1.tính nhanh
\(\dfrac{2009\cdot2010+2000}{2011\cdot2010+2020}\)
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\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
\(\frac{2008.2009+2000}{2009.2010-2018}\)
\(=\frac{2008.\left(2010-1\right)+2010}{\left(2008+1\right).2010-2018}\)
\(=\frac{2008.2010-2008+2010}{2008.2010+2010-2018}\)
\(=\frac{2008.2010+2}{2008.2010-18}\)
Mình nghĩ bài này sai đề, nếu đề là 2018 -> 2008 thì bảo mình, mình làm lại cho
\(\frac{2011.2010-1}{2009.2011+2010}=\frac{2011.\left(2009+1\right)-1}{2009.2011+2010}\)
\(=\frac{2011.2009+2011-1}{2009.2011+2010}\)
\(=\frac{2011.2009+2010}{2009.2011+2010}\)
\(=1\)
Nhớ k vs kp với mik nhé,mấy man!
\(\frac{2011\cdot2010-1}{2009\cdot2011+2010}=\frac{2011\cdot\left(2009+1\right)-1}{2009\cdot2011+2010}=\frac{2011\cdot2009+2011\cdot1-1}{2009\cdot2011+2010}=\frac{ }{ }\)\(\frac{2011\cdot2009+2011-1}{2009\cdot2011+2010}=\frac{2011\cdot2009+2010}{2009\cdot2011+2010}=1\)
= 2009 * ( 2011 - 1 ) - 1000 / 2011 * 2009 - 1009
= 2009 * 2011 - 2009 -1000 / 2011 * 2009 - 1009
= 2009 * 2011 - 1009 / 2011 * 2009 - 1009
= 1
A=1⋅2⋅3⋅...⋅2010(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{2010}\))
= 1⋅2⋅3⋅...⋅2010[(1+\(\dfrac{1}{2010}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{2009}\))+(\(\dfrac{1}{3}\)+\(\dfrac{1}{2008}\))+...+(\(\dfrac{1}{1005}\)+\(\dfrac{1}{1006}\))]
= 1⋅2⋅3⋅...⋅2010(\(\dfrac{2011}{2010}\)+\(\dfrac{2011}{2009\cdot2}\)+\(\dfrac{2011}{2008\cdot3}\)++...+\(\dfrac{2011}{1006\cdot1005}\))
= 2011*(\(\dfrac{2010!}{2010}\)+\(\dfrac{2010!}{2009\cdot2}\)+\(\dfrac{2010!}{2008\cdot3}\)++...+\(\dfrac{2010!}{1006\cdot1005}\))
=> A⋮2011 (dpcm)
\(=\frac{2008+2009.2010}{2010.\left(2009+2\right)-2012}\)
\(=\frac{2009.2010+2008}{2010.2009+2010.2-2012}\)
\(=\frac{2008+2009.2010}{2008+2009.2010}=1\)
M=1/6+1/12+1/20+..+1/2009.2010
M=1/2.3+1/3.4+1/4.5+...+1/2009.2010
M=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2009-1/2010
M=1/2-1/2010
M=1004/2010
Tự rút gọn bn nha
M=1/6+=1/12+1/20+......+1/2009.2010]
M=1/2.3+1/3.4+1/4.5+.........+1/2009.2010
M=1/2-1/3+1/3-1/4+1/4-1/5+........+1/2009-1/2010
M=1/2-1/2010
=502/1005
\(=\dfrac{2019.2010+1,005.2010}{2011.2010+2.2010}=\dfrac{2010\left(2019+1,005\right)}{2010\cdot\left(2011+2\right)}\)
\(=\dfrac{2020,005}{2013}=\dfrac{2020005}{2013000}=\dfrac{404001}{402600}\)