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a) \(\frac{1}{8}.16^n=2^n\)
\(\frac{2^n}{16^n}=\frac{1}{8}\)
\(\left(\frac{2}{16}\right)^n=\frac{1}{8}\)
\(\left(\frac{1}{8}\right)^n=\frac{1}{8}\)
=> n = 1
ta có 8^9>7^9>6^9>5^9>4^9>3^9>2^9>1^9
->8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9<8^9×8=8^10<9^10
k mk nha
Ta có :
\(A=3^{2008}-3^{2007}+3^{2006}-...+3^2-3+1\)
\(3A=3^{2009}-3^{2008}+3^{2007}-...+3^3-3^2+3\)
\(3A+A=\left(3^{2009}-3^{2008}+3^{2007}-...+3^3-3^2+3\right)+\left(3^{2008}-3^{2007}+3^{2006}-...+3^2-3+1\right)\)
\(4A=3^{2009}+1\)
\(A=\frac{3^{2009}+1}{4}>\frac{1}{4}\)
Vậy \(A>\frac{1}{4}\)
Chúc bạn học tốt ~
Ta có \(3A=3^{2009}-3^{2008}+...-3^2+3\)
\(A=3^{2008}-3^{2007}+...-3+1\)
=> \(4A=3A+A=3^{2009}+1\)
=> \(A=\frac{3^{2009}+1}{4}\)= \(\frac{3^{2009}}{4}+\frac{1}{4}>\frac{1}{4}\)
\(4^5\cdot9^4-2.6^9=\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9=2^{10}.3^8-2.2^9.3^9=2^{10}.3^8-2^{10}.3^9=3^8-3^9=-13122\)
\(2^{10}.3^8+6^8.20=2^{10}.3^8+\left(2.3\right)^8.2^2.5=2^{10}.3^8+2^8.3^8.2^2.5=2^{10}.3^8+2^{10}.3^8.5=2^{10}.3^8.\left(1+5\right)=2^{10}.3^8.6=2^{10}.3^8.2.3=2^{11}.3^9\)