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1, Đặt \(A=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(A=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)\(A=\frac{2^{28}\left(5.2^2.3^{18}-2.3^{20}\right)}{2^{28}\left(5.3^{19}-7.2.3^{18}\right)}\)
\(A=\frac{5.2^2.3^{18}-2.3^{20}}{5.3^{19}-7.2.3^{18}}\)\(A=\frac{3^{18}\left(5.2^2-2.3^2\right)}{3^{18}\left(5.3-7.2\right)}\)
\(A=\frac{5.2^2-2.3^2}{5.3-7.2}\)\(A=2\)
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
Ta có : \(\frac{5.4^{15}.9^9+4.3^{20}.\left(-8\right)^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2^{29}.3^{18}}{2^{28}.3^{18}}=2\)
a) 378
b) 3
c) 2
d) 2
e) \(\frac{8748}{1715}\)
Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?
\(\frac{5.4^{15}.\left(-9\right)^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.\left(2^2\right)^{15}.\left(-3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{-5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}=\frac{-5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)\(=\frac{-5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{-2^{29}.3^{18}.\left(5.2+3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}=\frac{-2^{29}.3^{18}.19}{2^{28}.3^{18}.1}=-2.19=-38\)
\(\frac{5.\left(2^2\right)^{15}.\left(-3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.2^{19}.3^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{5.2^{30}.-3^{18}-2^2.3^{20}.2^{27}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{-5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{-2^{29}.3^{18}.\left(5.2+3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}=\frac{-2.29}{1}=-58\)
c) G = \(\frac{636363.37-373737.63}{1+2+3+...+2017}\)
G = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2017}\)
G = \(\frac{0}{1+2+3+...+2017}\)
=> G = 0
Vậy G = 0
a) \(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\right)\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(\Rightarrow E=\frac{1}{2}.\frac{612}{1225}\)
\(\Rightarrow E=\frac{306}{1225}\)
Vậy...
b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)
d) Bạn xem lại đề nhé