Tìm x:
\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9=5\right|\)
\(\left|x-2\right|+\left|x-3\right|+\left|2x-8\right|=9\\ \left|x+2\right|+\left|x+3\right|+\left|x+1\right|=4\\ \left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.13}\right|+...+\left|x+\dfrac{1}{397.401}\right|=101x\)
\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
