Thực hiện phép tính
(1 - \(\frac{1}{2^2}\)) (1 - \(\frac{1}{4^2}\)) (1 - \(\frac{1}{5^2}\)) ...(1 -\(\frac{1}{2017^2}\)) = ?
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Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)
\(=\frac{2015}{2}+1+\frac{2014}{3}+1+...........+\frac{1}{2015}+1\)
\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)
\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)
Thay A vào biểu thức ta dc
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)
\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)
CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
1) tự làm (thực hiện từ dưới lên)
2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)
= \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12
1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)
\(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}\)\(=5+\frac{15}{7}=\frac{50}{7}\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)