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đề bài \(\frac{5\cdot4^{15}\cdot\left(-9\right)^9-4\cdot3^{20}\cdot8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
Giải:
\(\frac{5\cdot4^{15}\cdot\left(-9\right)^9-4\cdot3^{20}\cdot8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)=\(\frac{5.\left(2^2\right)^{15}.\left(-3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{-5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
=\(\frac{-5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
=\(\frac{-2^{29}.3^{18}.\left(5.2+3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}\)
=\(\frac{-2^{29}.3^{18}.19}{2^{28}.3^{18}.1}\)
=\(=-2.19=-38\)
=\(\frac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot3^{18}}\)=\(\frac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)=\(\frac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}\)=2
nek cậu:
A=5.415.99-4.320.89và B=5.29.619-7.229.276
Tính A,B
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)
ko bít