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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) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
\(\begin{array}{l}a){( - 2)^3}.{( - 2)^4} = {( - 2)^{3 + 4}} = {( - 2)^7}\\b){(0,25)^7}:{(0,25)^3} = {(0,25)^{7 - 3}} = {(0,25)^4}\end{array}\)
a) \(\frac{6}{5}.{\left( {1,2} \right)^8} = 1,2.{(1,2)^8} = {(1,2)^{1 + 8}} = {(1,2)^9}\)
b) \({\left( {\frac{{ - 4}}{9}} \right)^7}:\frac{{16}}{{81}} = {\left( {\frac{{ - 4}}{9}} \right)^7}:{\left( {\frac{{ - 4}}{9}} \right)^2} = {\left( {\frac{{ - 4}}{9}} \right)^{7 - 2}} = {\left( {\frac{{ - 4}}{9}} \right)^5}\)
ta có : \(4^3.3^7.12^6.9^3.8^2=\left(2^2\right)^3.3^7.\left(2^2.3\right)^6.\left(3^2\right)^3.\left(2^3\right)^2\)
= \(2^6.3^7.2^{12}.3^6.3^6.2^6=2^{24}.3^{19}\)
\(4^3.3^7.12^6.9^3.8^2=2^6.3^7.2^{12}.3^6.3^6.2^6=2^{24}.3^{19}\)
Đề này là chịu rồi
a) \(7^5:343\)
\(=7^5:7^3\)
\(=7^{5-3}\)
\(=7^2\)
b) \(a^{12}:a^{18}\)
\(=\dfrac{a^{12}}{a^{18}}\)
\(=\dfrac{a^0}{a^6}\)
\(=\dfrac{1}{a^6}\)
c) \(x^7\cdot x^4\cdot x\)
\(=x^{7+4+1}\)
\(=x^{12}\)
\(\left(\dfrac{1}{7}\right)^7\cdot7^7\)
\(=\left(\dfrac{1}{7}\cdot7\right)^7\)
\(=\left(\dfrac{7}{7}\right)^7\)
\(=1^7\)
\(=1\)
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\(\dfrac{3^7}{\left(0,375\right)^7}\)
\(=\left(\dfrac{3}{0,375}\right)^7\)
\(=8^7\)
\(=\left(2^3\right)^7\)
\(=2^{21}\)