Question

Tìm x nguyên biết ; |x-1|+|x-3|+|x-5|+|x-7|=8

Answer 1

Ta có: |x-1|+|x-3|+|x-5|+|x-7| = (|x-1|+|7-x|)+(|x-3|+|5-x|) \(\ge\) |x-1+7-x| + |x-3+5-x| = 6+2 = 8 (1)

Mà |x-1|+|x-3|+|x-5|+|x-7|=8 suy ra (1) xảy ra dấu "=" khi:

\(\hept{\begin{cases}\left(x-1\right)\left(7-x\right)\ge0\\\left(x-3\right)\left(5-x\right)\ge0\end{cases}\Rightarrow\hept{\begin{cases}1\le x\le7\\3\le x\le5\end{cases}\Rightarrow}3\le x\le5}\)

Do x nguyên nên \(x\in\left\{3;4;5\right\}\)