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TH
0
J
1
15 tháng 11 2018
\(C=\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(2013^2-2014^2\right)+2015^2\)
\(C=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+....+\left(2013-2014\right)\left(2013+2014\right)+2015^2\)
\(C=-\left(1+2\right)-\left(3+4\right)-....-\left(2013+2014\right)+2015^2\)
\(C=-\left(1+2+3+4+...+2014\right)+2015^2\)
\(C=-\dfrac{\left(2014+1\right)2014}{2}+2015^2\)
\(C=-2015.1007+2015^2\)
\(C=2015\left(2015-1007\right)=2015.1008\)
2 tháng 1 2016
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
G
11 tháng 5 2023
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
DT
1
22 tháng 5 2022
A=20182+20162+20142+...+42 +22-(20172 +20152+20132+...+ 32 + 1)
A=(2018²-2017²)+(20162-20152)+(2014²-2013²)+...+(2² −1²)
A=2018+2017+2016+2015+2014+2013+...+2+1
\(A=\dfrac{2018\left(2018+1\right)}{2}=\text{2 037 171}\)
KE
2
8 tháng 3 2018
pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
8 tháng 3 2018
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
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