Question

Rút gọn

\(A=\frac{21.99+21}{35.201-35}\)

\(B=\frac{\left(-2\right)^{36}.3^{17}}{3^{18}.2^{35}}\)

\(C=\frac{7^{100}-7^{99}}{7^{98}-7^{97}}\)

Answer 1

\(A=\frac{21.99+21}{35.201-35}=\frac{21\left(99+21\right)}{35\left(201-1\right)}=\frac{21.\left(99+1+20\right)}{35.200}=\frac{21.120}{35.200}=\frac{\left(3.7\right).\left(2.2.3.\right)10}{\left(5.7\right).\left(2.2.5\right).10}=\frac{3.3}{5.5}=\left(\frac{3}{5}\right)^2\)

\(B=\frac{\left(-2\right)^{36}.3^{17}}{3^{18}.2^{35}}=\frac{2^{36}.3^{17}}{2^{35}.3^{18}}=2^{\left(36-35\right)}.3^{17-18}=2.3^{-1}=\frac{2}{3}\)

\(C=\frac{7^{100}-7^{99}}{7^{98}-7^{97}}=\frac{7^{99}\left(7^1-7^0\right)}{7^{97}\left(7^1-7^0\right)}=7^{99-97}.\frac{\left(7-1\right)}{7-1}=7^2.1=49\)

Answer 2

mấy bn giải chi tiết giúp mk nhé !