\(\frac{2^{27}.9^4}{6^9.8^5}\)
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a ) \(\frac{-5}{9}:\frac{-7}{18}+1\frac{2}{7}\)
\(=\frac{-5}{9}.-\frac{18}{7}+\frac{9}{7}\)
\(=\frac{-5.-18}{9.7}+\frac{9}{7}\)
\(=\frac{10}{7}+\frac{9}{7}\)
\(=\frac{10+9}{7}\)
\(=\frac{19}{7}\)
\(\frac{2^{27}\times9^4}{6^9\times8^5}=\frac{2^{27}\times\left(3^2\right)^4}{\left(2\times3\right)^9\times\left(2^3\right)^5}=\frac{2^{27}\times3^8}{2^9\times3^9\times2^{15}}=\frac{2^3}{3}=\frac{8}{3}\)
\(\sqrt{13^2}-5^2+\sqrt{3^2+4^2}-\sqrt{\left(-7\right)^2}=13-25+\sqrt{9+16}-\sqrt{49}=13-25+5-7=-14\)
\(C=\frac{9^9.8^2}{27^5.6^3}\)
\(=\frac{\left(3^2\right)^9.2^6}{\left(3^3\right)^5.\left(2.3\right)^3}\)
\(=\frac{3^{18}.2^6}{3^{15}.2^3.3^3}\)
\(=\frac{3^{18}.2^3.2^3}{3^{18}.2^3}=2^3=8\)
a)\(\dfrac{27^4.4^3}{9^5.8^2}\)
=\(\dfrac{3^{12}.2^6}{3^{10}.2^6}\)
=3\(^2\)=9
b)\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)
=\(\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)
=\(\dfrac{9}{2}\)
\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=\dfrac{3^{12}}{3^{10}}=3^2=9\)
_________
\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}=\dfrac{1}{3^2.2}=\dfrac{1}{9.2}=\dfrac{1}{18}\)
\(a)\left(6\frac{4}{11}+3\frac{7}{9}\right)-4\frac{4}{11}\)
\(=(6\frac{4}{11}-4\frac{4}{11})+3\frac{7}{9}\)
\(=2\frac{4}{11}+3\frac{7}{9}\)
\(=\frac{26}{11}+\frac{34}{9}\)
\(=\frac{608}{99}\)
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
Bài giải
Ta có :
\(\frac{2^{27}\cdot9^4}{6^9\cdot8^5}=\frac{2^{27}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^9\cdot\left(2^3\right)^5}=\frac{2^{27}\cdot3^8}{3^9\cdot2^9\cdot2^{15}}=\frac{2^{27}\cdot3^8}{3^9\cdot2^{24}}=\frac{2^{24}\cdot3^8\cdot2^3}{2^{24}\cdot3^8\cdot3}=\frac{2^3}{3}=\frac{8}{3}\)
Bài giải
Ta có :
\(\frac{2^{27}\cdot9^4}{6^9\cdot8^5}=\frac{2^{27}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^9\cdot\left(2^3\right)^5}=\frac{2^{27}\cdot3^8}{3^9\cdot2^9\cdot2^{15}}=\frac{2^{27}\cdot3^8}{3^9\cdot2^{24}}=\frac{2^3}{3}=\frac{8}{3}\)