A = 2¹⁹ x 3⁹ x 12151

cho mình biết cách làm đươc ko

\(\frac{2^{19}.27^3+15^4.4^9.9^4}{6^9.2^{10}+12^{10}}\)

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

\(=\frac{2^{19}.3^9+3^{12}.5^4.2^{18}}{2^{19}.3^9+2^{20}.3^{10}}\)

\(=\frac{2^{18}.3^9.\left(2+3^3+5^4\right)}{2^{19}.3^9.\left(1+2+3\right)}\)

\(=\frac{654}{2.6}\)

\(=\frac{109}{2}\)

Chúc bn học tốt !!!!

giúp mk ik, năn nỉ đó.

\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)

mình biết làm câu D thôi còn câu còn lại chắc bạn ghi sai đề

\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}=\frac{2^{14}.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}=\frac{2^{22}}{3.2^{15}.2^8-5.2^22^{20}}=\frac{2^{22}}{3.2^{22}.2-5.2^{22}}\)

\(=\frac{2^{22}}{2^{22}\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\) bạn ơi dấu . là nhân nhé

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐