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\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
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\(A=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-\frac{2011}{2010}\right)=0\)
so sánh : cho A\(\frac{2010^{2011}+1}{2010^{2012}+1}\)
cho B =\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
Ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\)
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}\)
\(2010A=1+\frac{2009}{2010^{2012}+1}\)
Lại có:
\(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}\)
\(2010B=1+\frac{2009}{2010^{2011}+1}\)
Vì \(1+\frac{2009}{2010^{2012}+1}< 1+\frac{2009}{2010^{2011}+1}\)
nên 2010A < 2010B
hay A < B
Vậy A < B
\(A=\frac{ }{ }sdadsad\text{đ}\text{s}gh\text{d}fg\text{d}\)sf
trong dãy tích A sẽ có phân số \(1-\frac{2010}{2010}=1-1=0\)
=>A=0
\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)
\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)
mà \(2010^{2012}+1>2010^{2011}+1\)
nên A<B
Ta có:
2010.A=\(\frac{2010^{2012}+2010}{2010^{2012}+1}\)
2010.B=\(\frac{2010^{2011}+2010}{2010^{2011}+1}\)
2010.A có phần thừa với 1 là:\(\frac{2009}{2010^{2012}+1}\)
2010.B có phần thừa với 1 là:\(\frac{2009}{2010^{2011}+1}\)
Vì \(\frac{2009}{2010^{2012}+1}<\frac{2009}{2010^{2011}+1}\)
=>2010.A<2010.B
=>A<B
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
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010}{2010^{2012}+1}\)
\(1-B=1-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010^{2011}+1}{2010^{2011}+1}-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010}{2010^{2011}+1}\)
Do \(\frac{2010}{2010^{2012}+1}<\frac{2010}{2010^{2011}+1}\)nên \(A>B\)
Do 20102011+1<20102012+1=>A<1
Tương tự với B;B<1
Theo đề bài ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}<\frac{2010^{2011}+1+2009}{2010^{2012}+1+2009}=\frac{2010^{2011}+2010}{2010^{2012}+2010}=\frac{2010.\left(1+2010^{2010}\right)}{2010.\left(1+2010^{2011}\right)}=\frac{2010^{2010}+1}{2010^{2011}+1}=B\)(*)
Từ (*)=> A<B
Trong tích của A có một thừa số bằng 1-2010/2010=0