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26 tháng 6 2021

\(A=\frac{4\cdot0,125\cdot20,2\cdot800\cdot0,25}{1,01\cdot75+0,26\cdot101-1,01}\)   

\(=\frac{4\cdot0,25\cdot0,125\cdot800\cdot20,2}{1,01\cdot75+0,26\cdot100\cdot1,01-1,01}\)   

\(=\frac{1\cdot100\cdot20,2}{1,01\cdot\left(75+26-1\right)}\)   

\(=\frac{100\cdot20,2}{100\cdot1,01}\)   

= 20 

\(B=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{80}{56}-\frac{15}{12}:\frac{7}{8}\right)\)   

\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{10}{7}-\frac{5}{4}\cdot\frac{8}{7}\right)\)   

\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot\left(\frac{10}{7}-\frac{10}{7}\right)\)   

\(=\left(\frac{178}{179}+\frac{179}{180}+\frac{180}{181}\right)\cdot0\)   

= 0 

18 tháng 8 2023

a) \(25^{\dfrac{1}{2}}=5\)

b) \(\left(\dfrac{36}{49}\right)^{-\dfrac{1}{2}}=\dfrac{7}{6}\)

c) \(100^{1,5}=1000\)

6 tháng 7 2018

Ta có \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)

\(=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)

Ta thấy \(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}>0\)suy ra \(3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)

Khi đó \(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 3\)

19 tháng 1 2020

A=2100-299-298-...-22-2-1

\(\Rightarrow\)2A=2101-2100-299-...-23-22-2

\(\Rightarrow\)2A+A=(2101-2100-299-...-23-22-2)+(2100-299-298-...-22-2-1)

\(\Rightarrow\)3A=2101+1

\(\Rightarrow\)A=\(\frac{2^{101}+1}{3}\)

Vậy A=\(\frac{2^{101}+1}{3}\).

19 tháng 1 2020

Ta có : A = 2100 - 299 - 298 - ... - 22 - 2 - 1 

=> 2A = 2101 - 2100 - 299 - ... - 23 - 22 - 2

Lấy A - 2A = (2100 - 299 - 298 - ... - 22 - 2 - 1) - (2101 - 2100 - 299 - ... - 23 - 22 - 2)

   => - A     = 2100 + 2100 - 2101 - 1

   => - A     = 2.2100 - 2101 - 1

   => - A     = 2101 - 2101 - 1

   => - A     = - 1

  => A = 1   

HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023

a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)

b)

\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)

10 tháng 9 2023

a) \(3\cdot4\cdot25\cdot8\cdot125\)

\(=\left(125\cdot8\right)\cdot\left(4\cdot25\right)\cdot3\)

\(=1000\cdot100\cdot3\)

\(=300000\)

b) \(36\cdot63+64\cdot663\)

\(=63\cdot\left(36+64\right)\)

\(=63\cdot100\)

\(=6300\)

c) \(45\cdot129+71\cdot129\)

\(=129\cdot\left(129+71\right)\)

\(=129\cdot200\)

\(=25800\)

d) \(26\cdot\left(43+57\right)+74\cdot\left(43+57\right)\)

\(=26\cdot100+74\cdot100\)

\(=2600+7400\)

\(=10100\)

10 tháng 9 2023

a) 3.4.25.8.125

=3(4.25).(8.125)

=3.100.1000

300000

b) 36.63+ 64.63

=(36+64).63

=100.63

=6300
c) 45.129+ 71.45 

=45.(129+71)

=45.200

=9000              

d) 26. (43+ 57)+ 74(57 +43)

=26.43+26.57 + 74.57 + 74.43

=(26.43+74.43)+(26.57 + 74.57 )

= [43.(26+74)]+[ (26+74).57]

=[43.100]+[100.57]

=4300+5700

= 10000

28 tháng 7 2016

\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)

\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)

\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)

mk lm chak vớ vẩn rồi

HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023

a)

 \(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)

b)

\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)

  

a: \(log_{\dfrac{1}{4}}8=log_{2^{-2}}2^3=\dfrac{-3}{2}\cdot log_22=-\dfrac{3}{2}\)

b: \(log_45\cdot log_56\cdot log_68\)

\(=log_45\cdot\dfrac{log_46}{log_45}\cdot\dfrac{log_48}{log_46}\)

\(=log_48=log_{2^2}2^3=\dfrac{3}{2}\)