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\(\frac{2^9.3^{10}+2^8.3^7.135}{2^{10}.3^{10}+2^9.3^{11}}\)\(=\frac{2^9.3^{10}+2^8.3^7.3^3.5}{2^{10}.3^{10}+2^9.3^{11}}=\frac{2^9.3^{10}+2^8.3^{10}.5}{2^{10}.3^{10}+2^9.3^{11}}=\frac{2^8.3^{10}.\left(2+5\right)}{2^8.3^{10}.\left(2^2+2.3\right)}=\frac{7}{10}\)
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\(=\frac{2^9.3^{10}+2^8.3^7.3^3.5}{2^{10}.3^{10}+2^9.3^{11}}=\frac{2^8.3^{10}.\left(2+5\right)}{2^8.3^{10}.\left(2^2+2.3\right)}=\frac{7}{10}\)
\(\dfrac{1}{9}.3^4.3^n=9^4\)
\(\dfrac{1}{9}.3^{4+n}=9^4\)
\(3^{4+n}=\dfrac{9^4}{\dfrac{1}{9}}\)
=> \(3^{4+n}=9^4.9\)
=> \(3^{4+n}=9^5\)
=> \(3^{4+n}=(3^2)^5\)
=> \(3^{4+n_{ }}=3^{10}\)
=> \(4+n=10\)
\(n=10-4\)
\(n=6\)
\(3^{-1}.3^x+9.3^x=28\\ \Leftrightarrow3^x\left(\dfrac{1}{3}+9\right)=28\\ \Leftrightarrow3^x.\dfrac{28}{3}=28\\ \Leftrightarrow3^x=3\\ \Leftrightarrow x=1\)
\(3^{n-1}+9.3^n=3^{n-1}\left(1+9.3^1\right)=28.3^{n-1}=28.3^5\Rightarrow n=6\)
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