\(Q=\dfrac{-2015}{2016}\cdot\left(-50\right)\cdot\dfrac{-153}{154}\cdot1\dfrac{1}{2015}\cdot20\%\)
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24 tháng 7 2017
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1-\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1}{2017}\)
H
24 tháng 7 2017
Giải:
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right).\left(1-\dfrac{1}{2017}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)
\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)
\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)
Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:
\(A=\dfrac{1}{2017}\)
Vậy \(A=\dfrac{1}{2017}\).
Chúc bạn học tốt!
7 tháng 8 2017
\(F=\left(-\dfrac{1}{2015}\right)^0-\left(\dfrac{13}{27}.\dfrac{162}{39}-1\right)^{2015}+\left(-\dfrac{1}{3}\right)^2\\ F=1-\left(2-1\right)^{2015}+\dfrac{1}{9}\\ F=1-1+\dfrac{1}{9}\\ F=\dfrac{1}{9}\)
Chúc bạn học tốt!!!
4 tháng 11 2018
1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)
=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)
=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)
=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)
=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)
=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)
=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)
=\(\dfrac{1}{2007}.\left(-1\right)+0\)
=\(\dfrac{-1}{2007}\)
4 tháng 11 2018
2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)
=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)
=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)
=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)
=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)
bí rồi
13 tháng 5 2022
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
NN
15 tháng 11 2017
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
15 tháng 11 2017
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
\(Q=\dfrac{-2015}{2016}\cdot\left(-50\right)\cdot\dfrac{-153}{154}\cdot1\dfrac{1}{2015}\cdot20\%\)
\(=\dfrac{-2015}{2016}\cdot\left(-50\right)\cdot\dfrac{-153}{154}\cdot\dfrac{2016}{2015}\cdot\dfrac{1}{5}\\ =\left(-\dfrac{2015}{2016}\cdot\dfrac{2016}{2015}\right)\cdot\left(-50\cdot\dfrac{1}{5}\right)\cdot-\dfrac{153}{154}\\ =\left(-1\right)\cdot\left(-10\right)\cdot\left(-\dfrac{153}{154}\right)\\ =10\cdot\left(-\dfrac{153}{154}\right)\\ =-\dfrac{1530}{154}\\ =-\dfrac{765}{77}\)