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\(x^{2016}=x^{2017}\\ \Leftrightarrow x^{2016}-x^{2017}=0\\ \Leftrightarrow x^{2016}\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^{2016}=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy S = {0;1}
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1+3-5-7+9+11-13-15+...+2013+2015-2017-2019
=(1+3-5-7)+(9+11-13-15)+...+(2013+2015-2017-2019)
=(-8)+(-8)+...+(-8)
=(-8)x252,5
=-2020
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A = 3+5-7-9+11+13-15-17+19+.....+2011+ 2013-2015-2017+2019+2020
A=(3 +5-7-9)+(11+13-15-17)+....+(2011+2013-2015-2017)+2019+2020
xét dãy số : 3; 11; .....;2011
Dãy số trên có số số hạng : (2011 -3):8+1 = 252
Vậy tổng A là tổng 252 nhóm và 2019 + 2020
Mỗi nhóm có giá trị: 3 + 5 - 7 - 9 = -8
A = -8.252 + 2019 + 2020
A = 2023
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A = 1 - 3 - 5 + 7 + 9 - 11 - 13 + ... - 2013 + 2015 + 2017 ( có 1009 số ; 1009 : 4 dư 1)
A = (1 - 3 - 5 + 7) + (9 - 11 - 13 + 15) + ... + (2009 - 2011 - 2013 + 2015) + 2017
A = 0 + 0 + ... + 0 + 2017
A = 2017
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S=5-7+9-11+...+2013-2015+2017-2019
=(5-7)+(9-11)+...+(2013-2015)+(2017-2019) (có 504 cặp)
=(-2)+(-2)+...+(-2)
=(-2).504=-1008
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100=25 .4 =5^2.4
5^2019:( 5^2013-5^2.4.5^2010)=5^2019: ( 5^2013-5^2012.4)=5^2019: (5^2012.(5-4))
=5^2019:5^2012=5^7(=78125)
Học tốt
5x = 52019 : (52013 -100*52010)
<=> 5x = 52019 : (52010 .(53-100))
<=> 5x = 52019 : (52010 . 25)
<=> 5x = 52019 : 52012
<=> 5x = 57
<=> x = 7
Vậy x = 7
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Bài 1
\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó
Ta tách :
\(\frac{2017}{\left(2018+2019\right)+2018}\)
đến đây ta tách
\(\frac{2017}{2018+2019}< \frac{2017}{2018}\)
vậy....
mấy câu khác tương tự
2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)
= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)
=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)
= \(\frac{1}{2}\)
3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)
= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)
= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)
=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(4026.\left(1-\frac{1}{2013}\right)\)
= \(4026.\frac{2012}{2013}\)
=\(4024\)