\( \left| x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+19\right|=20x\)
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\..................\\\left|x+19\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+19\right|\ge0\)
\(\Rightarrow20x\ge0\)
\(\Rightarrow x+1+x+2+...+x+19=20x\)
\(\Rightarrow19x+\left(1+2+...+19\right)=20x\)
\(\Rightarrow19x+190=20x\)
\(\Rightarrow x=190\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+19\right|\ge0\)
\(\Rightarrow20x\ge0\Rightarrow x\ge0\)
\(\left\{{}\begin{matrix}x+1>0\\x+2>0\\..........\\x+19>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\...........\\\left|x+19\right|=x+19\end{matrix}\right.\)
Thay vào ta có:
\(x+1+x+2+x+3+...+x+19=20x\)
\(\Rightarrow19x+\left(1+2+3+....+19\right)=20x\)
\(\Rightarrow x=\dfrac{\left(1+19\right).19}{2}=190\)
Vậy..................
Chúc bạn học tốt!!!