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LT
27 tháng 6 2021
a) (−3,1597)+(−2,39)= -5,5497
b) (−0,793)−(−2,1068)= 1.3138
c) (−0,5).(−3,2)+(−10,1).0,2= -0,42
d) 1,2.(−2,6)+(−1,4):0,7=-5,12
17 tháng 10 2018
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 \(81^{10}-27^{13}-9^{21}⋮225\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)\({\left( { - 2} \right)^2}.{\left( { - 2} \right)^3} = {\left( { - 2} \right)^{2 + 3}} = {\left( { - 2} \right)^5}\);
b)\({\left( { - 0,25} \right)^7}:{\left( { - 0,25} \right)^5} = {\left( { - 0,25} \right)^{7 - 5}} = {\left( { - 0,25} \right)^2} = {\left( {0,25} \right)^2}\);
c)\({\left( {\frac{3}{4}} \right)^4}.{\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^{4 + 3}} = {\left( {\frac{3}{4}} \right)^7}.\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)
8 tháng 5 2019
\(A=\frac{\left(0,4-\frac{2}{\sqrt{81}}+\frac{2}{11}\right)}{1,4-\frac{7}{\sqrt{81}}+\frac{7}{11}}=\frac{2\left(0,2-\frac{1}{\sqrt{81}}+\frac{1}{11}\right)}{7\left(0,2-\frac{1}{\sqrt{81}}+\frac{1}{11}\right)}=\frac{2}{7}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}0,6 + \left( {\frac{3}{{ - 4}}} \right) = \frac{6}{{10}} + \left( {\frac{{ - 3}}{4}} \right)\\ = \frac{{12}}{{20}} + \left( {\frac{{ - 15}}{{20}}} \right) = \frac{{12 + \left( { - 15} \right)}}{{20}}\\ = \frac{{ - 3}}{{20}}\end{array}\)
b)
\(\begin{array}{l}\left( { - 1\frac{1}{3}} \right) - \left( { - 0,8} \right) = \frac{{ - 4}}{3} + \frac{8}{{10}}\\ = \frac{{ - 4}}{3} + \frac{4}{5} = \frac{{ - 20}}{{15}} + \frac{{12}}{{15}} = \frac{{ - 8}}{{15}}.\end{array}\)
19 tháng 9 2023
a: =0,6-0,75=-0,15
b: \(=-\dfrac{4}{3}+\dfrac{4}{5}=\dfrac{-20+12}{15}=-\dfrac{8}{15}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);
b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)
c)
\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)
d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a) Ta có:
\({\left( {\frac{1}{3}} \right)^2}.{\left( {\frac{1}{3}} \right)^2} = \frac{1}{3}.\frac{1}{3}.\frac{1}{3}\frac{1}{3} = {\left( {\frac{1}{3}} \right)^4}\)
b)
\({\left( {0,2} \right)^2}.{\left( {0,2} \right)^3} = \left( {0,2.0,2} \right).\left( {0,2.0,2.0,2} \right) = {\left( {0,2} \right)^5}\)
8 tháng 12 2023
Đặt \(A=\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\left(\dfrac{3^3}{6}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(\dfrac{3^6}{9}-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(81-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
=0
A = 1,4(51)-0,2(3)+0,7(81)
=479/330-7/30+43/55
=479/330-77/330+258/330
=2
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