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a) \(T=\frac{9^{14}\times25^6\times8^7}{18^{12}\times625^3\times24^3}\)
\(=\frac{\left(3^2\right)^{14}\times25^6\times\left(2^3\right)^7}{\left(2\times3^2\right)^{12}\times\left(25^2\right)^3\times\left(3\times2^3\right)^3}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{12}\times3^{24}\times25^6\times3^3\times2^9}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{\left(2^{12}\times2^9\right)\times\left(3^{24}\times3^3\right)\times25^6}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{21}\times3^{27}\times25^6}=3\)
b) \(A=\frac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^9\times6^{19}-7\times2^{29}\times27^6}\)
\(=\frac{5\times\left(2^2\right)^{15}\times\left(3^2\right)^9-2^2\times3^{20}\times\left(2^3\right)^9}{5\times2^9\times\left(2\times3\right)^{19}-7\times2^{29}\times\left(3^3\right)^6}\)
\(=\frac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^9\times2^{19}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{28}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{28}\times3^{18}\times\left(5\times3-7\times2\right)}\)
\(=\frac{2\times\left(10-9\right)}{15-14}=\frac{2\times1}{1}=2\)
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)
\(=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(5^4\right)^3.3^3.\left(2^3\right)^3}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.3^3.2^9}\)
\(=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{27}.5^{12}}\)
\(=\frac{3}{5^2}\)
\(=\frac{3}{25}\)
Ta có abab/cdcd=abab:101/cdcd:101=ab/cd
ababab/cdcdcd=ababab:10101/cdcdcd:10101=ab/cd
Vì ab/cd=ab/cd nên abab/cdcd= ababab/cdcdcd
\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd}\)
\(\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)
VẬY \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
\(\frac{52}{75}=\frac{52.101}{75.101}=\frac{5252}{7575};\frac{52}{75}=\frac{52.10101}{75.10101}=\frac{525252}{757575}\)
\(\frac{13}{15}=\frac{13.101}{15.101}=\frac{1313}{1515};\frac{13}{15}=\frac{13.10101}{15.10101}=\frac{131313}{151515}\)
\(\frac{ab}{cd}=\frac{101ab}{101cd}=\frac{abab}{cdcd};\frac{ab}{cd}=\frac{10101ab}{10101cd}=\frac{ababab}{cdcdcd}\)
ai k minh minh k lai
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
Cho e hỏi cái này. Ở câu 1 ý, cuối đề là \(-\frac{1}{7}\) sao xuống dưới phải đổi thành -1 thế ạ ? E chưa hiểu lắm :<
1)
a) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{24}.5^{12}.3^3.2^9}=\frac{3}{5^2}=\frac{3}{25}\)
Bài 2:
\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd};\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)