\(\frac{4^9.9^4-10.6^9}{2^{10}.3^8+6^8-28}=\frac{2^{18}.3^8-2.5.2^9.3^9}{2^{10}.3^8+2^8.6^8-2^2.7}\)

Nhóm các tích trên tử và mẫu sao cho bằng nhau rồi tính.

\(=\frac{\left(2^2\right)^9.\left(3^2\right)^4-2.5.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8-28}\)

\(=\frac{2^{18}.3^8-2.5.2^9.3^9}{2^{10}3^8+2^8.3^8-28}\)

\(=\frac{2^{18}.3^8-2^{10}.3^9.5}{2^8.3^8.\left(2^2+1\right)-28}\)

\(=\frac{2^{10}.3^8.\left(2^8-3.5\right)}{2^8.3^8.5-28}\)

\(=\frac{2^2.\left(2^8-3.5\right)}{5-28}\)

\(=\frac{2.\left(256-15\right)}{-23}\)

\(=\frac{-482}{23}\)

lamf như thế nào

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)

\(B=\frac{7}{2.7}\)

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

\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)

\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)

\(C=\frac{-2}{9}\)

\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)

\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)

\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)

\(D=1\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)

\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)

Vậy A = \(\frac{1}{2}\)

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)

Vậy B = \(\frac{1}{2}\)

A = \(\frac{2}{3}+\frac{5}{6}:6-\frac{1}{8}.\left(-3\right)^2\)

A = \(\frac{2}{3}+\frac{5}{6}.\frac{1}{6}-\frac{1}{8}.9\)

A = \(\frac{2}{3}+\frac{5}{36}-\frac{9}{8}\)

A =\(\frac{-23}{72}\)

\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)

\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)

\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)

\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)

\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)

\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)

\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)