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31 tháng 7 2020

\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)

\(\Leftrightarrow2x-4,36=1\)

\(\Leftrightarrow2x=5,36\)

\(\Leftrightarrow x=2,68\)

b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)

\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)

\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)

\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)

Bài 1:

a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)

\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)

\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)

\(\frac{2\cdot x-4,36}{0,125}=8\)

\(2\cdot x-4,36=8\cdot0,125\)

\(2\cdot x-4,36=1\)

\(2\cdot x=1+4,36\)

\(2\cdot x=5,36\)

\(x=\frac{5,36}{2}=2,68\)

b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)

\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)

\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)

\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)

\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)

Bài 2:

a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )

\(x+5,2=4,7\cdot3,2+0,5\)

\(x+5,2=15,54\)

\(x=15,54-5,2=10,34\)

b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)

\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)

\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)

Bài 3:

a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)

\(x\cdot\left(104,5-14,1+9,6\right)=25\)

\(x\cdot100=25\)

\(x=\frac{25}{100}=\frac{1}{4}=0,25\)

b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)

\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)

\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)

20 tháng 6 2019

#)Giải :

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)

\(=0\)

20 tháng 6 2019

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)

\(=0\)

7 tháng 6 2019

Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.

7 tháng 6 2019

Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)

\(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(1-\frac{1}{2011}\)

\(\frac{2010}{2011}\)

18 tháng 5 2019

\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)+\left(\frac{1}{2010}-\frac{1}{2010}\right)-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

~ Hok tốt ~

18 tháng 5 2019

\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1}:\frac{2}{1}+\frac{1}{2}:\frac{3}{1}+\frac{1}{3}:\frac{4}{1}+...+\frac{1}{2009}:\frac{2010}{1}+\frac{1}{2010}:\frac{2011}{1}\)

\(=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2009}\cdot\frac{1}{2010}+\frac{1}{2010}\cdot\frac{1}{2011}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2010}+\frac{1}{2010\cdot2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

Dấu " . " là dấu nhân nhé

15 tháng 8 2015

Bài 2 : 

\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2008}\)và 4