Question

Tìm các số nguyên x thỏa mãn :

\(2004=|x-4|+|x-10|+|x+101|+|x+990|+|x+1000|\)

Answer 1

2004=(x-x-x+x+x)+(4+10+101+990+1000)

còn tiếp theo thì mk ko biết

Answer 2

Ta có: 

\(\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|990\right|+\left|x+1000\right|\)

\(=\left|x+101\right|+\left|x-4\right|+\left|x-10\right|+\left|x+990\right|+\left|x+1000\right|\)

\(\ge\left|x+101\right|+\left|x-4+x-10+x+990+x+1000\right|x+101\left|+\right|2004\)

\(\Rightarrow2004\ge\left|x+101\right|+2004\Rightarrow\left|x+101\right|\le0\)

Mà: \(\left|x+101\right|\ge0\)nên \(\left|x+101\right|=0\Rightarrow x=-101\)

Vậy: \(x=-101\)