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\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\dfrac{5.3^{20}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}\)
\(=2\)
\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\dfrac{2^{29}.3^{18}}{2^{28}.3^{18}}=\dfrac{2^{29}}{2^{28}}=2^1=2\)
Sửa đề:\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^9\cdot3^{19}\cdot2^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}\cdot\left(5\cdot2-9\right)}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}=2\)
\(A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{7.2^{29}.27^6-5.2^9.6^{19}}\)
\(A=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{7.2^{29}.3^{18}-5.2^9.2^{19}.3^{19}}\)
\(A=\dfrac{2^{29}.3^{18}.\left(5.2-3^2\right)}{2^{28}.3^{18}.\left(7.2-5.3\right)}\)
\(A=\dfrac{2.\left(10-9\right)}{14-15}=\dfrac{2}{-1}=-2\)
Chúc bạn học tốt!!!
\(A=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{7.2^{29}.27^6-5.2^9.6^{19}}=\dfrac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^2\right)^9}{7.2^{29}.27^6-5.2^9.\left(2.3\right)^{19}}\)
\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{7.2^{29}.3^{18}-5.2^9.2^{19}.3^{19}}\)
\(=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(7.2-5.3\right)}\)
\(=\dfrac{2\left(10-9\right)}{14-15}=\dfrac{2}{-1}=-2\)
\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\dfrac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^2.\left(3.2\right)^{19}-7.2^{29}.\left(3^3\right)^6}\)
\(=\dfrac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^2.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{2^{29}.3^{18}.\left(5.2-3\right)}{2^{28}.3^{18}.\left(5.1.3-7.2.1\right)}\)
\(=\dfrac{2^{29}.3^{18}.1}{2^{28}.3^{18}.1}\)
\(=\dfrac{2^{29}}{2^{28}}\)
\(=2\)
=\(\dfrac{5.2^{30}.3^{27}-2^2.3^{20}.3^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
=\(\dfrac{5.2^{30}.3^{27}-2^2.3^{47}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
=\(\dfrac{2^2.3^{27}.\left(5.2^{28}-3^{20}\right)}{2^{28}.3^{18}.\left(5.3-7.2\right)}\)
chắc bạn chép sai đầu bài nhưng bạn cứ tách ra theo thừa số nguyên tố rồi rút gọn sau khi đã đưa cả tử và mẫu về một tích nhé
a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100