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mãi ko thấy ai làm tớ làm giúp nì =)
\(\text{ta có:}\hept{\begin{cases}\frac{2002}{2003}< 1\\\frac{2005}{2004}>1\end{cases}}\Rightarrow\frac{2005}{2004}>\frac{2002}{2003}\Rightarrow-\frac{2005}{2004}< -\frac{2002}{2003}\)
\(\text{ta có: }\hept{\begin{cases}-\frac{1}{10^5}< 0\\\frac{-9}{-10}>0\end{cases}}\Rightarrow\frac{-1}{10^5}< \frac{-9}{-10}\)
a, \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(=\frac{1\frac{25}{108}.320\frac{1}{25}+46\frac{3}{4}}{4\frac{16}{21}:\left(-1\frac{20}{21}\right)}=\frac{330\frac{1}{25}}{-2\frac{18}{41}}=\)\(-135,3164\)
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,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)
\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)
\(=\frac{-8}{2005}\)
b,Ta có: \(81^{10}-27^{13}-9^{21}\)
\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)
\(=3^{40}-3^{39}-3^{42}\)
\(=3^{39}.3-3^{39}-3^{39}.3^3\)
\(=3^{39}.\left(3-1-3^3\right)\)
\(=3^2.3^{37}.\left(-25\right)\)
\(=3^{37}.\left(-225\right)⋮225\)
Vậy \(81^{10}-27^{13}-9^{21}⋮225\)
\(A=\left[0,8\cdot7+(0,8)^2\right]\cdot\left[1,25\cdot7-\frac{4}{5}\cdot1,25\right]-47,86\)
\(=0,8\cdot(7+0,8)\cdot1,25\cdot(7-0,8)-47,86\)
\(=0,8\cdot7,8\cdot1,25\cdot6,2-47,86\)
\(=48,36-47,86=0,5\)
\(B=\frac{(1,09-0,29)\cdot\frac{5}{4}}{(18,9-16,65)\cdot\frac{8}{9}}=\frac{0,8\cdot1,25}{2,25\cdot\frac{8}{9}}=\frac{1}{2}\)
\(A:B=0,5:\frac{1}{2}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}\cdot2=1\)
A gấp 1 lần B