Tính:
\(\frac{\left(0,125\right)}{\left(-0,3\right)^5}^5.\frac{\left(2,4\right)}{\left(0.01\right)^3}^5\)
So sánh:
a) \(12^8\)và \(8^{12}\)
b) \(\left(-5\right)^{39}\)và \(\left(-2\right)^{91}\)
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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 \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
\(\frac{\left(-0,125\right)^5\times\left(2,4\right)^5}{\left(-0,3\right)^5\times\left(0,01\right)^3}=\frac{\left(-0,12\times2,4\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{\left(-0,3\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{1}{0,000001}\)
\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)
\(=\dfrac{\left(0.125\cdot2.4\right)^5}{\left(-0.3\cdot0.01\right)^3\cdot\left(-0.3\right)^2}=\dfrac{0.3^5}{0.3^2\cdot\left(-0.003\right)^3}=\dfrac{0.3^3}{-0.003^3}=-1000000\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)