(x+1)+(x+2)+(x+3)...+(x+2013)=6079260
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
( x + 1 ) + ( x + 2 ) + .... + ( x + 2013 ) = 6079260
( x + x + ... + x ) + ( 1 + 2 + ....+ 2013 ) = 6079260
2013x + 2027091 = 6079260
2013x = 6079260 - 2027091
2013x = 4052169
x = 4052169 : 2013
x = 2013
( x + 1 ) + ( x + 2 ) + .... + ( x + 2013 ) = 6079260
x + 1 + x + 2 + x + 3 + .... + x + 2013 = 6079260
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 2013) = 6079260
2013x + 2027091 = 6079260
2013x =6079260-2027091 = 4052169
x = 4052169÷2013=2013
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow3\left(x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=2\\6x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-2013\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{4}\end{matrix}\right.\\ d,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 1 x 2 x 3 x ..... x 2012 x 2013 - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2013) x (2 x 4 x 6 x ..... x 2012) - 1 x 3 x 5 x ..... x 2011 x 2013
= (1 x 3 x 5 x ..... x 2011 x 2013) x (2 x 4 x 6 x ..... x 2012 - 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
(1-2+3-4+.....-98+99)x(2013x6-2013-2013x5)
=(1-2+3-4+....-98+99)x[ 2013x(6-1-5)]
= (1-2+3-4+....-98+99)x(2013x0)
= (1-2+3-4+....-98+99)x0 = 0
(x+1)+(x+2)+(x+3)+...+(x+2013)=6079260
x+1+x+2+x+3+...+x+2013=6079260
(x+x+x+...+x)+(1+2+3+...+2013)=6079260
2013x+2027091=6079260
2013x=6079260-2027091
2013x=4052169
x=4052169:2013
x=2013
Vậy x=2013