So sánh các số sau
\(99^{20}\) và \(9999^{10}\)
\(\left|\dfrac{-25}{46}\right|\) và \(\left(\dfrac{-25}{46}\right)^{2005}\)
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d, ta có :(-32)9=-(329) ;(-18)13=-(1813)
329=32\(\times\)328=32\(\times\)(322)4=32\(\times\)10244=32\(\times\)1024\(\times\)10243
1813=18\(\times\)1812=18\(\times\)(183)4=18\(\times\)58324=18\(\times\)5832\(\times\)58323
18\(\times\)5832 >16\(\times\)5832=32\(\times\)2916>32\(\times\)1024 =58323>10243
nên 1813>329
vậy (-18)13 <(-32)9
(-32)9=-(329)
(-18)13=-(1813)
329<369
ta có :369=(2\(\times\)18)9=29\(\times\)189
vì 184>164mà 164=(24)4=216
mà 216>29
\(\Rightarrow\)184>29
\(\Rightarrow\)184\(\times\)189>29\(\times\)189
\(\Rightarrow\)1813>369mà 369 >329
\(\Rightarrow\)1813>329
\(\Rightarrow\)(-18)13<(-32)9
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
\(=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot230.04+46.75}{\left(\dfrac{13}{10}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{53.25+46.75}{\dfrac{139}{30}:\dfrac{-41}{21}}=100:\left(\dfrac{139}{30}\cdot\dfrac{-21}{41}\right)\)
\(=100\cdot\dfrac{-400}{973}=\dfrac{-40000}{973}\)
\(E=\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)
\(E=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{100}{-\dfrac{100}{41}}\)
\(E=-41\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).203\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
= \(\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5076}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
= \(\dfrac{\dfrac{25}{108}.\dfrac{5076}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
=\(\dfrac{\dfrac{5076}{108}+\dfrac{187}{4}}{\dfrac{-100}{4}}\)
= \(\dfrac{47+\dfrac{187}{4}}{-25}\)
= \(\dfrac{375}{4}:\left(-25\right)\)
= \(\dfrac{-15}{4}\)
\(\left\{{}\begin{matrix}99^{29}=\left(99^2\right)^{10}=9801^{10}\\9999^{10}=9999^{10}\end{matrix}\right.\)
\(9801^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
\(\left\{{}\begin{matrix}\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\\\left(-\dfrac{25}{46}\right)^{2005}< 0\end{matrix}\right.\)
\(\Rightarrow\left(-\dfrac{25}{46}\right)^{2005}< \left|-\dfrac{25}{46}\right|\)
a/ Ta có :
\(99^{20}=\left(99^2\right)^{10}=9081^{10}\)
Vì \(9081^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
b/ Ta có :
\(\left|\dfrac{-25}{46}\right|=\dfrac{25}{46}>0\)
\(\left(\dfrac{-25}{46}\right)^{2005}< 0\)
\(\Leftrightarrow\left|\dfrac{-25}{46}\right|>\left(\dfrac{-25}{46}\right)^{2005}\)