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1) Tính :
a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).0\)
\(=0\)
2) Tìm x
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
khi ko mún tích thì tích 1 tích
khi mún tích thì tích 50 tích
Có thể có nhiều hơn mà =.= 2 s của mik là 2 nick mik k cho người trả lời dc =.=
a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
I don't now
mik ko biết
sorry
......................
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\left(\frac{x-1}{2x+2}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2011}{2013}\)
\(\Rightarrow x-1=2011\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2013-1\)
\(\Rightarrow x=2012\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\) (1/3=2/6;...)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(1-\frac{2}{x+1}=\frac{2011}{2013}\)
\(\frac{2}{x+1}=\frac{2}{2013}\)
=> x + 1 = 2013
x = 2012
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{n\left(n+1\right)}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{n\left(n+1\right)}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{n+1}\right)\)
\(=1-\frac{2}{n+1}\)
\(=\frac{n+1}{n+1}-\frac{2}{n+1}\)
\(=\frac{n-1}{n+1}\)
2011/2013