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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.
Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)
= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)
a)(3/2-0,5)/x=7/2+1/4
(3/2-1/2)/x=14/4+1/4
1/x=15/4
x=1:15/4
x=4/15
b)(x*0,25+2010)*2011=(53+2010)*(2012-1)
(x*0,25+2010)*2011=2063*2011
=>0,25x+2010=2063
0,25x=2063-2010
0,25x=53
x=53/0,25
x=212
a ) y : 2 + y + y : 3 + y : 4 = 25
y : ( 2 + 1 + 3 + 4 ) = 25
y : 10 = 25
y = 25 x 10
y = 250
b ) 100%: y - 50% : y + 40% : y = 18 + 30% : y
100%: y - 50% : y + 40% : y - 30% : y = 18
( 100% - 50% + 40% - 30% ) : y = 18
60% : y = 18
y = 60% : 18
y = \(\frac{1}{30}\)
Nếu mình đúng thì các bạn k mình nhé
a) \(y:2+y+y:3+y:4=25\)
\(y:\left(2+3+1+4\right)=25\)
\(y:10=25\)
\(y=250\)
b) \(100\%:y-50\%:y+40\%:y=18+30\%:y\)
\(100\%:y-50\%:y+40\%:y-30\%:y=18\)
\(\left(100\%-50\%+40\%-30\%\right):y=18\)
\(60\%:y=18\)
\(y=60\%:18\)
\(y=\frac{1}{30}\)
Trả lời:
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times\left(2009+1\right)-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2011-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2010}{2009\times2011+2010}\)
\(=1\)
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times2010-1}{2009\times2011+2011-1}\)
\(=\frac{2011\times2010-1}{2010\times2011-1}\)
\(=1\)
\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)+\left(\frac{1}{2010}-\frac{1}{2010}\right)-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
~ Hok tốt ~
\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)
\(=\frac{1}{1}:\frac{2}{1}+\frac{1}{2}:\frac{3}{1}+\frac{1}{3}:\frac{4}{1}+...+\frac{1}{2009}:\frac{2010}{1}+\frac{1}{2010}:\frac{2011}{1}\)
\(=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2009}\cdot\frac{1}{2010}+\frac{1}{2010}\cdot\frac{1}{2011}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2010}+\frac{1}{2010\cdot2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
Dấu " . " là dấu nhân nhé
