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a: 27^4=(3^3)^4=3^12<3^18
b: 49^4=(7^2)^4=7^8
c: 9^16=(9^2)^8=81^8>27^8
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
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1218 = 318 . 236
276 . 169 = 318 . 336
=> 1218 = 276 . 169
a) \(32^{50}\)và \(27^{51}\)
\(32^{50}=\left(32^2\right)^{25}=1024^{25}\)
\(27^{51}=\left(27^2\right)^{25}.27=729^{25}.27\)
Vì \(1024>729\)nên \(1024^{25}>729^{25}.27\)hay \(32^{50}>27^{51}\)
b) \(31^9\)và \(9^{16}\)
\(31^9=\left(91^3\right)^2=273^2\)
\(9^{16}=\left(9^2\right)^4=81^4=\left(81^2\right)^2=6561^2\)
Vì \(6561>273\)nên \(273^2< 6561^2\)hay \(31^9< 9^{16}\).
\(a,81^3=\left(9^2\right)^3=9^6\)
Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)
\(b,5^{14}=\left(5^2\right)^7=25^7\)
Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)
\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)
a) 2^27=(2^3)^9=8^9
3^18=(3^2)^9=9^9
Vì 8^9 bé hơn 9^9 nên 2^27 bé hơn 3^18