Tính các luỹ thừa sau:
\(\left(-\frac{1}{2}\right)^4=\frac{\left(-1\right)^4}{2^4}=\frac{1}{16}\)
Giải hộ mik bài này ik ạ!
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\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)
b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)
c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)
d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)
\(=\frac{45}{4}-\frac{223}{28}\)
\(=\frac{315}{28}-\frac{223}{28}\)
\(=\frac{23}{7}\)
b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)
\(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)
\(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)
\(=\frac{69}{8}\)
A=\(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=>2A=1+\(\frac{1}{2}+...+\frac{1}{2^{98}}\)
=>2A-A=A=\(\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
=>A=\(1-\frac{1}{2^{99}}\)
mình chịu thua vì mình cũng gặp câu này mà ko có lời giải
\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}.\)
\(\frac{\left(-1\right)^4}{2^4}=\frac{1}{16}.\)
\(\frac{1}{16}=\frac{1}{16}.\)
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