Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2\)
\(=\frac{3^2\cdot3^3\cdot3^2}{3^4}\)
\(=3^3=27\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)\)
\(=\frac{2^2\cdot2^2\cdot2^4}{2^3}\)
\(=2^5=32\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2\)
\(=\frac{3^2\cdot2^5\cdot2^4}{3^2}\)
\(=2^9=512\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2\)
\(=\frac{1^2\cdot1\cdot3^4}{3^2}\)
\(=3^2=9\)
\(a,9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^{\left(-4\right)}.3^2=3^{2+3-4+2}=3^3.\)
\(b,4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.2^{-4}\right)=2^{2+5}:2^{3-4}=2^7:2^{-1}=2^{7-\left(-1\right)}=2^8.\)
\(c,3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=\left(\frac{3^2}{3^2}\right).\left(2^5.2^2\right)=1.2^{5+2}=2^7\)
\(d,\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^2.\frac{1}{3}.\left(3^2\right)^2=\left(\frac{1}{3}\right)^{2+1}.3^4=\left(\frac{1}{3}\right)^3.\left(\frac{1}{3}\right)^{-4}=\left(\frac{1}{3}\right)^{3-4}=\left(\frac{1}{3}\right)^{-1}=3^1\)
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)
\(\Leftrightarrow3n-1=2\)
\(\Leftrightarrow3n=3\)
\(\Leftrightarrow n=1\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)
\(\Leftrightarrow n+2=-1\)
\(\Leftrightarrow n=-3\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)
\(\Leftrightarrow-n+1=-3\)
\(\Leftrightarrow n=-4\)
c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)
\(\Leftrightarrow3n+1=-3\)
\(\Leftrightarrow3n=-4\)
\(\Leftrightarrow n=-\frac{4}{3}\)
a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2=3^2\cdot3^3\cdot\left(\frac{1}{3}\right)^43^2=3^7\cdot\frac{1}{3^4}=3^3\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)=2^2\cdot2^5:\left(2^3\cdot\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2=3^2\cdot2^5\cdot\frac{2^2}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2=\frac{1}{3^2}\cdot\frac{1}{3}\cdot3^4=\frac{1}{3^3}\cdot3^4=3\)
a)9.33.\(\frac{1}{81}\).32
=32.33.34.\(\frac{1}{3^4}\).32
=311.\(\frac{1}{3^4}\)
=37
b)4.25:(\(2^3.\frac{1}{16}\))
=22.25:(\(2^3.\frac{1}{2^4}\))
=27:\(\frac{2^3}{2^4}\)
=27.\(\frac{2^4}{2^3}\)
=\(\frac{2^{11}}{2^3}\)
=28
c)32.25.\(\left(\frac{2}{3}\right)^2\)
=32.25.\(\frac{2^2}{3^2}\)
=\(\frac{3^2.2^5.2^2}{3^2}\)
=27
d)\(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.\left(3^2\right)^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.3^4\)
=\(\frac{1^2}{3^2}.\frac{3^4}{3}\)
=\(\frac{1^2}{3^2}.3^3\)
=3