a, Ta có : \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|=10x-100\)
Ta thấy : \(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\\left|...\right|\ge0\\\left|x-9\right|\ge0\end{matrix}\right.\)
=> \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|\ge0\)
=> \(10x-100\ge0\)
=> \(x\ge10\)
=> \(\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|...\right|=...\end{matrix}\right.\)
=> \(x-1+x-2+...+x-9=10x-100\)
=> \(9x-45=10x-100\)
=> \(10x-9x=100-45\)
=> \(x=55\) ( TM )
Vậy ....
b, Ta có : \(\left|x-2\right|+...+\left|x-9\right|=1-x\)
Ta thấy : \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|...\right|\ge0\end{matrix}\right.\)
=> \(\left|x-2\right|+...+\left|x-9\right|\ge0\)
=> \(1-x\ge0\)
=> \(x\le1\)
=> \(\left\{{}\begin{matrix}\left|x-2\right|=2-x\\\left|...\right|=-...\end{matrix}\right.\)
=> \(2-x+...+9-x=1-x\)
=> \(44-8x=1-x\)
=> \(8x-x=44-1\)
=> \(x=\frac{43}{7}\) ( KTM )
Vậy phương trình vô nghiệm .
