`@TH1: x < 1`
Ta có: `Bth=1-x+3-x+5-x+7-x=16-4x`
`@TH2: 1 <= x < 3`
Ta có: `Bth=x-1+3-x+5-x+7-x=14-2x`
`@TH3: 3 <= x <= 5`
Ta có: `Bth=x-1+x-3+5-x+7-x=8`
`@TH4: 5 < x <= 7`
Ta có: `Bth=x-1+x-3+x-5+7-x=2x-2`
`@TH5: x > 7`
Ta có: `Bth=x-1+x-3+x-5+x-7=4x-16`