Question

Giảu bất phương trình sau: |x-1| + |x-5| >8

Answer 1

\(\left|x-1\right|+\left|x-5\right|>8\\ TH1:x< 1\\ BPT\Leftrightarrow\left(1-x\right)+\left(5-x\right)>8\\ \Leftrightarrow-2x+6>8\\ \Leftrightarrow-2x>2\\ \Leftrightarrow x< -1\left(tm\right)\\ TH2:1\le x< 5\\ \left(x-1\right)+\left(5-x\right)>8\\ \Leftrightarrow4>8\left(vôlý\right)\\ TH3:x\ge5\\ BPT\Leftrightarrow\left(x-1\right)+\left(x-5\right)>8\\ \Leftrightarrow2x-6>8\\ \Leftrightarrow2x>14\\ \Leftrightarrow x>7\left(tm\right)\)

Vậy \(x< -1;x>7\)