\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)
VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)
Ta có: \(\left|x+\frac{1}{101}\right|\ge0\) \(\forall x\)
\(\left|x+\frac{2}{101}\right|\ge0\) \(\forall x\)
\(\left|x+\frac{3}{101}\right|\ge0\) \(\forall x\)
\(............\)
\(\left|x+\frac{100}{101}\right|\ge0\) ∀\(x\)
\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\) \(\forall x\)
\(\Leftrightarrow101x\ge0\)
\(\Leftrightarrow x\ge0\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 hạng tử x 100 số hạng
\(\Leftrightarrow100x+\left(\frac{\left(100+1\right)\cdot100:2}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{101\cdot50}{101}=101x\)
\(\Leftrightarrow50=101x-100x\)
\(\Rightarrow x=50\)