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a, (1/5)5x(3)5
=1/3125.243
=243/3125
b, (0,125)^3.512
=1/512.512
=1
c,(0,25)^4.1024
=1/256.1024
=4
2,hầu như chịu hết

\(=\dfrac{3^6\cdot3^8\cdot5^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+3^{12}\cdot5^6}=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{3^{12}\cdot5^6\cdot2}\)
\(=\dfrac{3^{13}\cdot5^4\cdot2}{3^{12}\cdot5^6\cdot2}=\dfrac{3}{25}\)

Ta có:
\(S=2^2+4^2+6^2+...+20^2\)
\(\Rightarrow S=\left(1.2\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+...+\left(2.10\right)^2\)
\(\Rightarrow S=1^2.2^2+2^2.2^2+2^2.3^2+...+2^2.10^2\)
\(\Rightarrow S=\left(1^2+2^2+3^2+...+10^2\right).2^2\)
\(\Rightarrow S=385.4\)
\(\Rightarrow S=1540\)
S=22+42+...+102
=(1*2)2+(2*2)2+...+(2*10)2
=12*22+22*22+...+22*102
=22*(12+22+...+102)
=4*385
=1540

\(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\le0\left(1\right)\)
Ta có:\(\hept{\begin{cases}\left(2a+1\right)^2\ge0;\forall a,b,c\\\left(b+3\right)^4\ge0;\forall a,b,c\\\left(5c-6\right)^2\ge0;\forall a,b,c\end{cases}}\)\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\ge0;\forall a,b,c\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(2a+1\right)^2=0\\\left(b+3\right)^4=0\\\left(5c-6\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{-1}{2}\\b=-3\\c=\frac{6}{5}\end{cases}}\)
Vậy \(\left(a,b,c\right)=\left(\frac{-1}{2};-3;\frac{6}{5}\right)\)

\(\frac{2^{15}.9^4}{6^3.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=1944\)
\(\frac{2^{15}\times9^4}{6^6\times8^3}\)
\(=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^{14}.3^6}=2.3^2=18\)