\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=10+x\)
Ta có:
\(\left|x+1\right|\ge0\forall x\)
\(\left|x+2\right|\ge0\forall x\)
............................
\(\left|x+100\right|\ge0\forall x\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|\ge0\forall x\)
Mà \(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=10+x\)
\(\Rightarrow10+x\ge0\)
\(\Rightarrow x\ge-10\)
\(\Rightarrow\left(-1-x\right)+\left(-2-x\right)+...+\left(-10-x\right)+\left(11+x\right)+...+\left(100+x\right)=10+x\)
\(90x-10x+\left(11+12+...+100\right)-\left(1+2+...+10\right)=10+x\)
\(79x+\frac{\left(100+11\right).90}{2}-\frac{\left(10+1\right).10}{2}-10=0\)
\(79x+4995-55-10=0\)
\(79x+4930=0\)
\(79x=-4930\)
\(x=-\frac{4930}{79}\)
Vậy \(x=-\frac{4930}{79}\)
Sai thì thôi nhé~
