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9 2 − 2 3 − x + 7 4 = − 5 4 2 3 − x + 7 4 = 9 2 − − 5 4 2 3 − x + 7 4 = 9 2 + 5 4 2 3 − x + 7 4 = 23 4 x + 7 4 = 2 3 − 23 4 x + 7 4 = 8 − 69 12 x + 7 4 = − 57 12 x + 7 4 = − 19 4 x = − 19 4 − 7 4 x = − 26 4 x = − 13 2
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\(M=54-\frac{1}{2}.\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)-....-\frac{1}{12}.\left(1+2+...12\right)\)
\(M=54-\left[\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{12}.\left(1+2+...+12\right)\right]\)
\(M=54-\left(\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+....+\frac{1}{12}\cdot\frac{12.13}{2}\right)\)
\(M=54-\left(\frac{3}{2}+\frac{4}{2}+...+\frac{13}{2}\right)=54-44=10\)
Vậy M=10
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Bài 3:
\(24^{54}\cdot54^{24}\cdot2^{10}\)
\(=\left(2^3\cdot3\right)^{54}\cdot\left(3^3\cdot2\right)^{24}\cdot2^{10}\)
\(=2^{108}\cdot3^{54}\cdot3^{72}\cdot2^{24}\cdot2^{10}\)
\(=2^{142}\cdot3^{78}\)
\(72^{63}=\left(2^3\cdot3^2\right)^{63}=2^{189}\cdot3^{126}⋮2^{142}\cdot3^{78}\)(ĐPCM)
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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}}