cos \tfrac{\pi}{3} \log_79-\log_76}+\left(\dfrac{1}{5}\right)^{\log_\sqrt{5} 4}\right]\\ =2017 \left[49^{\tfrac{1}{2} \log_79-\log_76}+\left(\dfrac{1}{5}\right)^{\log_{5^\frac{1}{2}} 4}\right]\\ =2017 \left[7^{\log_79-2\log_76}+\left(\dfrac{1}{5}\right)^{2\log_5 4}\right]\\ =2017 \left[7^{\log_79-\log_736}+\left(5^{-1}\right)^{\log_5 16}\right]\\ =2017 \left[7^{\log_7 \tfrac{9}{36}}+5^{-\log_5 16}\right]\\ =2017 \left[ \dfrac{1}{4}+5^{\log_5 \tfrac{1}{16}}\right]\\ =2017 \left[ \dfrac{1}{4} +\dfrac{1}{16}\right]\\ =\dfrac{10085}{16}$