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a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}
a) Ta có : 4<5
=> 453<553
=> (22)53<553
=> 2106<553
Mà 291<2106 nên 291<553
Vậy 291<553.
b) Ta có : 544=544
2112=(213)4=92614
Mà 53<9261 nên 544<92614
=> 544<2112
Vậy 544<2112.
Bài dễ mà you ko tự suy nghĩ được, đúng là lười suy nghĩ
a) 2561=(52)61=52.61=5122
Vì 122>120 nên 5122>5120 hay 2561>5120
b) 1680 = (42)80= 42.80=4160
Vì 160>65 nên 4160>465 hay 1680>465
Mấy câu khác tự làm
2) a) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\)
\(\frac{21}{81^8}=\frac{21}{\left(3^4\right)^8}=\frac{21}{3^{32}}=\frac{21.3}{3^{33}}=\frac{63}{3^{33}}>\frac{1}{3^{33}}\)
=> \(\frac{21}{81^8}>\frac{1}{27^{11}}\)
b) Rõ ràng : 399 < 1121 => \(\frac{1}{399}>\frac{1}{11^{21}}\)
a) \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}\)=> \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=1-\sqrt[3]{\frac{4}{9}}\)
=> x = \(\frac{6}{5}-\frac{6}{5}.\sqrt[3]{\frac{4}{9}}\)
b) => \(\frac{1}{13}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\)
=> \(\left(\frac{1}{2}x-1\right)^4=\frac{13}{48}\)
=> \(\frac{1}{2}x-1=\sqrt[4]{\frac{13}{48}}\) hoặc \(\frac{1}{2}x-1=-\sqrt[4]{\frac{13}{48}}\)
=> \(x=2+2\sqrt[4]{\frac{13}{48}}\) hoặc \(x=2-2\sqrt[4]{\frac{13}{48}}\)
g, Ta có :
\(54^8=\left(54^2\right)^4=2916^4\)
\(21^{12}=\left(21^3\right)^4=9261^4\)
Vì 2916 < 9261 nên \(2916^4< 9261^4\)
Vậy \(58^8< 21^{12}\) .
h, Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì 8192 > 3125 nên \(8192^7>3125^7\)
Vậy \(2^{91}>5^{35}\) .
\(3^{81}=3^{3.27}=\left(3^3\right)^{27}=27^{27}\)
\(5^{54}=5^{2.27}=\left(5^2\right)^{27}=25^{27}\)
Vì 2527 < 2727 nên 381 < 554