giá trị của biểu thức \(\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2007}\right)}\) là.....
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Ta có: \(A=\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2017}\right)}=\frac{\frac{2017+1}{1}\frac{2017+2}{2}...\frac{2017+1009}{1009}}{\frac{1009+1}{1}\frac{1009+2}{2}...\frac{1009+2017}{2017}}\)
\(\Leftrightarrow A=\frac{\frac{2018.2019...3026}{1.2...1009}}{\frac{1010.1011...3026}{1.2...2017}}=\frac{2018.2019...3026}{1.2...1009}.\frac{1.2...2017}{1010.1011...3026}\)
\(\Leftrightarrow A=\frac{1.2...2017.2018.2019...3026}{1.2...1009.1010.1011...3026}=\frac{1.2.3...3026}{1.2.3...3026}=1.\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
=> A= \(\frac{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}\right).23.7.1009}{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right).23.7.1009}\) + \(\frac{1}{30.1009-160}\)
=> A= \(\frac{7.1009+23.1009-23.7}{7.1009+23.1009-23.7+1}\) + \(\frac{1}{7.1009+23.1009-23.7+1}\) = \(\frac{7.1009+23.1009-23.7+1}{7.1009+23.1009-23.7+1}\) = 1.
tr`
pái bn lun đó đỗ văn thành
tự đăng tự giải
haizzz
1
tick mình nha thank