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) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)

Các bạn vào trang cá nhân của mik đi, có cái này hay lắm!!!

\(a)\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)

\(b)\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)

\(c)\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^6\left(3^{25}+2^{30}\right)}{2^{11}.3^6\left(3^{25}+2^{30}\right)}=\frac{1}{2}\)

\(d)a\left(-b\right)\left(-a\right)^2\left(-b\right)^3\left(-a\right)^3\left(-b\right)^4\)

\(=\left[a\left(-a\right)^2\left(-a\right)^3\right]\left[\left(-b\right)\left(-b\right)^3\left(-b\right)^4\right]\)

\(=\left[a.a^2\left(-a\right)^3\right]\left[-b.b^4\left(-b\right)^3\right]=\left(-a^3.a^3\right)\left[\left(-b\right)^4.b^4\right]=-b^8\)

\(e)\left[-\left(-a\right)^3\right]\left(-a^2\right)^3\left[\left(-b\right)^2\right]^3\left[-\left(-b\right)^4\right]\)

\(=a^3\left(-a^6\right).b^6\left(-b^4\right)=-a^3.\left(-b^2\right)=a^3b^2\)

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

\(=\frac{3^4}{3^3}\)

\(=3\)

\(\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}\)

\(=\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2\cdot\left(2^{10}\cdot3^{31}+2^{40}\cdot3^6\right)}\)

\(=\frac{1}{2}\)

\(\frac{6:\frac{3}{5}-1\frac{1}{6}.\frac{6}{7}}{4\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}}=\frac{10-\frac{7}{6}.\frac{6}{7}}{\frac{21}{5}.\frac{10}{11}+\frac{57}{11}}=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}=\frac{9}{9}=1\)

\(A=\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}=\dfrac{100}{300}\)

\(B=\dfrac{25}{11}\cdot\dfrac{13}{12}\cdot\dfrac{-11}{5}=\dfrac{-65}{12}=\dfrac{-1625}{300}\)

\(C=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\cdot\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\cdot\dfrac{-2}{5}=\dfrac{-22}{100}=\dfrac{-11}{50}=\dfrac{-66}{300}\)

Vì -1625<-66<100

nên B<C<A