C= \(\frac{9^9.8^2}{27^5.6^3}\)
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Bài giải
Ta có :
\(\frac{2^{27}\cdot9^4}{6^9\cdot8^5}=\frac{2^{27}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^9\cdot\left(2^3\right)^5}=\frac{2^{27}\cdot3^8}{3^9\cdot2^9\cdot2^{15}}=\frac{2^{27}\cdot3^8}{3^9\cdot2^{24}}=\frac{2^{24}\cdot3^8\cdot2^3}{2^{24}\cdot3^8\cdot3}=\frac{2^3}{3}=\frac{8}{3}\)
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a)\(\dfrac{27^4.4^3}{9^5.8^2}\)
=\(\dfrac{3^{12}.2^6}{3^{10}.2^6}\)
=3\(^2\)=9
b)\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)
=\(\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)
=\(\dfrac{9}{2}\)
\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=\dfrac{3^{12}}{3^{10}}=3^2=9\)
_________
\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}=\dfrac{1}{3^2.2}=\dfrac{1}{9.2}=\dfrac{1}{18}\)
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a ) \(\frac{-5}{9}:\frac{-7}{18}+1\frac{2}{7}\)
\(=\frac{-5}{9}.-\frac{18}{7}+\frac{9}{7}\)
\(=\frac{-5.-18}{9.7}+\frac{9}{7}\)
\(=\frac{10}{7}+\frac{9}{7}\)
\(=\frac{10+9}{7}\)
\(=\frac{19}{7}\)
\(\frac{2^{27}\times9^4}{6^9\times8^5}=\frac{2^{27}\times\left(3^2\right)^4}{\left(2\times3\right)^9\times\left(2^3\right)^5}=\frac{2^{27}\times3^8}{2^9\times3^9\times2^{15}}=\frac{2^3}{3}=\frac{8}{3}\)
\(\sqrt{13^2}-5^2+\sqrt{3^2+4^2}-\sqrt{\left(-7\right)^2}=13-25+\sqrt{9+16}-\sqrt{49}=13-25+5-7=-14\)
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\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(=\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(=\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)
\(=\frac{121}{81}\)
37,52 + 4,7 x 2,3 - 9,8
= 37,52 + 10,81 - 9,8
= 26,71 - 9,8
= 16,91
(9^9.8^2)/(3^15.2^3.2^3)
(3^18.2^6)/(3^18.2^3)
2^3
\(C=\frac{9^9.8^2}{27^5.6^3}\)
\(=\frac{\left(3^2\right)^9.2^6}{\left(3^3\right)^5.\left(2.3\right)^3}\)
\(=\frac{3^{18}.2^6}{3^{15}.2^3.3^3}\)
\(=\frac{3^{18}.2^3.2^3}{3^{18}.2^3}=2^3=8\)