\(3^{x+1}+3^{x+2}+3^{x+3}=39.3^{30}\\ =>3^x.\left(3^1+3^2+3^3\right)=39.3^{30}\\ =>3^x.39=3^{30}.39\\ =>3^x=3^{30}\\ =>x=30\)

\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}=\left(\pm\dfrac{3}{5}\right)^2\\ =>\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=\dfrac{2}{5}-\dfrac{3}{5}=-\dfrac{1}{5}\\3x=\dfrac{2}{5}-\left(-\dfrac{3}{5}\right)=1\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2009\right)+\left(x+2010\right)=2011.1009\\ =>\left(x+x+x+...+x+x\right)+\left(1+2+...+2009+2010\right)=2011.1009\\ =>2011x+\dfrac{\left(2010+1\right).2010}{2}=2011.1009\\ =>2011x+2011.1005=2011.1009\\ =>2011.\left(x+1005\right)=2011.1009\\ =>x+1005=1009\\ =>x=4\)

\(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+3\right)^2}\\ =\left|x-1\right|+\left|x+3\right|\\ =\left|x+3\right|+\left|-x+1\right|>=\left|x+3-x+1\right|=\left|4\right|=4\\ \)

\(Dấu\ ''=''\ xảy\ ra \ khi : \) \(\left(x+3\right)\left(-x+1\right)\ge0< =>\left(x+3\right)\left(x-1\right)\le0\\ =>\left\{{}\begin{matrix}x+3\ge0\\x-1\le0\end{matrix}\right.< =>-3\le x\le1\)

\(Vậy \ GTNN \ của \ biểu \ thức \ trên \ là : 4\ khi\ -3=<\ x\ =<1\)

\(7.2^x=2^9+5.2^8\\ =>7.2^x=2^8.\left(2^1+5\right)\\ =>7.2^x=7.2^8\\ =>x=8\)

\(\dfrac{2^{2x-3}}{4^{10}}=8^3.16^5\\ =>2^{2x-3}=8^3.16^5.4^{10}=\left(2^3\right)^3.\left(2^4\right)^5.\left(2^2\right)^{10}=2^{3.3}.2^{4.5}.2^{2.10}\\ =>2^{2x-3}=2^9.2^{20}.2^{20}=2^{9+20+20}=2^{49}\\ =>2x-3=49\\ =>2x=52\\ =>x=26\)

\(x-\dfrac{10}{3}=\dfrac{7}{15}.\dfrac{3}{5}=\dfrac{21}{75}\\ =>x=\dfrac{21}{75}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\\ \\ x+\dfrac{3}{22}=\dfrac{27}{121}.\dfrac{11}{9}=\dfrac{9.3.11}{11.11.9}=\dfrac{3}{11}\\ =>x=\dfrac{3}{11}-\dfrac{3}{22}=\dfrac{6-3}{22}=\dfrac{3}{22}\\ \\ \dfrac{8}{23}.\dfrac{46}{24}-x=\dfrac{1}{3}\\ =>\dfrac{8.23.2}{23.8.3}-x=\dfrac{1}{3}\\ =>\dfrac{2}{3}-x=\dfrac{1}{3}\\ =>x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

\(9.5^x=6.5^6+3.5^6\\ =>9.5^x=5^6.\left(6+3\right)=9.5^6\\ =>x=6\)

\(\dfrac{2^{4-x}}{16^5}=32^6\\ =>2^{4-x}=32^6.16^5=\left(2^5\right)^6.\left(2^4\right)^5=2^{5.6}.2^{4.5}\\ =>2^{4-x}=2^{30}.2^{20}=2^{30+20}=2^{50}\\ =>4-x=50\\ =>x=-46\)

\(A=-5=>x-\left|x-2\right|=-5\) \((*)\)

\(+) TH1:x >=2\)

\(=>x-2>=0=>\left|x-2\right|=x-2\)

\((*)\)\(=>x-\left(x-2\right)=-5< =>2=-5\) \(( Vô\ lí )\)

\(=>x\in\varnothing\)

\(+)TH2:x<2\)

\(=>x-2< 0=>\left|x-2\right|=-\left(x-2\right)\)

\((*)\)\(=>x-\left[-\left(x-2\right)\right]=-5< =>x+x-2=-5< =>2x=-3< =>x=-\dfrac{3}{2}\left(TMDK\right)\)

\(Vậy\ để\ A=-5\ thì\ x=-3/2\)

\(A=2=>x-\left|x-2\right|=2\)\((**)\)

\(+)TH1:x>=2\)

\((**)\)\(=>x-\left(x-2\right)=2< =>2=2\)\((Luôn\ đúng)\)

\(=>x>=2\)

\(+)TH2:x<2\)

\(=>x-\left[-\left(x-2\right)\right]=2< =>2x-2=2< =>x=2\left(KTMDK\right)\)

\(=>x\in\varnothing\)

\(Vậy\ để\ A=2\ thì \ x>=2\)