A = \(\frac{2^{13}.5^2.2^6.3^4}{8.2^{18}.81.5}\)

   = \(\frac{2^{19}.5^2.3^4}{2^3.2^{18}.3^4.5}\)

   = \(\frac{2^{19}.5^2.3^4}{2^{21}.3^4.5}\) 

   = \(\frac{5}{2^2}\) = \(\frac{5}{4}\)

\(A=\frac{2^{13}.5^2.2^6.3^4}{8.2^{18}.81.5}\)

\(A=\frac{2^{19}.5^2.3^4}{2^{21}.3^4.5}=\frac{5}{2^3}=\frac{5}{8}\)

\(A=.................=\frac{2^3.2^{10}.5^2.2^6.3^4}{2^3.2^{10}.2^6.2^2.3^4.5}=\frac{5}{2^2.5}=\frac{1}{2^2}=\frac{1}{4}\)

chắc như thế này , k chắc

\(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}\\ =\frac{2^8.3^8.2^4-2^{2.5}.3^{2.4}.2^4}{3^{3.3}.2^{3.4}-3^9.2^{13}}\\ =\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{13}}\\ =\frac{2^{12}.3^8\left(1-2^2\right)}{3^9.2^{12}.\left(1-2\right)}\\ =\frac{3^8.\left(-3\right)}{3.\left(-1\right)}=3^8\)

a: \(=1995^2-\left(1995^2-1\right)=1995^2-1995^2+1=1\)

b: \(=18^8-18^8+1=1\)

c: \(=\left(163+37\right)^2=200^2=40000\)

a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)

1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)

\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)

CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E

a) \(\frac{5}{9}:\frac{13}{7}+\frac{5}{9}:\frac{13}{9}-1\frac{2}{3}\\ =\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{3}\\ =\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}\right)-\frac{5}{3}\\ =\frac{5}{9}\cdot\frac{16}{13}-\frac{5}{3}\\ =\frac{80}{117}-\frac{5}{3}\\ =\frac{80}{117}-\frac{195}{117}=\frac{-115}{117}\)

b) \(\left(15-6\frac{13}{18}\right):11\frac{1}{27}-2\frac{1}{8}:1\frac{11}{40}\\ =\left(\frac{270}{18}-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\\ =\frac{149}{18}\cdot\frac{27}{298}-\frac{17}{8}\cdot\frac{40}{51}\\ =\frac{3}{4}-\frac{5}{3}\\ =\frac{9}{12}-\frac{20}{12}=\frac{-11}{12}\)

Đặt \(A=\frac{9+\frac{9}{11}+\frac{18}{23}-\frac{27}{37}}{8+\frac{8}{11}+\frac{16}{23}-\frac{24}{37}}-\frac{2+\frac{16}{29}-\frac{24}{13}-\frac{32}{11}}{3+\frac{24}{29}-\frac{36}{13}-\frac{48}{11}}\)\(=\frac{9\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}{8\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}-\frac{2\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}{3\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}\)

\(=\frac{9}{8}-\frac{2}{3}\)(do \(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37};1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\ne0\))

\(=\frac{27}{24}-\frac{16}{24}=\frac{11}{24}.\)

Vậy A = \(\frac{11}{24}.\)