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a) 63
36 = 32.3 = ( 32)3 = 93
Do 6 < 9 nên 63 < 93 hay 63 < 36
^^
a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)
\({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)
Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);
b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)
\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)
Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)
c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)
\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)
Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).
d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)
Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
\(\begin{array}{l}{\left( {0,25} \right)^8} = {\left[ {{{\left( {0,5} \right)}^2}} \right]^8}=(0,5)^{2.8} = {\left( {0,5} \right)^{16}};\\{\left( {0,125} \right)^4} = {\left[ {{{\left( {0,5} \right)}^3}} \right]^4} =(0,5)^{3.4}= {\left( {0,5} \right)^{12}};\\{\left( {0,0625} \right)^2} = {\left[ {{{\left( {0,5} \right)}^4}} \right]^2} =(0,5)^{4.2}= {\left( {0,5} \right)^8}\end{array}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
\(\dfrac{8^2.6^3}{9^2.16^2}=\dfrac{\left(2^3\right)^2.2^3.3^3}{\left(3^2\right)^2.\left(2^4\right)^2}=\dfrac{2^{3.2+3}.3^3}{3^4.2^8}=\dfrac{3^3.2^8.2}{3.3^3.2^8}=\dfrac{2}{3}\\ ---\\ \dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\left(\dfrac{0,15}{0,5}\right)^4.\dfrac{1}{0,5}=\left(\dfrac{3}{10}\right)^4.2=\dfrac{81}{10000}.2=\dfrac{81}{5000}\\ ---\\ d,\left(\dfrac{3}{4}\right)^3.\left(\dfrac{16}{9}\right)^3=\left(\dfrac{3}{4}.\dfrac{16}{9}\right)^3=\left(\dfrac{48}{32}\right)^3=\left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}\)
b) \(\dfrac{8^2.6^3}{9^2.16^2}=\dfrac{2^6.2^3.3^3}{3^4.2^8}=\dfrac{2^9.3^3}{3^4.2^8}=\dfrac{2}{3}\)
c) \(\dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\dfrac{\left(0,5\right)^4.\left(0,3\right)^4}{\left(0,5\right)^5}=\dfrac{0,3^4}{0,5}\)
d) \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{16}{9}\right)^3=\dfrac{3^3}{4^3}.\dfrac{4^6}{3^6}=\dfrac{4^3}{3^3}=\left(\dfrac{4}{3}\right)^3\)
Bài 31 Viết các số và
dưới dạng các lũy thừa của cơ số 0,5
Lời giải:
Ta có:
Giải : Ta có:
(0,25)8 =[(0,5)2]8=(0,5)16
(0,125)4=[(0,5)3]4=(0,5)12
a) ( 0,5 )6 = \(\frac{5^6}{10^6}=\left(\frac{5}{10}\right)^6=\left(\frac{1}{2}\right)^6\)
( 0,5 )9 = \(\frac{5^9}{10^9}=\left(\frac{5}{10}\right)^9=\left(\frac{1}{2}\right)^9\)
vì \(\left(\frac{1}{2}\right)^6>\left(\frac{1}{2}\right)^9\)nên \(\left(0,5\right)^6>\left(0,5\right)^9\)
b) vì ( -0,125)8 = ( 0,125 )8 = ( 0,5 )24
=> ( -0,125 )8 = ( 0,5 )24