\(3^{2x-1}=27\\\Rightarrow3^{2x-1}=3^3\\ \Rightarrow2x-1=3\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

\(-\left(x^2-4\right)=-x^2+4\)

\(\dfrac{x+1}{x+4}-\dfrac{x^2-4}{x^2-16}\\ =\dfrac{x+1}{x+4}-\dfrac{x^2-4}{\left(x-4\right)\left(x+4\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)-x^2+4}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{x^2+x-4x-4-x^2+4}{\left(x+4\right)\left(x-4\right)}\\ =\dfrac{-3x}{\left(x+4\right)\left(x-4\right)}\)

\(a,=12,3\times\left(3,12+6,88\right)\\ =12,3\times10\\ =123\\ b,=\left(2,5\times2\right)\times\left(5,5\times4\right)\\ =5\times22\\ =110\\ c,=7,89\times100\times2\times0,5\\ =\left(7,89\times100\right)\times1\\ =789\)

\(=-625-\left\{\left(-547\right)-352-\left[\left(-147\right)-\left(-735\right)+2850\right]\right\}\\ =-625-\left[-899-\left(588+2850\right)\right]\\ =-625-\left[-899-3438\right]\\ =-625-\left(-4337\right)\\ =3712\)

\(a,x:4,28=9,5\\ =>x=9,5\times4,28\\ =>x=40,66\\ b,102,66+x=80,8\\ =>x=80,8-102,66\\ =>x=-21,86\\ c,x-38,85=50,21\\ =>x=50,21+38,85\\ =>x=89,06\\ d,79,27-x=38,08\\ =>x=79,27-38,08\\ =>x=41,19\)

\(=10\times\left(2,5+6,7+1\right)\\ =10\times10,2\\ =102\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}+\left|3-\sqrt{5}\right|\\ =\sqrt{5}+2+3-\sqrt{5}\\ =2+3=5\)

Ta có \(CD=DB=4\left(cm\right)\)

\(\Rightarrow D\) là trung điểm \(CB\)

\(\Rightarrow AD\) là đường trung tuyến

\(\Rightarrow AD=CD=DB=4\left(cm\right)\)