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S
19 tháng 3 2020
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
PN
2
S
27 tháng 7 2016
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
\(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
27 tháng 7 2016
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
=> \(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
=> \(\left(\frac{1}{27}\right)^{23}< \left(\frac{1}{81}\right)^{16}\)
27 tháng 12 2016
a) \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
\(=\frac{15}{6}-\frac{1}{6}.9+\frac{3}{8}\)
\(=\frac{15}{6}-\frac{9}{6}+\frac{3}{8}\)
\(=1+\frac{3}{8}\)
\(=\frac{11}{8}\)
b) \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=2^2.3=12\)
AT
27 tháng 12 2016
a/ \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
= \(\frac{15}{6}-\frac{3}{18}.9+\frac{8}{8}\)
= \(\frac{15}{6}-\frac{3}{2}+\frac{3}{8}\)
= \(\frac{60-36+9}{24}=\frac{33}{24}=\frac{11}{8}\)
b/ \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}\) \(=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=\frac{2^2.3^{35}}{3^{34}}=\frac{4.3}{1}=12\)
30 tháng 6 2015
Bạn để ý: 81 = 3^4 (34) 27 = 3^3
64 = 2^6 256 = 2^8
Vậy \(\left(\frac{27}{64}\right)^{15}=\left(\frac{3^2}{8^2}\right)^{15}=\left(\frac{3}{8}\right)^{30};\left(\frac{81}{256}\right)^{10}=\left(\frac{3^4}{4^4}\right)^{10}=\left(\frac{3}{4}\right)^{40}\)
Vì 3/8 <3/4 ; 30<40 nên \(\left(\frac{3}{8}\right)^{30}
17 tháng 10 2018
a,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)
\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)
\(=\frac{-8}{2005}\)
b,Ta có: \(81^{10}-27^{13}-9^{21}\)
\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)
\(=3^{40}-3^{39}-3^{42}\)
\(=3^{39}.3-3^{39}-3^{39}.3^3\)
\(=3^{39}.\left(3-1-3^3\right)\)
\(=3^2.3^{37}.\left(-25\right)\)
\(=3^{37}.\left(-225\right)⋮225\)
Vậy \(81^{10}-27^{13}-9^{21}⋮225\)
27 tháng 1 2019
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
27 tháng 1 2019
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
23 tháng 8 2018
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)
\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)
Do \(\left(\frac{3}{2^2}\right)^{45}