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\(\dfrac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^{10}\times6^{19}-7\times2^{29}\times27^6}\\ =\dfrac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^{10}\times3^{19}\times2^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{29}\times3^{19}-7\times2^{29}\times3^{18}}\\ =\dfrac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{29}\times3^{18}\times\left(5\times3-7\right)}\\ =\dfrac{10-9}{15-7}\\ =\dfrac{1}{8}\)
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\(A=\dfrac{2^{19}\cdot3^9-3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot2^{10}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{19}\cdot3^9-2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\left(2-5\right)}{2^{19}\cdot3^9\cdot7}=\dfrac{1}{2}\cdot\dfrac{-3}{7}=\dfrac{-3}{14}\)
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\(-29\times165+29\times65=29\times\left(65-165\right)=29\times\left(-100\right)=-2900\\ ---\\ \left\{215-\left[5\times\left(5\times16-25\times2\right)\right]-60\right\}:5^3\\ =\left\{215-\left[5\times\left(80-50\right)\right]-60\right\}:125\\ =\left\{215-\left[5\times30\right]-60\right\}:125\\ =\left\{215-150-60\right\}:125=5:125=\dfrac{1}{25}\)
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\(\left(1-\frac{3}{10}-x\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
\(\Rightarrow\left(\frac{7}{10}-x\right):\frac{1}{2}=\frac{1}{5}\\ \frac{7}{10}-x=\frac{1}{5}.\frac{1}{2}\\ \frac{7}{10}-x=\frac{1}{10}\\ \Rightarrow x=\frac{7}{10}-\frac{1}{10}=\frac{3}{5}\)
\(2^x\times2^{x+1}=10^{19}\div5^{19}\)
\(\Rightarrow2^{2x+1}=2^{19}\)
\(\Rightarrow2x+1=19\)
\(\Rightarrow2x=19-1\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=18\div2\)
\(\Rightarrow x=9\)
\(2^x.2^{x+1}=10^{19}:5^{19}\)
\(2^{x+x+1}=\left(10:5\right)^{19}\)
\(2^{2x+1}=2^{19}\)
\(\Rightarrow2x+1=19\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
Vậy \(x=9\)