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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:
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times\left(2009+1\right)-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2011-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2010}{2009\times2011+2010}\)
\(=1\)
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times2010-1}{2009\times2011+2011-1}\)
\(=\frac{2011\times2010-1}{2010\times2011-1}\)
\(=1\)
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
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 :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)
= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)