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Ta có:
134012 = 132.2006 = (132)2006 = 1692006
Vì 1313 > 169 và 2006 > 0 nên 13132006 > 1692006 hay 13132006 > 134012
Ta có :
\(1313^{2006}=13^{2006}.101^{2006}>13^{2006}.13^6=13^{2012}\)
\(\Rightarrow1313^{2006}>13^{2012}\)
a/b và a+2006/b+2006
=> a/b và a+2006/b/2006
==> a/b < a+2006/b+2006
a, \(\frac{1313}{2727}=\frac{13\cdot101}{27\cdot1001}=\frac{13}{27}\)
b,\(\frac{151515}{232323}=\frac{15.10101}{23.10101}=\frac{15}{23}\)
a. \(\frac{7}{15}< \frac{7}{14}=\frac{1}{2};\frac{15}{23}>\frac{15}{30}=\frac{1}{2}\text{ hay }\frac{7}{15}< \frac{1}{2}< \frac{15}{23}\)
Vậy \(\frac{7}{15}< \frac{15}{23}\).
b. \(x=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13x=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
\(y=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13y=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
Vì \(13^{17}+1>13^{16}+1\) nên \(\frac{12}{13^{17}+1}< \frac{12}{13^{16}+1}\)
Mà 1 = 1 => \(1+\frac{12}{13^{17}+1}< 1+\frac{12}{13^{16}+1}\text{ hay }13x< 13y\)
=> x < y.
a)
\(\frac{13}{27}=\frac{13.101}{27.101}=\frac{1313}{2727}\)
=> \(\frac{13}{27}=\frac{1313}{2727}\)
b)
\(-\frac{15}{23}=-\frac{15.10101}{23.10101}=-\frac{151515}{232323}\)
=>\(-\frac{15}{23}=-\frac{151515}{232323}\)
a) \(\frac{1313}{2727}=\frac{1313:101}{2727:101}=\frac{13}{27}\)
Vậy \(\frac{13}{27}=\frac{1313}{2727}\)
b) \(-\frac{151515}{232323}=\frac{-151515:10101}{232323:10101}=-\frac{15}{23}\)
Vậy \(-\frac{15}{23}=-\frac{151515}{232323}\)
Ta có :
\(\left(20^{2006}+11^{2006}\right)^{2007}=20^{2006.2007}+2.20^{2006}.11^{2006}+11^{2006.2007}\)
\(\left(20^{2007}+11^{2007}\right)^{2006}=20^{2007.2006}+2.20^{2007}.11^{2007}+11^{2007.2006}\)
Vì \(2.20^{2006}.11^{2006}< 2.20^{2007}.11^{2007}\) nên \(\left(20^{2006}+11^{2006}\right)^{2007}< \left(20^{2007}+11^{2007}\right)^{2006}\)
Chúc bạn học tốt ~
Ta có:\(13^{4012}=\left(13^2\right)^{2006}=169^{2006}\)
Vì 169<1313 nên\(169^{2006}< 1313^{2006}\)
Vậy\(13^{4012}< 1313^{2006}\)