Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5-7.2\right)}\)\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-14\right)}\)
\(=\frac{2^{29}.3^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}=\frac{2.9}{-9}=-2\)
Vậy A=-2
=\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
=\(\frac{2^{29}\left(5.3^{18}-3^{18}\right)}{2^{28}\left(5.3^{18}-3^{18}\right).7}\)
=\(\frac{2}{7}\)
a,\(5.2^2\)\(-18:3^2\)
\(=5.4-18:9\)
\(=20-2\)
\(=18\)
b, \(\text{17.85+15.17-120}\)
\(=17.\left(85+15\right)\)\(-120\)
= \(17.100-120\)
\(=1700-120\)
\(=1580\)
c, \(2^3\)\(.17-2^3\)\(.14\)
= \(2^3\)\(.\left(17-14\right)\)
\(=8.3\)
\(=24\)
Hok tốt !