6,887 : 9.3
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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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\(=\frac{2^9.3^{10}+2^8.3^7.3^3.5}{2^{10}.3^{10}+2^9.3^{11}}=\frac{2^8.3^{10}.\left(2+5\right)}{2^8.3^{10}.\left(2^2+2.3\right)}=\frac{7}{10}\)
\(a.=9.3\times\left(6.7+3.3\right)\)
\(=9.3\times10\)
\(=93\)
\(b.=0.35\times\left(7.8+2.2\right)\)
\(=0.35\times10\)
\(=3.5\)
10+2x=45:9.3
=>10+2x=5.3
=>10+2x=15
=>2x=15-10
=>2x=5
=>x=5:2
=>x=2,5
Vậy x=2,5
\(3^x+9\cdot3^{-x}=10\)
=>\(3^x+\dfrac{9}{3^x}=10\)
=>\(\dfrac{\left(3^x\right)^2+9}{3^x}=10\)
=>\(\left(3^x\right)^2+9=10\cdot3^x\)
=>\(\left(3^x\right)^2-10\cdot3^x+9=0\)
=>\(\left(3^x-1\right)\left(3^x-9\right)=0\)
=>\(\left[{}\begin{matrix}3^x-1=0\\3^x-9=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3^x=1\\3^x=9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
9.3^x+1= 81
3^2.3^x+1= 81
3^x+3=81
3^x+3=3^4
Suy ra x+3=4
x= 4-3
x=1
\(9.3^{x+1}=81\)
\(\Rightarrow3^{x+1}=81:9\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
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