(-2013)^0+2013^-1.√(-4)^2
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15 tháng 11 2017
a)A=20130+20131+20132+...+20132011
2013A=2013+20132+20133+...+20132012
2013A-A=2012A=20132012-20130
A=20132012-1/2012
k tao đi tao làm phần b cho
25 tháng 3 2018
b này : Chép cái đề bài vào
=>(2013+20131)+(20132+20133)+.....+(20132010+20132011)
=>2013.(1+2013)+20132.(1+2013)+.....+20132010.(1+2013)
=>2013.2014+20132.2014+......+20132010+.2014
=>2014.(2013+20132+.....+20132010) chia hết cho 2014
Vậy A chia hết cho 2014
24 tháng 7 2016
\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(A=\frac{2013}{2014}\)
S
25 tháng 7 2016
\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
CN
3
20 tháng 10 2020
a) ( 2x - 1 )( 2x + 1 ) - 4( x2 + x ) = 16
⇔ 4x2 - 1 - 4x2 - 4x = 16
⇔ -4x - 1 = 16
⇔ -4x = 17
⇔ x = -17/4
b) 5x( x - 2013 ) - x + 2013 = 0
⇔ 5x( x - 2013 ) - ( x - 2013 ) = 0
⇔ ( x - 2013 )( 5x - 1 ) = 0
⇔ \(\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)
20 tháng 10 2020
a) \(\left(2x-1\right)\left(2x+1\right)-4.\left(x^2+x\right)=16\)
\(4x^2-1-4x^2-4x=16\)
\(-1-4x=16\)
\(-4x=17\)
\(x=-\frac{17}{4}\)
b) \(5x\left(x-2013\right)-x+2013=0\)
\(\left(x-2013\right)\left(5x-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-2013=0\\5x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}}\)
YG
1
12 tháng 10 2017
TĐB:P=20130+20131+20132+.....+20132010
P=1+20131+20132+.....+20132010
2013xP= 20131+20132+.....+20132010+20132011
- P=1+ 20131+20132+.....+20132010
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Px2012=20132011-1
Px2012+1=20132011
Vậy Px2012+1=20132011
4 tháng 5 2016
=> B=2013. (1+\(\frac{1}{1+2}\) +\(\frac{1}{1+2+3}\) +...+ \(\frac{1}{1+2+3+...+2012}\))
=>B= 2013.(\(\frac{2}{2}\) + \(\frac{2}{2.3}\) +\(\frac{2}{3.4}\) +...+\(\frac{2}{2012.2013}\))
=>B= 2013.2.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) +...+\(\frac{1}{2012.2013}\))
=>B=4026. (1-\(\frac{1}{2}\) +\(\frac{1}{2}\) -\(\frac{1}{3}\) + ...+\(\frac{1}{2012}\) - \(\frac{1}{2013}\))
=>B=4026.(1-\(\frac{1}{2013}\))
=>B=4026.\(\frac{2012}{2013}\) => B=2.2012=4024 Vậy B=4024
Q6
0
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 5 2017
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
T
2
19 tháng 3 2018
B=2013.(1+
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
B=2013(\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
B=2013.2(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)