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Ta có
\(\left(x+2x+3x+..+2014x\right)+\left(1+...+2013\right)=8114406\)
\(2029105x+2027091=8114406\)
\(2029105x=8114406-2027091\)
\(x=\frac{6087315}{2029105}=3\)
\(Vậyx=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1x+2x+3x+...+10x = 165
x ( 1+2+3+4+....+10) = 165
55x = 165
x = 165 : 55
x = 3
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![](https://rs.olm.vn/images/avt/0.png?1311)
1x+2x+3x+4x+.........+100x=10100
<=>100x+(1+2+3+4+...+100)=10100
<=>100x+5050=10100
<=>100x=5050
<=>x=100/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
1x+2x+3x+...+2014x=4058210
x(1+2+3+...+2014)=4058210
2029105x=4058210
x=2
Tk cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!