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a ) \(\frac{-5}{9}:\frac{-7}{18}+1\frac{2}{7}\)
\(=\frac{-5}{9}.-\frac{18}{7}+\frac{9}{7}\)
\(=\frac{-5.-18}{9.7}+\frac{9}{7}\)
\(=\frac{10}{7}+\frac{9}{7}\)
\(=\frac{10+9}{7}\)
\(=\frac{19}{7}\)
\(\frac{2^{27}\times9^4}{6^9\times8^5}=\frac{2^{27}\times\left(3^2\right)^4}{\left(2\times3\right)^9\times\left(2^3\right)^5}=\frac{2^{27}\times3^8}{2^9\times3^9\times2^{15}}=\frac{2^3}{3}=\frac{8}{3}\)
\(\sqrt{13^2}-5^2+\sqrt{3^2+4^2}-\sqrt{\left(-7\right)^2}=13-25+\sqrt{9+16}-\sqrt{49}=13-25+5-7=-14\)
\(C=\frac{9^9.8^2}{27^5.6^3}\)
\(=\frac{\left(3^2\right)^9.2^6}{\left(3^3\right)^5.\left(2.3\right)^3}\)
\(=\frac{3^{18}.2^6}{3^{15}.2^3.3^3}\)
\(=\frac{3^{18}.2^3.2^3}{3^{18}.2^3}=2^3=8\)
i) \(\frac{25^9}{5^{16}}-5^3:5\)
\(=\frac{\left(5^2\right)^9}{5^{16}}-5^2\)
\(=\frac{5^{18}}{5^{16}}-25\)
\(=5^2-25\)
\(=25-25\)
\(=0.\)
k) \(\frac{5}{7}-\left|\frac{2}{-7}\right|\)
\(=\frac{5}{7}-\left|\frac{-2}{7}\right|\)
\(=\frac{5}{7}-\frac{2}{7}\)
\(=\frac{3}{7}.\)
l) \(\frac{3^6.3^4}{9^3}\)
\(=\frac{3^{6+4}}{\left(3^2\right)^3}\)
\(=\frac{3^{10}}{3^6}\)
\(=3^4.\)
\(=81.\)
Chúc bạn học tốt!
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
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 :
\(\frac{2^{27}\cdot9^4}{6^9\cdot8^5}=\frac{2^{27}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^9\cdot\left(2^3\right)^5}=\frac{2^{27}\cdot3^8}{3^9\cdot2^9\cdot2^{15}}=\frac{2^{27}\cdot3^8}{3^9\cdot2^{24}}=\frac{2^3}{3}=\frac{8}{3}\)