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\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
\(=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...0\left(1-\frac{2011}{2010}\right)\)
\(=0\)
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)
ta có \(\left(1-\frac{1}{2010}.\right).\left(1-\frac{2}{2010}\right)....\left(1-\frac{2010}{2010}\right).\left(1-\frac{2011}{2010}\right)\)\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).......0.\left(1-\frac{2011}{2010}\right)=0\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
\(A=\frac{ }{ }sdadsad\text{đ}\text{s}gh\text{d}fg\text{d}\)sf
Suy ra : A = ( 1 - 1 / 2010 ) . ( 1 - 2 / 2010 ) .... 0 . ( 1 - 2011 / 2010 ) = 0
Suy ra A = 0
A = 1. ( 1/2010 + 2/2010 ) - ( 3/2010 + 4/2010 ) - ... - ( 2010/2010 + 2011/2010 )
= 1/2010 - 2011/2010
= -2010/2010
Ta có: A=\(\frac{1}{2011}+\frac{2}{2010}+\frac{3}{2009}+...+\frac{2009}{3}+\frac{2010}{2}+\frac{2011}{1}\)
=> A=\(\frac{2012-2011}{2011}+\frac{2012-2010}{2010}+...+\frac{2012-2}{2}+\frac{2012-1}{1}\)
=>A=\(\frac{2012}{2011}-1+\frac{2012}{2010}-1+...+\frac{2012}{2}-1+2012-1\)
=>A=\(2012\cdot\left(\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{2}\right)+1\)
=> A= \(2012\cdot\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{2}\right)\)
ko biết có đúng hay ko nựa sai thì bỏ qua nha ^^
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
Ta có:\(1-\frac{2010}{2010}=1-1=0\)
Tích\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)....\left(1-\frac{2011}{2010}\right)\)có chứa thừa số \(1-\frac{2010}{2010}=0\)
Vậy tích\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-\frac{2011}{2010}\right)=0\)