Ta có:
\(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
............................
\(\left|x+9\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+....+\left|x+9\right|\ge0\)
hay \(10x\ge0\) hay \(x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+9\right|=x+1+x+2+...+x+9=10x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(1+2+...+9\right)=10x\)
\(\Rightarrow9x+45=10x\)
\(\Rightarrow x=45\)
Vậy \(x=45\)
Đúng 0
Bình luận (0)
