(x+1)+(x+2)+(x+3)+...+(x+2013)=20132014
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\((x+1)+(x+2)+(x+3)+...+(x+2013)=20132014\)
\(x+1+x+2+x+3+...+x+2013=20132014\)
\(2013x+(1+2+3+...+2013)=20132014\)
\(2013x+\dfrac{2013\left(2013+1\right)}{2}=20132014\)
\(2013x+2027091=20132014\)
\(2013x=20132014-2027091\)
\(2013x=18104923\)
\(x=18104923:2013\)
\(x=8994\)(kq đã làm tròn)
Ta có : \(\left(x+1\right)+\left(x+2\right)+...+\left(x+2013\right)=20132014\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+2013\right)=20132014\)
\(\Leftrightarrow2013x+\dfrac{2013\left(2013+1\right)}{2}=20132014\)
\(\Leftrightarrow2013x+2027091=20132014\)
\(\Leftrightarrow2013x=18104923\)
\(\Leftrightarrow x=\) (chỗ nãy bẫm mãy tính ra số ngộ quá @@ nên ko ghi vào )
![](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)
(x + 1) + (x + 2) + (x + 3) + .... + (x + 2013) = 20132014
(x + x + x + ... + x) + (1 + 2 + 3 + ... + 2013) = 20132014
2013x + 2027091 = 20132014
2013x = 20132014 - 2027091
2013x = 18104923
=> x =8994,000497