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a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)
\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)
\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)
\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)
\(=\frac{3^{61}}{3^{60}}\)
\(=3\)
\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)
\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)
\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)
\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)
\(=\frac{3^4}{3^3}\)
\(=3\)
\(A=\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}=\dfrac{100}{300}\)
\(B=\dfrac{25}{11}\cdot\dfrac{13}{12}\cdot\dfrac{-11}{5}=\dfrac{-65}{12}=\dfrac{-1625}{300}\)
\(C=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\cdot\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\cdot\dfrac{-2}{5}=\dfrac{-22}{100}=\dfrac{-11}{50}=\dfrac{-66}{300}\)
Vì -1625<-66<100
nên B<C<A
thank bạn 
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.\left(2.3\right)^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^6.\left(3^{25}+2^{30}\right)}{2^{11}.3^6.\left(3^{25}+2^{30}\right)}=\frac{1}{2}\)
d) \(a.\left(-b\right).\left(-a\right)^2\left(-b\right)^3.\left(-a\right)^3.\left(-b\right)^4=-a^6b^8\)